Explicit zero-free regions for the Riemann zeta-function

Author:
Yang, Andrew
Publication Date:
2025
DOI:
https://doi.org/10.26190/unsworks/31825
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Explicit zero-free regions for the
         Riemann zeta-function


                      Andrew Yang


A thesis in fulfilment of the requirements for the degree of
                   Doctor of Philosophy




                     School of Science
      The University of New South Wales, Canberra

                       August 2025
Abstract

Over the course of the last century, the Riemann zeta-function has played a pivotal role in
furthering our understanding of the distribution of prime numbers. Of particular impor-
tance are the locations of its complex-valued zeroes — a topic which is currently poorly
understood and a hurdle to establishing many results in analytic number theory. It is
known that all such zeroes lie inside the critical strip, the set of complex numbers s for
which 0 < Re s < 1. The famous Riemann hypothesis asserts that these zeroes in fact lie
on the vertical line Re s = 1/2; however, at present the best available techniques can only
exclude zeroes from certain subsets of the critical strip, known as zero-free regions. In
this thesis, we develop a collection of analytic and computational tools to refine existing
approaches to construct zero-free regions. These tools allow us to prove a series of explicit
zero-free regions larger than those previously known.




                                             iii
Acknowledgement

I would like to express my deepest gratitude to my supervisor Tim for your unwavering
support, insightful guidance, and constant encouragement throughout every stage of my
candidature. Without you this thesis would not have been possible. Your mentorship
has been invaluable both academically and personally, and I am grateful for the many
opportunities to learn under your supervision.

To my parents, thank you for your lifelong encouragement and belief in me. Your support
lays the foundation for all that I will achieve.

To Chiara, my partner in work and in life, thank you for your endless patience, love, and
understanding. Your support has sustained me through the most difficult moments and
brought joy to the best ones.

I am profoundly thankful to my co-authors and research collaborators, whose contribu-
tions, ideas, and discussions have significantly shaped the direction of my work. Their
intellectual generosity and commitment to excellence have been a constant source of in-
spiration.

To other past and present members of the ever-expanding research group: Adrian, Aleks,
Ali, Brochette, Bryce, Daniel, Dion, Ethan, Forrest, Giulia, Gustav, James, Jérémie,
Jordy, Liang, Matteo, Marrium, Michaela, Neea, Nicol, Riddhi, Saunak, Sebastian, Shàshi,
Shehzad, Valeriia and Whisky, thank you for the camaraderie, thoughtful conversations,
and shared challenges over the years. Your support has made the journey all the more
rewarding.

Last but not least, many thanks to the thesis examiners for a thorough review of the
manuscript: your insightful suggestions have greatly improved the quality and readability
of this thesis.




                                           iv
Publications

The following papers were either published or accepted for publication in journals during
my PhD candidature. Ideas from some of these publications appear in Chapter 3 and 4
of this thesis and are cited where appropriate.


   • D. Johnston and A. Yang. “Some explicit estimates for the error term in the prime
     number theorem”. In: J. Math. Anal. Appl. 527.2, 127460 (2023).

   • G. A. Hiary, D. Patel, and A. Yang. “An improved explicit estimate for ζ(1/2 + it)”.
     In: J. Number Theory 256 (2024), pp. 195–217.

   • M. J. Mossinghoff, T. S. Trudgian, and A. Yang. “Explicit zero-free regions for the
     Riemann zeta-function”. In: Res. Number Theory 10, 11 (2024).

   • A. Yang. “Explicit bounds on ζ(s) in the critical strip and a zero-free region”. In:
     J. Math. Anal. Appl. 534.2, 128124 (2024).

   • D. Patel and A. Yang. “An explicit sub-Weyl bound for ζ(1/2 + it)”. In: J. Number
     Theory 262 (2024), pp. 301–334.

   • C. Bellotti and A. Yang. “On the generalised Dirichlet divisor problem”. In: Bull.
     Lond. Math. Soc. 56.5 (2024), pp. 1859–1878.

   • T. S. Trudgian and A. Yang. “Toward optimal exponent pairs”. In: Math. Comput.
     94 (2025), pp. 1467–1502.

   • G. A. Hiary, N. Leong, and A. Yang. “Explicit bounds for the Riemann zeta-function
     on the 1-line”. In: Funct. Approx. Comment. Math. 73.1 (2025), pp. 53–89.

   • T. Tao, T. S. Trudgian, and A. Yang. “New exponent pairs, zero density estimates,
     and zero additive energy estimates: a systematic approach”. In: Math. Comput.
     (to appear) (2025).




                                           v
Contents


Abstract                                                                                      iii


Acknowledgement                                                                               iv


Publications                                                                                   v


1 Introduction                                                                                 1

  1.1   Classical zero-free regions . . . . . . . . . . . . . . . . . . . . . . . . . . . .    3

  1.2   Asymptotically larger zero-free regions . . . . . . . . . . . . . . . . . . . . .      8

  1.3   New results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

  1.4   Other approaches and conditional results       . . . . . . . . . . . . . . . . . . . 12


2 The classical zero-free region                                                              15

  2.1   Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

  2.2   Outline of argument . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

        2.2.1   An optimization problem . . . . . . . . . . . . . . . . . . . . . . . . 20

  2.3   The test function f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

        2.3.1   Properties of the functions w and f      . . . . . . . . . . . . . . . . . . 23

  2.4   The trigonometric polynomial P       . . . . . . . . . . . . . . . . . . . . . . . . 33

  2.5   Proof of Lemma 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37




                                              vi
3 Littlewood’s zero-free region                                                             45

  3.1   Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

  3.2   The zero detector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

  3.3   Proof of Lemma 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

        3.3.1   Bounding the main terms . . . . . . . . . . . . . . . . . . . . . . . . 53

        3.3.2   Bounding the sums over zeroes . . . . . . . . . . . . . . . . . . . . . 62

        3.3.3   Other estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

        3.3.4   Putting it all together . . . . . . . . . . . . . . . . . . . . . . . . . . 76


4 The Vinogradov–Korobov zero-free region                                                   79

  4.1   Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

  4.2   An explicit estimate of zeta . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

  4.3   Proof of Lemma 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86


References                                                                                  97




                                             vii
viii
Chapter 1


Introduction

Let s = σ + it denote a complex number, and let ζ(s) denote the Riemann zeta-function,
defined as
                                                 ∞
                                                 X     1
                                        ζ(s) =
                                                       ns
                                                 n=1

for σ > 1, and by its analytic continuation in the rest of the complex plane, except at
s = 1 where there is a simple pole. We refer to the set of complex numbers s for which
0 < Re s < 1 as the critical strip. The behaviour of ζ(s) in the critical strip is both
mysterious and intimately connected to the distribution of prime numbers. In particular,
an important open problem in analytic number theory is the location and distribution of
complex numbers ρ for which ζ(ρ) = 0, collectively known as the zeroes of the Riemann
zeta-function.

The zeroes of ζ(s) fall into two categories. There are infinitely many real trivial zeroes
occurring at s = −2n for any positive integer n; these zeroes are well-understood. This
thesis instead focuses on the locations of the less-understood non-trivial zeroes, which are
complex valued. One motivation for studying such zeroes is their astonishing connection
to the prime numbers, as demonstrated through the “explicit” formula [Man95]
                    X                  X xρ                 1
                          Λ(n) = x −           − log 2π −     log(1 − x−2 ),
                                       ρ
                                           ρ                2
                    n≤x




                                               1
CHAPTER 1. INTRODUCTION


for x > 0 not coinciding with a prime power, where the sum on the right is over all non-
trivial zeroes of ζ(s), and Λ(n) is the von Mangoldt function, defined as log p if n = pm
for some prime number p and positive integer m, and 0 otherwise.

However, despite over a century of effort initiated by Riemann’s famous memoir [Rie59],
relatively little is known about the precise locations of the non-trivial zeroes. By the
classical Euler product formula
                                       Y      1
                              ζ(s) =                        (Re s > 1)
                                       p
                                           1 − p−s

where p runs through the prime numbers, one can see that there are no zeroes ρ with
Re ρ > 1. In addition, by the functional equation (see e.g. Titchmarsh [Tit86, Ch. II])
                                                 πs 
                          ζ(s) = 2s π s−1 sin            Γ(1 − s)ζ(1 − s),
                                                  2

it is evident that there are also no zeroes with Re ρ < 0 other than those arising from the
poles of the gamma function, which are precisely the trivial zeroes. Therefore, all non-
trivial zeroes reside inside the strip 0 ≤ Re s ≤ 1. We can also observe from the functional
equation that if ρ is a non-trivial zero, then so is 1 − ρ, i.e. non-trivial zeroes exist in pairs
that are symmetric about the vertical line Re s = 1/2, commonly known as the critical
line. Furthermore, since ζ(s) = ζ(s), the set of non-trivial zeroes is also symmetric about
the real axis. It is also known that there are infinitely many such zeroes [Har14]; in fact
precise growth rates are known: the number of non-trivial zeroes ρ with Im ρ ∈ [−T, T ] is
(1/π + o(1))T log T as T → ∞ [Man05].

Riemann’s paper contained a conjecture that all non-trivial zeroes in fact lie on the critical
line; this later became known as the Riemann hypothesis, a famous open problem in pure
mathematics. The truth of the Riemann hypothesis will have far-reaching consequences
in many fields, including for instance the prime number estimate [Sch76]
                                                                         Z x
                                       1 1/2                                    dt
                     |π(x) − li(x)| <    x log x,             li(x) :=               ,
                                      8π                                  2    log t

for x ≥ 2657, where π(x) denotes the number of primes no greater than x. While a
proof of the Riemann hypothesis appears far out of reach, a number of partial results



                                                  2
                                                    1.1. CLASSICAL ZERO-FREE REGIONS


have been established. It is known that a positive proportion of zeroes lie on the critical
line [Sel42], and that the proportion is at least 5/12 [Pra+20]. Computationally, the first
1013 zeroes have been verified to lie on the critical line [PT21]. A number of zero-density
estimates have been proved, which bound the quantity N (σ, T ) counting the number of
zeroes ρ = β + iγ with β ≥ σ > 1/2 and |γ| ≤ T . Lastly, it is known that there are no
zeroes inside certain regions within the critical strip; we refer to these regions as zero-free
regions.

The goal of this thesis is to enlarge the known explicit zero-free regions by refining a
number of methods that we will review over the next few chapters. Here we refer to a
result as “explicit” if all implied constants are specified (including, say, those constants
implied by requiring a parameter to be “sufficiently large”). In the next few sections, we
review the historical progression of results in this area and then present the new results
that will be proved in this thesis.



1.1        Classical zero-free regions

In 1896, Hadamard [Had96] and de la Vallée Poussin [VP96] showed independently that
ζ(s) has no zeroes on the vertical line Re s = 1 (and thus also ζ(s) ̸= 0 on Re s = 0). This
represented the first “non-trivial” result regarding the location of zeta zeroes, and led to
the proof of the celebrated prime number theorem that

                                          π(x) ∼ li(x)

as x → ∞. In 1899, de la Vallée Poussin [VP99] adapted the method to show that if t is
sufficiently large, then in fact ζ(σ + it) ̸= 0 for

                                                        A
                                         σ >1−
                                                      log t

for some absolute constant A > 0. We shall refer to this as the classical zero-free region,
and this implies the estimate
                                                              
                             π(x) − li(x) ≪ x exp −c(log x)1/2



                                                3
CHAPTER 1. INTRODUCTION


for some constant c > 0. Here, we provide a sketch of the proof of de la Vallée Poussin’s
result, which will be useful to facilitate a comparison with the various refinements to the
method that followed. One begins by writing
                                                               ζ ′ (s)
                                                       −Re
                                                               ζ(s)
in two ways: first as a Dirichlet series (for Re s > 1), and second as an expansion obtained
via the Hadamard product formula (see e.g. Davenport [Dav00, Ch. 12]). This gives
               X Λ(n)          1           1 Γ′  s          1       X 1
          Re                = − log π + Re          + 1 + Re     − Re       ,              (1.1)
                      ns       2           2Γ 2              s−1      ρ
                                                                        s−ρ
               n≥1

for Re s > 1, where the sum is over non-trivial zeroes of zeta (here and thereafter, zeroes are
summed in conjugate pairs and counted with multiplicity). Suppose now that ρ0 = β + it
is a non-trivial zero off the critical line. For some σ > 1 to be chosen later, we evaluate
(1.1) at s = σ, σ + it and σ + 2it. Routine calculations give (as t → ∞)
                                            X Λ(n)          1
                                                       =       + O(1),
                                                  nσ       σ−1
                                            n≥1


                                 X Λ(n)                  X           1
                           Re              σ+it
                                                = −Re                       + O(log t),    (1.2)
                                       n                   ρ
                                                                 σ + it − ρ
                                 n≥1

                                X Λ(n)                   X            1
                           Re             σ+2it
                                                = −Re                        + O(log t),   (1.3)
                                      n                    ρ
                                                                 σ + 2it − ρ
                                n≥1

where all implied constants are absolute. Meanwhile, in light of the non-negative trigono-
metric polynomial
                                 3 + 4 cos θ + cos 2θ = 2(1 + cos θ)2 ≥ 0,                 (1.4)

one has
                     X Λ(n)
               0≤                 (3 + 4 cos(t log n) + cos(2t log n))
                            nσ
                     n≥1
                      3        X     1           X      1
                =        − 4Re              − Re               + O(log t).
                     σ−1       ρ
                                 σ + it − ρ      ρ
                                                   σ + 2it − ρ

Next we make use of the fact that the terms appearing in the sums are one-signed, since
σ > 1 > Re ρ and
                                                  1
                                            Re      >0           (Re z > 0).               (1.5)
                                                  z


                                                           4
                                                  1.1. CLASSICAL ZERO-FREE REGIONS


We use this to drop the entire second sum, and also all terms except for ρ = ρ0 in the first
sum. This gives
                                     3   4
                               0≤      −    + O(log t),                                 (1.6)
                                    σ−1 σ−β
where the implied constant can be made explicit. At this point we choose
                                                     c
                                        σ =1+
                                                   log t
where c is a suitable positive constant, which, upon rearranging, gives
                                                     A
                                        β ≤1−
                                                   log t
for some absolute constant A > 0 if t is sufficiently large.

In fact in [VP99, Theorem 31] it was shown that A = (30.46)−1 = 0.03282 . . . is permissible
for t sufficiently large. Subsequent refinements over the course of the last century steadily
increased the size of A but most still followed essentially the same argument. The historical
progression in this constant is recorded in Table 1.1, which we review next in chronological
order.

                     Table 1.1: Historical progress in the constant A.
         Year                  Reference                              A
         1899         de la Vallée Poussin [VP99]         (30.46)−1 = 0.0328 . . .
         1909               Landau [Lan09]                (18.524)−1 = 0.0539 . . .
         1938               Westphal [Wes38]              (17.537)−1 = 0.0570 . . .
         1941                Rosser [Ros41]                (17.72)−1 = 0.0564 . . .
         1962          Rosser–Schoenfeld [RS62]           (17.517)−1 = 0.0570 . . .
         1970               Stechkin [Ste70]                 (9.65)−1 = 0.103 . . .
         1975          Rosser–Schoenfeld [RS75]             (9.646)−1 = 0.103 . . .
         1977             Kondrat’ev [Kon77]                (9.548)−1 = 0.104 . . .
         2002                Ford [For02b]                  (8.463)−1 = 0.118 . . .
         2005                Kadiri [Kad05]                 (5.697)−1 = 0.175 . . .
         2014              Jang–Kwon [JK14]                 (5.684)−1 = 0.175 . . .
         2014        Mossinghoff–Trudgian [MT14]            (5.573)−1 = 0.179 . . .
         2022     Mossinghoff–Trudgian–Yang [MTY24]         (5.559)−1 = 0.179 . . .
         2025                 Theorem 1.1                   (4.862)−1 = 0.205 . . .


The first refinement, due to Landau (1909) [Lan09, page 321], was based on the observation
that one may replace the trigonometric polynomial (1.4) with any polynomial of the form
                                         X
                                P (θ) =       ak cos kθ
                                           0≤k≤K



                                              5
CHAPTER 1. INTRODUCTION


provided that P (θ) ≥ 0, ak ≥ 0 and a1 > a0 . Landau chose the polynomial 5 + 8 cos θ +
4 cos 2θ + cos 3θ to obtain A = (18.524)−1 . This was subsequently improved to A =
(17.537)−1 by Westphal (1938) [Wes38] and A = (17.517)−1 by Rosser–Schoenfeld (1962)
[RS62], using different polynomials.

The next idea, due to Stechkin (1970) [Ste70], was to make use of the fact that non-trivial
zeroes of ζ(s) exist in pairs symmetric about the critical line. That is, if ρ is a non-
trivial zero, then so is 1 − ρ. Central to Stechkin’s approach was the observation that if
                                          √                    √
F (z) = z −1 − κ(z + δ)−1 , where κ = 1/ 5 and δ = (1 − σ + 1 + 4σ 2 )/2, then for any
pair of non-trivial zeroes ρ and ρ′ = 1 − ρ one has

                                      ReF (s − ρ) + ReF (s − ρ′ ) ≥ 0.                 (1.7)

This achieves the same outcome as the inequality (1.5); however, one may now repeat the
classical argument with (1.1) replaced by

                                                  1 Γ′  s           Γ′ s + δ
                                                                                
       X Λ(n)              κ     1−κ                         
  Re                   1−      = −     log π + Re          + 1   − κ          + 1
             ns             nδ      2             2 Γ 2              Γ    2
       n≥1
                                                         X                             (1.8)
                                      + ReF (s − 1) − Re     F (s − ρ).
                                                                ρ

The partial cancellation of the Γ′ /Γ terms ultimately increases A by a factor of (1 − κ)−1
for sufficiently large t. Using this method Stechkin obtained A = (9.65)−1 . This was later
improved to A = (9.646)−1 by Rosser–Schoenfeld (1975) [RS75] and to A = (9.548)−1 by
Kondrat’ev (1977) [Kon77], using different trigonometric polynomials.

From the modern perspective, both (1.1) and (1.8) can be viewed as examples of zero-
detectors. Here, a zero-detector is simply a complex function Q(s) that is large when s is
close to a non-trivial zero, and which can be represented both as a Dirichlet series as well
as an expansion involving the zeroes of ζ(s). In 1992, Heath-Brown [HB92a] considered
the problem of choosing the most favourable zero-detector of the form
            X Λ(n)                    f (0) Γ′  s                    X
       Re             s
                        f (log n) =        Re      + 1 + ReF (s − 1) −   ReF (s − ρ)
                  n                     2     Γ 2                      ρ
            n≥1                                                                        (1.9)
                                            + error terms,



                                                     6
                                                 1.1. CLASSICAL ZERO-FREE REGIONS


where f (x) is any non-negative, compactly supported function satisfying certain regularity
conditions, and F is now the Laplace transform of f . Ignoring temporarily the requirement
that f be compact, this zero-detector generalizes both (1.1) and (1.8) (which take f (x) = 1
and f (x) = 1 − κe−δx respectively). Since such zero-detectors possess a Hadamard-type
expansion (commonly known as Weil’s explicit formula [Wei52]), one may use them directly
in place of (1.1). A calculus-of-variations argument is then used to derive specific choices
of f . Importantly, such zero-detectors have a larger radius of convergence than (1.1),
which allows one to take s within the critical strip (instead of σ > 1). Although [HB92a]
did not derive an explicit zero-free region for zeta (the focus was on other L-functions),
this marked a theoretical breakthrough and the same choices of f appear in all subsequent
works.

In 2002, Ford [For02b] developed a new zero-detector with the same Dirichlet series rep-
resentation as Heath-Brown’s zero-detector; however it replaces the usual Hadamard ex-
pansion with a Jensen-type formula:

                          f (0) ∞ log |ζ(s − λ + 2λiu/π)| − log |ζ(s + λ + 2λiu/π)|
     X Λ(n)                    Z
  Re          f (log n) =                                                           du
          ns               4λ −∞                      (cosh u)2
     n≥1
                                         X
                        + ReF (s − 1) −        ReF (s − ρ) + error terms,
                                            ρ
                                        Re ρ≥σ−λ

                                                                                     (1.10)

for suitable λ > 0. This formula is characteristic of Landau’s “local” approach (see §1.2
below), which in principle only uses properties of zeta close to the point s (as opposed to
de la Vallée Poussin’s “global” approach that involve e.g. sums over all zeta zeroes). As
with most Jensen-type formulas, Ford’s zero detector allows one to pass from bounds on
the size of a function to knowledge about its zeroes, and requires no information about
the specific analytic properties of ζ(s). When combined with an explicit subconvexity
estimate of ζ(s) on the critical line, Ford obtained A = (8.463)−1 .

Neither of the two previous treatments made substantial use of the fact that non-trivial
zeroes exist in pairs about the critical line — this is a limitation of the “local” approach
since one cannot apply (1.7) as the sum in (1.10) contains at most one zero from each



                                             7
CHAPTER 1. INTRODUCTION


pair. Nevertheless, Kadiri (2005) [Kad05] was able to combine the techniques of Stechkin
and Heath-Brown by using a zero-detector of the form
                 X Λ(n)         κ′
                                    
                            1 − δ′ f (log n) = R(s) − κ′ R(s + δ ′ ),                (1.11)
                      ns        n
                    n≥1

where R(s) is the expression on the right side of (1.9), κ′ and δ ′ are positive constants
and f is a smoothing function from [HB92a]. Along with some other refinements, Kadiri
obtained A = (5.697)−1 . Later, Jang–Kwon (2014) [JK14] revisited Kadiri’s approach
(1.11) with various other choices of f found in [HB92a], as well as a function studied by
Xylouris [Xyl11], to obtain A = (5.684)−1 .

The next source of improvement was computational in nature. In 2014, Mossinghoff–
Trudgian [MT14] used simulated annealing, a probabilistic optimization algorithm, to
find favourable non-negative trigonometric polynomials of high degree (previous treat-
ments used polynomials of degree at most four, with the exception of [Kon77] who used a
degree eight polynomial). This search algorithm was particularly well-suited for (approx-
imate) global optimization problems when the objective function has many suboptimal
local maxima. A large scale computation found a 16th degree polynomial, which, together
with other numerical refinements, produced A = (5.573)−1 . Figure 1.1 compares this poly-
nomial to some earlier polynomials. Later, Mossinghoff–Trudgian–Yang (2024) [MTY24]
slightly modified the argument to obtain A = (5.559)−1 .

In this thesis, we will combine the computational and analytic tools of previous works to
show that A = (4.862)−1 is admissible (Theorem 1.1 below). This is the goal of Chapter 2.



1.2    Asymptotically larger zero-free regions

A different approach to deriving zero-free regions was developed in 1924 by Landau [Lan24]
(see Titchmarsh [Tit86, pages 56–59] for an exposition). Roughly speaking, one first relates
the zeroes of ζ(s) on some domain D ⊂ C to the modulus of ζ(s) on D. Then, using known
upper-bounds on |ζ(s)|, one arrives at a contradiction if any zero is too close to the line



                                              8
                            1.2. ASYMPTOTICALLY LARGER ZERO-FREE REGIONS


                  P (θ)
             1




           0.5




                                                                                           θ
                                                    π                                      2π

                                    3 + 4 cos θ + cos 2θ (classical)
                          5 + 8 cos θ + 4 cos 2θ + cos 3θ (Landau [Lan09])
                      16th degree polynomial (Mossinghoff–Trudgian [MT14])

Figure 1.1: Comparison of trigonometric polynomials P (θ) (normalized so that P (0) = 1).

σ = 1. The first step can be accomplished using Jensen’s formula or another similar device
(such as Ford’s zero-detector (1.10)). The second step typically requires an estimate of
ζ(s) inside the critical strip, which can be obtained via the (often difficult) analysis of
certain exponential sums.

One limitation of de la Vallée Poussin’s approach is that Re Γ′ /Γ(s/2 + 1) ≫ log t, so one
can only obtain zero-free regions of classical type, i.e. σ > 1 − A/ log t for some constant
A. On the other hand, by appealing to certain non-trivial estimates of the exponential
sum
                                                    X
                                   S(N, t) :=               n−it ,
                                                N <n≤2N
one can obtain bounds on ζ(s) that are sharp enough to generate asymptotically larger
zero-free regions using Landau’s method. Littlewood [Lit22] in 1922 used the Weyl–van
der Corput method to derive an estimate of the form
                                    k       k                2−k +k/(2k −2)        k
                 S(N, t) ≪ N 1−k/(2 −2) t1/(2 −2) + N 1−2                     t−1/(2 −2)

uniformly for integers k ≥ 3, which was used to show that
                                                        k
                     ζ(1 − k/(2k − 2) + it) ≪ t1/(2 −2) log t          (t → ∞)


                                                9
CHAPTER 1. INTRODUCTION


and consequently that ζ(σ + it) ̸= 0 for t sufficiently large and

                                                      log log t
                                      σ >1−A                                          (1.12)
                                                        log t

where A > 0 is an absolute constant. This also implies the prime number estimate
                                          p
                   π(x) − li(x) ≪ x exp(−c log x log log x)             (x → ∞)

for some constant c > 0.

In 1935, Vinogradov [Vin35] developed a method for estimating averages of certain expo-
nential sums, which became known later as Vinogradov’s mean value theorem. The method
was successively refined by multiple authors, including Linnik [Lin43], Karatsuba [Kar73],
Stechkin [Ste75], Wooley [Woo12; Woo16] and Bourgain–Demeter–Guth [BDG16]. In its
modern form, Vinogradov’s method implies (amongst other consequences) an estimate of
the form
                                                                   2
                                 S(N, t) ≪ N 1−c(log N/ log t)

for some absolute constant c > 0, and also the following bound (due to Richert [Ric67])

                                          3/2
                     ζ(σ + it) ≪ tB(1−σ)        (log t)2/3        (1/2 ≤ σ ≤ 1)       (1.13)

where B > 0 is an absolute constant. In 1938, Chudakov [Chu38] used a early form of
Vinogradov’s mean value theorem to show that ζ(σ + it) ̸= 0 for

                                                        1
                                    σ >1−
                                                (log t)3/4+o(1)
as t → ∞. Then, in 1958, Vinogradov [Vin58] and Korobov [Kor58] independently proved1
that ζ(σ + it) ̸= 0 for t sufficiently large and

                                                        A
                                σ >1−                                                 (1.14)
                                         (log t)2/3 (log log t)1/3
which is currently the form of the largest known zero-free region as t → ∞. This region
also implies the prime number estimate

              π(x) − li(x) ≪ x exp(−c(log x)3/5 (log log x)−1/5 )           (x → ∞)

   1 Both proofs in fact claim the larger zero-free region σ > 1 − A(log t)−2/3 ; however

their methods appear to only produce a region of strength (1.14).


                                                 10
                                                                           1.3. NEW RESULTS


for some constant c > 0.

A number of explicit versions of (1.12) and (1.14) have been derived, which we will review
in Chapters 3 and 4. In the explicit setting, the values of A attainable for such zero-free
regions are often substantially smaller than those found in the classical zero-free region.
For many applications in explicit number theory, the classical zero-free region remains
the largest for ranges of t of interest. Nevertheless, in Chapter 3 and 4 of this thesis, we
derive improved explicit values of A for (1.12) and (1.14) respectively (Theorem 1.2 and
Theorem 1.3 below).



1.3     New results

The aim of this thesis is to establish the following zero-free regions, which, to the best of
the author’s knowledge, are new:

Theorem 1.1. One has ζ(σ + it) ̸= 0 if t ≥ 2 and

                                                  1
                                     σ >1−                .
                                              4.862 log t

Theorem 1.2. One has ζ(σ + it) ̸= 0 if t ≥ 3 and

                                               log log t
                                     σ >1−                .
                                              19.62 log t

Theorem 1.3. One has ζ(σ + it) ̸= 0 if t ≥ 3 and

                                                   1
                            σ >1−                                      .
                                     51.34 (log t)2/3 (log log t)1/3


The proofs of these theorems are the subject of Chapters 2, 3 and 4 respectively. In Platt–
Trudgian (2021) [PT21], all non-trivial zeroes ρ with |Im ρ| ≤ 3·1012 were computationally
verified to lie on the critical line. The above bounds represent the current largest known
zero-free regions in the range t > 3 · 1012 . Of these, Theorem 1.1 is sharpest in the interval
3 · 1012 < t < exp(56.5), Theorem 1.2 is sharpest for exp(56.6) < t < exp(5.4 · 105 ) and
Theorem 1.3 is sharpest for t > exp(5.5 · 105 ).



                                              11
CHAPTER 1. INTRODUCTION


1.4     Other approaches and conditional results

There exist other (often creative) methods of proving zero-free regions, although these
approaches have not yet been applied in an explicit context. We conclude this chapter
by briefly reviewing some of these methods, as well as recording some stronger zero-free
regions that are possible assuming certain hypotheses.

It is well-known that zero-free regions of ζ(s) imply certain error estimates in the prime
number theorem. Turán [THP84, Theorem 40.1]2 showed that the following converse
relationship also holds: if one has π(x) − li(x) ≪ x exp(−A(log x)θ ) for some fixed θ ∈
(1/2, 1), then there are no zeroes in the region σ > 1−A/(log t)1/θ−1 for t sufficiently large.
This provides a way of obtaining zero-free regions using an elementary proof of the prime
number theorem (first established by Selberg [Sel49] and Erdős [Erd49]). Presently, the
sharpest known prime number estimate obtained via elementary means is π(x) − li(x) ≪
x exp(−c(log x)1/6 ) for a constant c > 0, proved by Srinivasan–Sampath [SS88]. This
implies a zero-free region of the form σ > 1 − A/(log t)5 if t is sufficiently large. Although
this region is smaller than the largest known zero-free region, the method is interesting
since it requires only arithmetic information about the primes.

The approaches reviewed in previous sections all rely on the existence of a non-negative
trigonometric polynomial whose coefficients satisfy a1 > a0 . In 1976, Balasubramanian–
Ramachandra [BR76] developed a method of analysis with no such dependence, relying
instead on the polynomial 1 + cos θ and the estimate
                  X      log p
                               (1 + cos(t log p)) ≫ X 1−σ     (1 ≪ t ≪ X 1/2 )
                          pσ
                X<p≤2X

as X → ∞, obtained via sieve-theoretic methods. Their approach also requires no knowl-
edge about the behaviour of ζ(s) at height kt (k ≥ 2). As discussed in [Tit86, page 68], this
method is capable of producing a zero-free region of classical strength, i.e. σ > 1−A/ log t.
Another unrelated sieve-theoretic approach due to Motohashi [Mot78] used the Selberg
sieve to pass from bounds on the size of ζ(s) to a zero-free region, thereby providing an

   2 A more precise form of this relationship was later obtained by Pintz [Pin80].




                                              12
                              1.4. OTHER APPROACHES AND CONDITIONAL RESULTS


alternative approach to Landau’s method of detecting zeroes. When combined with (1.13),
Motohashi recovered the Vinogradov–Korobov zero-free region.

There is another loosely-related group of ideas that involve analyzing the interactions
between two or more complex zeroes of ζ(s). Levinson [Lev69] first showed that if the
non-trivial zeroes of ζ(s) are well-spaced close to the line σ = 1, then there are no zeroes
in the region
                                                         A
                                              σ >1−                                            (1.15)
                                                      log log t
for some constant A > 0. Here, “well-spaced” means there is a fixed δ > 0 such that, if ρ, ρ′
are zeroes with Re ρ, Re ρ′ ≥ 1 − δ, then Im (ρ − ρ′ ) ≥ δ. This spacing condition certainly
holds on average (for δ sufficiently small), since the number of zeroes in the rectangle
[1 − δ, 1] × [0, T ] is known to be O(T 2δ ) (see e.g. [Bou00]). However, local zero-density
estimates are presently not strong enough to ensure this spacing condition always holds,
and thus (1.15) remains a conditional result. One might then ask what types of zero-free
regions are possible given what we currently know about the local density of zeroes. In
this direction, Montgomery [Mon71, Chapter 11] showed that if ρ0 = β0 + it is a zero with
β0 > 1/2, 1 − β0 ≤ δ ≤ 1 and t > 0, then
                          Z 1Z ∞
                                   n(t, w, h) + n(2t, w, h)
                  δ   2
                                                            dh dw ≫ (1 − β0 )−1 ,              (1.16)
                          0   0         (h + δ)5 ew/δ

where n(t, h, w) is the number of zeroes in the rectangle [1 − w, 1] × [t − h/2, t + h/2].
Here, the left side represents a weighted average of the number of zeroes close to 1 + it and
1 + 2it. Bounding the left side with the best available zero-density estimates (obtained via
Jensen’s formula and (1.13)), Montgomery recovered the Vinogradov–Korobov zero-free
region. Later, Ramachandra [Ram78] obtained a pointwise version of (1.16), where the left
side is replaced with a sum over the zeroes close to 1 + it and 1 + 2it. A further refinement
in Balasubramanian–Ramachandra [BR82] showed that one may ignore the zeroes near
1 + 2it at the cost of a small amount of averaging at height t. One consequence of their
result is
                                  Z δ/2 Z δ+X
                              1
        n(t, δ, δ)e−δ/X +                       n(t + v, u, u)e−u/X du dv ≫ δ 6 X(1 − β0 )−1
                              X    −δ/2   0




                                                    13
CHAPTER 1. INTRODUCTION


for sufficiently small δ > 0 and all δ −4 (1 − β0 ) ≪ X ≪ (log log t)−1 . Each of these devices
can be used in conjunction with (1.13) to recover the Vinogradov–Korobov zero-free region.
While these methods currently have no explicit counterparts, it appears possible (at least
in principle) that some combination of these ideas can be used to further enlarge the
zero-free region.




                                              14
Chapter 2


The classical zero-free region


2.1     Introduction

In the previous chapter, we reviewed two ideas that were successfully used to enlarge the
classical zero-free region of ζ(s). The first idea, due to Stechkin [Ste70], was to make use of
the fact that the complex zeroes of the zeta-function exist in pairs about the critical line.
The second idea, due to Heath-Brown [HB92a], was to smooth the classical zero-detector
using a weight function f . Recall that the zero-detectors they used have the Dirichlet
series representations
                     X Λ(n)         κ                X Λ(n)
                                1−            and                 f (log n)
                           ns        nδ                      ns
                     n≥1                               n≥1

respectively, where f is a compactly supported, twice-differentiable real-valued function
that decays steadily to 0.

Remarkably, the two ideas appear to be largely orthogonal, despite their both involving
weighting the classical zero-detector n≥1 Λ(n)n−s by a function f (log n) (with f (x) =
                                     P

1−κe−δx in Stechkin’s case). This is because the weight functions are used in two different
ways. In Stechkin’s approach, the weight function allows one to reduce the key quantity
f (0) from 1 in the classical case to 1 − κ. Normally, doing so invalidates an important



                                              15
CHAPTER 2. THE CLASSICAL ZERO-FREE REGION


positivity condition (appearing as (1.5) in the classical proof); however, this problem was
avoided by exploiting the existence of zeroes in pairs. Meanwhile, Heath-Brown’s approach
makes no critical use of the fact that zeroes exist in pairs; instead, the improvement comes
from using f to attenuate the zero-detector’s tail oscillations so that one may move s
inside the critical strip. In terms of shape, the two weight functions are polar opposites:
Stechkin’s function attenuates terms in the zero-detector corresponding to small n, while
Heath-Brown’s function attenuates terms corresponding to large n.

One may therefore wonder if the two ideas can be used concurrently. Indeed, Kadiri
[Kad05] showed that this can be done profitably by using the zero detector

                                                     κ′
                             X Λ(n)                    
                                      f (log n) 1 − δ′                                 (2.1)
                                  ns                n
                                   n≥1

for some constants κ′ , δ ′ > 0, and where f is the same test function considered by
Heath-Brown. Although the paper contains several other innovations, the combination
of smoothing functions explains most of the improvement from A = (8.463)−1 [For02b] to
A = (5.69693)−1 [Kad05].

The modest goal of this chapter is to explore how the two ideas can be combined more
efficiently. Our main result is:

Lemma 2.1. If 2 ≤ t ≤ exp(56.6) then ζ(σ + it) ̸= 0 in the region

                                                     1
                                         σ >1−               .
                                                 4.862 log t


Theorem 1.1 then follows from combining this with Lemma 3.1 in the next chapter (since
Lemma 3.1 implies a result of the same strength for t ≥ exp(56.6)). Our improvement
                                                                 ′
comes from generalizing the “Stechkin factor” 1 − κ′ /nδ into a function h(log n), then
choosing h appropriately using a numerical approach.




                                                 16
                                                                2.2. OUTLINE OF ARGUMENT


2.2      Outline of argument

In this section we will roughly describe the argument, and use this to motivate our choice
of a smoothing function f . Throughout, we draw heavy inspiration both from the classical
proof in the previous chapter and from the argument of Heath-Brown [HB92a]. We also
take this opportunity to state plainly a number of observations that are implicit in several
works in the literature. First, since non-trivial zeroes are symmetric about the critical line
and the real axis, we will just focus on the quadrant {z ∈ C : Re z ≥ 1/2, Im z ≥ 0}. In
fact, we can further restrict our focus to

                                        Im z ≥ H := 3 · 1012                             (2.2)

since all zeroes ρ = β + iγ with 0 < γ ≤ H have been computationally verified to lie on
the critical line [PT21]. The non-trivial zeroes at heights just above H pose the greatest
difficulty — indeed, asymptotically wider zero-free regions are possible at sufficiently large
heights. Therefore, we optimize our arguments for zeroes whose imaginary part lies in a
finite interval, which explains the form of Lemma 2.1.

Suppose f is a non-negative, bounded and compactly supported function with a well-
defined Laplace transform
                                                Z ∞
                                     F (z) :=         e−zu f (u)du.
                                                 0

Under suitable assumptions, the associated zero-detector has an “explicit formula” of the
form
                      X Λ(n)                    f (0) Γ′  s    
                 Re             s
                                  f (log n) =        Re      + 1 + ReF (s − 1)
                            n                     2     Γ 2
                      n≥1
                                                         X                               (2.3)
                                                      −     ReF (s − ρ) + o(1),
                                                           ρ

where s = σ + it and the error term is taken with respect to t → ∞. This formula may
be viewed as a natural generalization of Hadamard’s product formula used in the classical
proof, with an important distinction that the formula remains valid for s inside the critical
strip.




                                                     17
CHAPTER 2. THE CLASSICAL ZERO-FREE REGION


It turns out that moving s inside the critical strip is accretive to the size of the zero-
free region, provided that s itself resides in a zero-free region. This inspires an interesting
inductive argument: one begins with a known zero-free region, then iteratively1 expands it
by taking s progressively deeper inside the critical strip as permitted by the new zero-free
region at each stage. This idea was introduced in [Kad05] and has since become standard
in the literature.

Specifically, suppose that there exists a simple2 , complex zero ρ0 = β0 + it lying just
outside a known zero-free region σ > 1 − A/ log t for some constant A > 0. We seek to
show that ρ0 actually satisfies a stronger inequality, of the form β0 ≤ 1 − (A + ε)/ log t
say, for some ε > 0. Assume for a contradiction that

                                    A ≤ (1 − β0 ) log t < A + ε.                                      (2.4)

We start by considering a non-negative trigonometric polynomial of the form
                                                X
                                     P (θ) =            ak cos(kθ),                                   (2.5)
                                               0≤k≤K

satisfying P (θ) ≥ 0, ak > 0 and a1 > a0 , which generalizes the polynomial 3+4 cos θ+cos 2θ
used in the classical proof. The existence and construction of such polynomials is non-
trivial, and we refer the reader to [MT14] for an account. Combined with (2.3) evaluated
at s = sk := σ + ikt for 0 ≤ k ≤ K, we obtain
           X Λ(n)                                   X           X Λ(n)
      0≤             σ
                       f (log n)P (t log n) = Re           ak               f (log n)
                 n                                                    nsk
           n≥1                                     0≤k≤K        n≥1
                                                                                        !             (2.6)
                 X           f (0) Γ′  sk                   X
        = Re            ak                 + 1 + F (sk − 1) −   F (sk − ρ)                  + o(1).
                               2 Γ 2                          ρ
               0≤k≤K

Using classical estimates, as t → ∞,
                        f (0) Γ′  sk
             X                                      
        Re         ak                 + 1 + F (sk − 1) < cf (0) log t + a0 F (σ − 1),
                          2 Γ 2
            0≤k≤K

   1 As usual, there is no free lunch — eventually the marginal progress made with each

iteration tends to 0 and the size of the zero-free region stabilizes.
   2 Since all zeroes in the sum (2.3) are counted with multiplicity, if ρ has multiplicity
                                                                          0
m > 1 then it is included in the sum m times, and we arbitrarily rename one of them ρ0 .
The remaining m − 1 terms are treated the same way as the other zeroes ρ ̸= ρ0 .


                                                   18
                                                             2.2. OUTLINE OF ARGUMENT


for an absolute constant c > 0. To bound the sum over zeroes we require an additional
assumption on f . Specifically, we assume the following positivity condition: for any non-
trivial zero ρ,
                           ReF (s − ρ) + ReF (s − (1 − ρ)) ≥ 0.                         (2.7)

Note that both ρ and 1 − ρ are (possibly distinct) complex zeroes of zeta, so (2.7) may
be viewed as a slightly “averaged” version of the pointwise condition ReF (s − ρ) ≥ 0
considered in Heath-Brown [HB92a]. Condition (2.7) is also satisfied by Kadiri’s [Kad05]
smoothing function, and it is at this point where the ideas of Stechkin and Heath-Brown
are combined. This condition allows us to isolate the contribution of just one pair of
zeroes, say β0 + it and 1 − β0 + it, since
                           X
                      Re       F (s1 − ρ) ≥ F (σ − β0 ) + F (σ − 1 + β0 )
                           ρ

and
                                    X
                               Re       F (sk − ρ) ≥ 0     (k ̸= 1).
                                    ρ

The F (σ − 1 + β0 ) term is o(1), so combining these estimates with (2.6), one has

                                    a1 F (σ − β0 ) − a0 F (σ − 1) + o(1)
                        c log t >                                        .              (2.8)
                                                    f (0)

Next we choose σ and f so that the right side is ≫ 1/η, where

                                           η := 1 − β0 .

This can be achieved, for instance, if one takes σ = 1 − (A + o(1))/ log t and f (u) = ηg(ηu)
for some fixed function g. With these choices the right side of (2.8) is

                                    a1 G(1 − µ) − a0 G(−µ) −1
                                                          η
                                             g(0)

where G is the Laplace transform of g and µ = (1 − σ)/η = 1 + O(ε). Taking ε → 0, the
coefficient of η −1 is a constant, as required. Therefore, one arrives at a contradiction to
(2.4) if A is small enough, which completes the argument.




                                                19
CHAPTER 2. THE CLASSICAL ZERO-FREE REGION


2.2.1    An optimization problem


Let us now consider the problem of choosing the best f for each η > 0 (we now regard η as
fixed). First, we review the constraints one needs to place on f . To control the various o(1)
error terms we require F (z) = f (0)/z +O(|z|−2 ) as |z| → ∞. This condition is surprisingly
mild3 — in [HB92a] this was enforced by requiring f to be compactly supported and having
a bounded and continuous second derivative. The second, more important condition comes
from (2.7) — for this it suffices to require that ReF (z)+ReF (2σ−1−z) ≥ 0 for all Re z ≥ 0.

Taking ε → 0 and ignoring o(1) terms, from (2.8) the optimization problem we are led to
consider is (for each fixed η)
                                         a1 F (0) − a0 F (−η)
                                  max                                                    (2.9)
                                    f            f (0)
subject to F (z) = f (0)/z + O(|z|−2 ) and

                      ReF (z) + ReF (2σ − 1 − z) ≥ 0            (Re z ≥ 0).            (2.10)

In [HB92a], Heath-Brown studied a similar optimization problem, which has the constraint
ReF (z) ≥ 0 in place of (2.10). In the next section we explore how solutions to Heath-
Brown’s optimization problem can be adapted to solve (2.9).



2.3     The test function f

In this section we choose the test function f . We emphasize that some of our arguments in
this section are not completely rigorous, and in particular we make no claims of optimality.
Rather, our goal is to heuristically motivate our choice of f .

First, suppose that w is a twice-differentiable, compactly-supported, real-valued function
and let W (z) denote its Laplace transform, which satisfies

                                 ReW (z) ≥ 0        (Re z ≥ 0).
   3 In fact, our eventual choice of f
                                     satisfies the stronger condition F (z) = f (0)/z +
O(|z|−3 ). The improved error term was also noted in [Kad05].


                                               20
                                                               2.3. THE TEST FUNCTION F


Examples of functions satisfying these conditions are studied by Heath-Brown [HB92a].
One function which is particularly relevant to our application is [HB92a, Lemma 7.1] with
λ = 1, given by:

Definition 2.1 (The functions w and W ). For fixed θ ∈ (0, π/2), let w be defined as

          w(u) := sec2 θ(sec2 θ(θ cot θ − u/2) cos(u tan θ) + 2θ cot θ − u

                          + sin(2θ − u tan θ) csc 2θ − 2(1 + sin(θ − u tan θ) csc θ))

                                                                                   R ∞ −zu
for 0 ≤ u ≤ 2θ cot θ, and w(u) := 0 elsewhere. Furthermore, let W (z) :=            0 e    w(u)du
denote its Laplace transform.


In our notation, one choice of f considered by Heath-Brown was

                                           ηw(ηu).                                         (2.11)

In [HB92a], it was shown via a calculus-of-variations argument that this choice of f solves,
within a large family of functions, the optimization problem

                 a1 F (0) − a0 F (−η)
           max                           subject to          ReF (z) ≥ 0     (Re z ≥ 0).
             f           f (0)

Consider now the function

                                                                   1
                      f (u) = ηw(ηu)h(u),          h(u) :=                   .             (2.12)
                                                             1 + e−(2σ−1)u

This choice of f satisfies (2.10), since for any z = iy (y ∈ R),
                                                 Z ∞
            ReF (z) + ReF (2σ − 1 − z) = Re             e−iyu (1 + e−(2σ−1)u )f (u)du      (2.13)
                                                 Z0 ∞
                                          = Re          e−iyu/η w(u)du                     (2.14)
                                                   0

                                          = ReW (iy/η) ≥ 0.

We can then extend this inequality to hold for all Re z ≥ 0 via a theorem of Heath-Brown
[HB92a, Lemma 4.1]. In particular, one might expect that this choice of weight function
h is uniquely efficient because there is no “loss” in passing from (2.13) to (2.14).




                                              21
CHAPTER 2. THE CLASSICAL ZERO-FREE REGION


Unfortunately, f does not satisfy the condition F (z) = f (0)/z+O(|z|−2 ). Furthermore, the
inverse Laplace transform of h does not converge, which makes analysis difficult. For these
reasons, we will instead use a discrete approximation of h(u) which produces numerically
similar results, and which possesses the desired properties. Roughly speaking, we choose
parameters κm (1 ≤ m ≤ M ) so that uniformly for x ∈ [0, 1], one has
                                     X                 1
                                            κm xm ≈       .                            (2.15)
                                                      1+x
                                   0≤m≤M

One may be inclined to find the coefficients κm in (2.15) using a Taylor expansion at say
x = 1/2; however, this leads to approximations that are catastrophically bad at e.g. x = 0.
Instead, we search for favourable coefficients computationally. As a first pass we choose
κm to minimize the L∞ norm

                                       1           X
                               max         −             κm xm .
                               0≤x≤1 1 + x
                                               0≤m≤M

Then, we perturb the coefficients so as to make the arguments that follow simpler (in
particular, we opt for slightly less favourable coefficients that can be more easily analyzed,
and whose rational approximations are “simple” to aid our presentation). In the end we
take M = 6 and
                                                                 
                                  851 780     525 171 28       29
                [κm ]0≤m≤M = 1, −    ,    , −    ,   ,    , −       .                  (2.16)
                                  859 859     859 859 859     859

Our choice of f is then given by

Definition 2.2 (The functions f and F ). For fixed η > 0 and 1/2 < σ < 1, define
                                              X
                           f (u) := ηw(ηu)            κm e−(2σ−1)mu                    (2.17)
                                             0≤m≤M

where M and κm are fixed parameters defined in (2.16).             Furthermore, let F (z) :=
R ∞ −zu
 0 e    f (u)du denote its Laplace transform.


In comparison, Kadiri’s [Kad05] test function (2.1) may be expressed in our notation as
                                             X
                                   ηw(ηu)           κm e−δm u                          (2.18)
                                            0≤m≤1




                                              22
                                                     2.3.1   Properties of the functions w and f


with κ0 = 1, δ0 = 0, κ1 = κ′ and δ1 = δ ′ . One can measure the “efficiency” of the test
functions (2.11), (2.12), (2.17) and (2.18) by evaluating the functional in (2.9), which in
each case gives

                      a1 F (0) − a0 F (−η)              a1 W (0) − a0 W (−1)
                                           = (c + O(η))                      ,
                              f (0)                            ηw(0)

where c = 859/433 = 1.98 . . . in the case of (2.17), c = 2 for the hypothetical function
(2.12), c = 1.78 . . . for (2.18) and c = 1 for (2.11). In this sense, the value attained by
our discrete approximation is within one percent of the hypothetical limit of the method.
Figure 2.1 provides a comparison of the respective test functions.

              f (u)
          1

        0.9

        0.8

        0.7

        0.6                                                  [HB92a, Lemma 7.4]
                                                                  [Kad05]
                                                                Definition 2.2
        0.5
                                                                                     u
                                5                 10                 15               20

Figure 2.1: Plot of different choices of test functions close to u = 0 (normalized so that
ηw(0) = 1) when σ = σ0 .



2.3.1   Properties of the functions w and f


Let us now record some properties of these functions, starting with:

Lemma 2.2. One has f (u) ≥ 0 for all u ≥ 0.


Proof. By inspection w(u) ≥ 0, so upon taking x = e−(2σ−1)u (and noting that the expo-




                                                23
CHAPTER 2. THE CLASSICAL ZERO-FREE REGION


nent is negative from the assumption σ > 1/2), it suffices to show that
                                X
                        p(x) =       κm x m > 0      (0 < x ≤ 1),
                                   0≤m≤M

which we verify by numerically isolating all M roots of the polynomial p, checking that
none reside on (0, 1], and noting that p(1) > 0.


Next, we record an explicit estimate for W (z), due to Ford [For02b; For22].

Lemma 2.3 (Ford [For02b] §7). Let θ ∈ (0, π/2) and r > tan θ be fixed, and let W be
defined in Definition 2.1. If we write
                                               w(0)
                                     W (z) =        + W0 (z),
                                                z
then for all Re z ≥ ν and |z| ≥ r one has |W0 (z)| ≤ C(ν, r)|z|−3 , where
                                     c2 (r + 1)2 (e−2νθ cot θ + 1) + c1 r + c3 r3
                   C(ν, r) := c0 r
                                                   (r2 − tan2 θ)2
and
                     c0 = csc θ sec2 θ,        c1 = (θ − sin θ cos θ) tan4 θ,

                    c2 = tan3 θ sin2 θ,         c3 = (θ − sin θ cos θ) tan2 θ.


Before proceeding further we fix
                                            θ = 1.1338                                 (2.19)

in Definition 2.1 and 2.2, the choice of which was determined via numerical experimenta-
tion. Thus f and F are now pure functions, and also

                 w(0) = sec2 θ(θ tan θ + 3θ cot θ − 3) = 5.672787598 . . . .

Recall H is defined in (2.2). For future use we also define the following parameters

                     A0 = (4.862)−1 = 0.2056766762 . . . ,

                     T0 = 1010 ,

                     δm = (2σ − 1)m,
                                                                                       (2.20)
                     η0 = A0 / log H = 0.0071590428 . . . ,

                     K = 16,

                     σ0 = 1 − A0 / log(KH + T0 ) = 0.9934710854 . . . .


                                                 24
                                                          2.3.1       Properties of the functions w and f


The parameter A0 is the constant in the zero-free region of Lemma 2.1. We defined σ0
so that our eventual choice of σ satisfies σ ∈ [σ0 , 1), and similarly one is guaranteed that
η ∈ (0, η0 ]. The parameter K is the degree of the non-negative trigonometric polynomial
which we discuss in §2.4.

Recall that ReW (z) ≥ 0 for Re z ≥ 0 (see [HB92a]). This property is used to show that f
satisfies the non-negativity condition (2.7).

Lemma 2.4 (Non-negativity property). Let σ ∈ [σ0 , 1), η ∈ (0, η0 ] and F (z) be as defined
in Definition 2.2. Then for all complex numbers z with 0 ≤ Re z ≤ 2σ − 1, one has

                                 ReF (z) + ReF (2σ − 1 − z) ≥ 0.                                   (2.21)

Furthermore, for all −η ≤ Re z ≤ 1 and |Imz| ≥ T0 , one has

                                                                            44
                         ReF (z) + ReF (2σ − 1 − z) > −                          η.                (2.22)
                                                                          |Imz|2

Proof. Since ReF (2σ −1−z) = ReF (2σ −1−z), we may show (2.21) by proving ReF (z)+
ReF (2σ − 1 − z) ≥ 0 on the boundary of the rectangle {z ∈ C : 0 ≤ Re z ≤ 2σ − 1, |Imz| ≤
T } for every sufficiently large T . The result then holds by the maximum modulus principle
(applied to the holomorphic function e−F (z)−F (2σ−1−z) ).

First we will show that the desired inequality holds on the line Re z = 0 (so that by
symmetry it also holds for Re z = 2σ − 1). Let us write
                                      X                           X
                         (1 + x)             κm x m =                      bm xm .
                                    0≤m≤M                   0≤m≤M +1

Then

            ReF (iy) + ReF (2σ − 1 − iy)
                          Z ∞                              X
                     = Re     e−iyu (1 + e−(2σ−1)u )ηw(ηu)   κm e−δm u du
                             0                                              0≤m≤M
                                             Z ∞
                                 X                                                                 (2.23)
                     = Re               bm           e−(δm +iy)u/η w(u)du
                            0≤m≤M +1         0
                                                                 
                                 X                   δm + iy
                     ≥ Re               bm W                          ,
                                                        η
                            1≤m≤M +1



                                                     25
CHAPTER 2. THE CLASSICAL ZERO-FREE REGION


where we dropped the term corresponding to m = 0 since ReW (iy/η) ≥ 0. If z =
(δm + iy)/η with m ≥ 1, then |z| ≥ Re z ≥ (2σ0 − 1)/η0 > 137. Thus, applying Lemma
2.3,

                                                 η02 C(137, 137) 1
                                       
             1                δm + iy                                 η        ε0    η
                     W0                     ≤                                <
        (2σ − 1)w(0)             η              (2σ0 − 1)2 w(0) m |δm + iy|2   m |δm + iy|2

with ε0 = 1/2000, and also

                            1          w(0)            η
                                   Re           =m            .
                       (2σ − 1)w(0) (δm + iy)/η    |δm + iy|2

Therefore,
                                                             
              1                X                    δm + iy
                      Re                bm W
        (2σ − 1)w(0)η                                  η
                           1≤m≤M +1
                               X        bm m − |bm |ε0 /m                      X
                       ≥                                  ≥ B(y) :=                    bm Bm (y)
                                           |δm + iy|2
                           1≤m≤M +1                                        1≤m≤M +1

where, since c0 m < (2σ0 − 1)m ≤ δm ≤ m, where c0 = 151/153,
                                     
                                       m + ε0 /m
                                                   , bm ≤ 0,
                                     
                                     
                                      2 2       2
                            Bm (y) := c0 m + y
                                      m − ε0 /m
                                     
                                                  , bm > 0.
                                        m2 + y 2
It suffices to verify that B(y) ≥ 0 for all y ∈ R. Substituting the values of κm (0 ≤ m ≤ M ),
one may explicitly derive (with the aid of computer assistance) that B(y) = p(y)/q(y),
where

   p(y) = 4061245152630328137981y 12 + 4077560173170236734684710y 10

         − 104378137212291977844887868y 8 − 4484512641017853031179075270y 6

         + 135673322742635307737680349343y 4 − 229732179325278720034298507440y 2

         + 112359769561546903428467326544,


               q(y) = 4012454647232553285540000(y 2 + 1)(y 2 + 9)(y 2 + 25)

                               × (y 2 + 4c20 )(y 2 + 16c20 )(y 2 + 36c20 )(y 2 + 49c20 ).

By inspection, q(y) has no real roots, and by computationally isolating all twelve complex
roots of p(y), we verify that none are real. Therefore, B(y) is one-signed on R which


                                                     26
                                                 2.3.1   Properties of the functions w and f


together with B(0) > 0 implies B(y) > 0 for all y ∈ R. Combined with (2.23), one obtains
(2.21) on Re z = 0.

Next we show (2.21) holds on the line segment Imz = T (0 ≤ Re z ≤ 2σ −1) for sufficiently
large T (and hence also on Imz = −T since ReF (z) = ReF (z)). For this we simply note
that for any fixed η and uniformly for 0 ≤ x ≤ 2σ − 1,
                                               
                           X             x + iT     ηw(0)
       ReF (x + iT ) = Re       κm W              =       (κ + o(1))           (T → ∞),
                                           η         T2
                          0≤m≤M
             P
where κ :=     0≤m≤M κm > 0, so ReF (x + iT ) > 0 for sufficiently large T .        Thus (2.21)
holds.

Now consider (2.22). In light of the bound (2.21) it suffices to show (2.22) for −η ≤ Re z <
0 and Imz ≥ T0 . First, suppose u = x + iy is a complex number and v a real number,
such that |v| ≤ η, x ≥ 0 and |y| ≥ T0 . Then, by Lemma 2.3 one has
                                                                  
             u+v           u                1      1          u+v         u
       W             −W         = ηw(0)         −      + W0         − W0     ,
               η           η               u+v u               η          η

where, since Re(u/η) ≥ 0, Re(u + v)/η ≥ −1 and |y/η| ≥ T0 /η0 one has

                          η02 C(−1, T0 /η0 ) η              η02 C(0, T0 /η0 ) η
                                                     
                u+v                                    u
          W0           ≤                        , W 0     ≤                      .
                 η               T0          y2        η           T0         y2

Meanwhile,
                                1   1       |v|      η
                                  −   =            ≤ 2.
                               u+v u    |u||u + v|  y
Combining everything,
                                                                 
         u+v           u    η        C(−1, T0 /η0 ) + C(0, T0 /η0 ) 2   5.7η
    W           −W        ≤ 2 w(0) +                               η0 < 2 .
           η           η   y                      T0                     y

Taking u = δm + iy (m ≥ 0) and v = Re z, one gets
                                         
                      X            z + δm               5.7η           X
          ReF (z) =         κm W            > ReF (iy) − 2                     |κm |.
                                      η                  y
                       0≤m≤M                                         0≤m≤M

Similarly, if we instead take u = 2σ − 1 + δm + iy and v = −Re z, then

                                                           5.7η    X
                 ReF (2σ − 1 − z) > ReF (2σ − 1 + iy) −                   |κm |.
                                                            y2
                                                                  0≤m≤M

Adding the two inequalities gives (2.22), since ReF (iy)+ReF (2σ−1+iy) ≥ 0 by (2.21).


                                            27
CHAPTER 2. THE CLASSICAL ZERO-FREE REGION


Lemma 2.4 is used to isolate the contribution of a pair of zeroes at height t, while discarding
all other zeroes (incurring a small loss).

Lemma 2.5. Suppose ζ(β0 + it) = 0 with β0 > 1/2 and t ≥ H. Let A ∈ (0, A0 ],
η ∈ (0, A0 / log t] and F be as defined in Definition 2.2. Furthermore suppose that ζ has no
zeroes β + iγ in the region β ≥ 1 − A/ log γ and γ ≥ H. Let sk = σ + ikt where 0 ≤ k ≤ K
and σ = 1 − A/ log(Kt + T0 ). Then
                    X
                 Re   F (s1 − ρ) ≥ F (σ − β0 ) + F (σ − 1 + β0 ) − 10−8 .
                          ρ

Furthermore, for any k ̸= 1 one has
                                  X
                               Re   F (sk − ρ) ≥ −10−8 .
                                       ρ


Proof. Throughout let us write
                                           T = Kt + T0 .

We divide the zeroes ρ = β + iγ of ζ in the critical strip into those with |γ| ≤ T and those
with |γ| > T .

Suppose first that |γ| ≤ T . Then

                              β ≤ 1 − A/ log |γ| ≤ 1 − A/ log T = σ

so that Re(sk − ρ) ≥ 0 for all such zeroes. A similar argument shows that Re(sk − ρ) ≤
2σ − 1. Also, if ρ is a complex zero off the critical line, then 1 − ρ is also a zero. On the
other hand, if ρ lies on the critical line then sk − ρ = sk − (1 − ρ). Thus
                X                    1      X
          Re           F (sk − ρ) = Re            (F (sk − ρ) + F (sk − (1 − ρ)))
                                     2
                 ρ=β+iγ                     ρ=1/2+iγ
                  |γ|≤T                       |γ|≤T
                                                  X
                                          + Re            (F (sk − ρ) + F (sk − (1 − ρ))).
                                                 ρ=β+iγ
                                                  β>1/2
                                                  |γ|≤T

First, suppose k ̸= 1. Since 0 ≤ Re(sk − ρ) ≤ 2σ − 1 we may apply (2.21) termwise with
z = s − ρ, so that both sums on the right side are non-negative. Therefore
                              X
                          Re        F (sk − ρ) ≥ 0     (k ̸= 1).                             (2.24)
                                 ρ=β+iγ
                                  |γ|≤T




                                                   28
                                                        2.3.1   Properties of the functions w and f


On the other hand if k = 1, then we isolate the term corresponding to ρ = ρ0 in the second
sum, and apply (2.21) to all other terms (recall that ρ0 = β0 + it is our hypothetical zero
off the critical line). We obtain
                             X
                     Re            F (s1 − ρ) ≥ F (σ − β0 ) + F (2σ − 1 + β0 ).              (2.25)
                          ρ=β+iγ
                           |γ|≤T

It remains to estimate the contribution of those zeroes with |γ| > T . These zeroes are far
from s so the corresponding terms are small, and we can afford to estimate them crudely.
Applying (2.22) termwise, one obtains
                     X                           X
               Re            F (sk − ρ) = Re            (F (sk − ρ) + F (sk − ρ))
                    ρ=β+iγ                     ρ=β+iγ
                     |γ|>T                      γ>T
                                                  X             1           1
                                                                                     
                                        ≥ −44η                       2
                                                                       +                 .
                                                             (γ − kt)    (γ + kt)2
                                                 ρ=β+iγ
                                                  γ>T

Here we have made use of the fact that |Im(sk − ρ)|, |Im(sk − ρ)| > T0 for each zero in the
sum. If k = 0 then we use [Leh66, Lemma 2] with n = 2 to get
                                           X 1     log T0
                                               2
                                                 <
                                             γ       T0
                                          γ>T0

so that, since η ≤ A0 / log t ≤ η0 by assumption,
                               X                            X 1
                        Re            F (σ − ρ) > −88η0          > −10−8 .                   (2.26)
                                                              γ2
                             ρ=β+iγ                         γ>T0
                              |γ|>T

If k ̸= 0, then we use the same treatment as [MT14]. Applying [Leh66, Lemma 1] with
ϕ(x) = (x − kt)−2 + (x + kt)−2 , we get
                       Z ∞                                     Z ∞        
         X           1                x       ∗                     ϕ(x)
            ϕ(γ) =          ϕ(x) log    dx + O 4ϕ(T ) log T + 2          dx ,
                    2π T             2π                          T   x
         γ>T

where, here and henceforth, O∗ (A) means a complex number whose modulus does not
exceed A. Recall that T = Kt + T0 and k ≤ K, so that
       Z ∞                    Z ∞                    
                     x               1         1            kt + x
           ϕ(x) log     dx ≤          2
                                        +           2
                                                        log        dx
        T           2π         T0   x      (x + 2kt)          2π
                    1      kt + T0        1        kt + T0 log(2kt + T0 ) log T0
                =      log         +           log          +            −       .
                   T0        2π       2kt + T0        2π            kt      kt



                                                  29
CHAPTER 2. THE CLASSICAL ZERO-FREE REGION


We drop the last term, and bound the second and third terms using kt ≥ H, since they
are both decreasing in kt. Finally, the first term may be bounded using log(1 + x) ≤ x,
so that                                           
                     kt + T0       Kt           T0             T0       K
                 log         ≤ log           1+      ≤ log t +    + log    .
                       2π          2π           Kt             KH       2π
Computing the constants explicitly, one finds
                        Z ∞
                                       x
                            ϕ(x) log      dx < 10−10 (log t + 1).                       (2.27)
                         T0           2π

Next, since kt ≥ H, we have
                                         
                           1      1
        ϕ(T0 ) log T0 ≤      +              log(kt + T0 ) < 10−10 (log t + 1),          (2.28)
                          T02 (2H + T0 )2

                   Z ∞                 Z ∞                  
                           ϕ(x)               1      1           dx
                                dx ≤            +                   < 10−10 ,           (2.29)
                      T0    x          T0     x2 (2H + x)2       x
where in deriving (2.28) we used log(kt + T0 ) ≤ log t + T0 /(KH) + log K. Combining
(2.27), (2.28) and (2.29), and since η ≤ A0 / log t ≤ η0 by assumption
                                        −10                 
                X                        10              −11
           Re        F (sk − ρ) > −44η           + 6 · 10      (log t + 1) > −10−8 .
                                          2π
             ρ=β+iγ
              |γ|>T

The result follows from combining the above inequality with (2.24), (2.25) and (2.26) (with
some room to spare).


One can use the assumed properties of f to obtain an “explicit formula”. First, recall the
following theorem from [Kad05].

Lemma 2.6 (Kadiri [Kad05] Proposition 2.1). Let d > 0 and suppose g is a real-valued
function supported on [0, d) and twice continuously differentiable on [0, d], satisfying g(d) =
                                             R∞                               R∞
g ′ (0) = g ′ (d) = g ′′ (d) = 0. Let G(z) := 0 e−zu g(u)du and G2 (z) := 0 e−zu g ′′ (u)du
denote respectively the Laplace transforms of g and g ′′ . Then, for all complex numbers s,

                                                1 Γ′  s
         X Λ(n)                                                
                                      1
     Re           g(log n) = g(0)   −   log π +   Re        + 1    + Re G(s − 1)
              ns                      2         2    Γ 2
         n≥1

                                                   G2 (s − z) Γ′  z 
                                         Z 1/2+i∞                                   
                 X                    1                                       G2 (s)
           − Re      G(s − ρ) + Re                            Re         dz +          .
                  ρ
                                     2πi 1/2−i∞ (s − z)2          Γ 2           s2



                                                 30
                                                         2.3.1   Properties of the functions w and f


The next lemma is used to control the error term in the case relevant to our application.

Lemma 2.7 (Bounding the error term). Let s = σ + it, σ ≥ σ0 and η ∈ (0, η0 ]. Assume
either t = 0 or t ≥ H and η ≤ A0 / log t. If W0 (z) = W (z) − w(0)/z then
                                                            
         1
           Z 1/2+i∞      s − z  Γ′  z           s     723η 3 ,
                                                            
                                                                                                t = 0,
                     W0          Re        dz + W0       ≤
        2πi 1/2−i∞          η       Γ 2              η      14η 2 + 425η 3 ,                   t ≥ H.
                                                            


Proof. Let E1 (s) denote the left side of the above inequality, so that
                    Z ∞
                                                    Γ′ 1/2 + iy
                                                               
                  1             s − (1/2 + iy)
        E1 (s) ≤          W0                     Re                  dy + |W0 (s/η)|
                 2π −∞                 η            Γ       2
               = T1 + T2 ,

say. For convenience, throughout we write x0 := σ0 − 1/2. By assumption, η ≤ η0 . If
ξ = (s − 1/2 − iy)/η then |ξ| ≥ Re ξ ≥ x0 /η0 . Applying Lemma 2.3 (and noting that
C(x0 /η0 , x0 /η0 ) < 52), one has

                                                       52η 3
                                             
                                 s − 1/2 − iy
                            W0                  ≤ 2                  .                                   (2.30)
                                      η          (x0 + (t − y)2 )3/2

Meanwhile, as in [Kad05, Lemma 3.6] we use
                                  
                                    1       16          2
                                   log            +          + 2,     |y| < 1/2,
                                  
        Γ′ 1/2 + iy
                                
                                                 2   1 + 4y 2
     Re                 ≤ U (y) := 2      1 + 4y
        Γ        2                      |y|        2        2      1
                                   log     −           +      +     , |y| ≥ 1/2.
                                  
                                  
                                         2    1 + 4y 2    3|y| 8y 2

If t = 0 then via direct computation,
                                    Z ∞
                                              U (y)
                                         2    2 3/2
                                                           dy < 36.                                      (2.31)
                                    −∞ (x0 + y )

On the other hand if t ≥ H then we divide the integral as
              Z ∞                         Z −(t−2)       Z t−2         Z ∞
                    U (t + y)
                     2 + y 2 )3/2
                                  dy =               +             +         = I1 + I2 + I3 .
               −∞ (x 0                     −∞             −(t−2)       t−2

One may verify that U (x) < log(100 − x) for x ≤ 2, so that
                                    Z ∞
                                         log(y + 100)
                             I1 <               3
                                                      dy < 10−10 .
                                     H−2      y




                                                 31
CHAPTER 2. THE CLASSICAL ZERO-FREE REGION


For x ≥ 2, one may check that U (x) ≤ log x so
                  Z t−2                              Z
                         log(t + y)                                     dy       2
            I2 ≤                     dy < log t                              <       log t
                          2    2 3/2
                 −(t−2) (x0 + y )
                                                               2    2 3/2
                                                     |y|≤t−2 (x0 + y )           x20

and also
                                        Z ∞
                                             log y
                                 I3 ≤           3
                                                   dy < 10−10 .
                                         H−2  y
Combining these estimates with (2.30) and (2.31), and using η ≤ A0 / log t ≤ η0 one has
                                                     
                       52 3
                            Z ∞
                                      U (y)          298η 3 , t = 0,
                                                     
                  T1 ≤    η         2     2 3/2
                                                dy <                               (2.32)
                       2π    −∞ (x0 + y )            14η 2 ,  t ≥ H.
                                                     


Next if ξ = s/η then |ξ| ≥ Re ξ ≥ σ0 /η0 so that by Lemma 2.3, for all k ≥ 0

                                     C(σ0 /η0 , σ0 /η0 ) 3
                              T2 ≤                      η < 425η 3 .                         (2.33)
                                            x30

The desired result follows from combining (2.32) and (2.33).

Lemma 2.8 (The explicit formula). Let s = σ + it with σ ∈ [σ0 , 1) and t ≥ H, and
suppose η ∈ (0, A0 / log t]. If f and F are as in Definition 2.2, then

                                                              1 Γ′ s + δm
                                                                            
          X Λ(n)                        X          1
       Re           f (log n) = f (0)       κm − log π + Re                +1
                ns                                 2          2    Γ     2
          n≥1                         0≤m≤M
                                                      X
                                   + ReF (s − 1) − Re     F (s − ρ) + E,
                                                           ρ

where                                 
                                      2730η 3 ,
                                      
                                                               t = 0,
                              |E| <
                                      53η 2 + 1605η 3 ,
                                      
                                                               t > H.


Proof. Note that if g(u) = ηw(ηu) then g(d) = g ′ (0) = g ′ (d) = g ′′ (d) and

                              G2 (z)          g(0)
                                 2
                                     = G(z) −      = W0 (z/η).
                               z               z




                                                32
                                               2.4. THE TRIGONOMETRIC POLYNOMIAL P


We apply Lemma 2.8 with this choice of g and with s replaced by s + δm (0 ≤ m ≤ M ):
             X Λ(n)                        X             X Λ(n)
        Re             s
                         f (log n) = Re             κm                   ηw(η log n)
                   n                                           ns+δm
             n≥1                          0≤m≤M          n≥1
                                                       1 Γ′ s + δm
                                                                      
                            X                 1
                   = Re         κm ηw(0) − log π +                   +1
                                              2        2Γ        2
                         0≤m≤M
                                              X                                   
                                   s + δm − 1             s + δm − ρ
                            +W                  −    W                 + E1 (s + δm )
                                        η         ρ
                                                               η
                                                    1 Γ′ s + δm
                                                                    
                               X           1
                   = f (0)Re        κm − log π +                   +1
                                           2        2Γ       2
                             0≤m≤M
                                               X                X
                            + ReF (s − 1) − Re    F (s − ρ) +         κm E2 (s + δm )
                                                         ρ                   0≤m≤M

where
                                                                         Γ′  z 
                                    Z 1/2+i∞                   
                               1                        s−z
               E2 (s) := Re                    W0                   Re            dz + ReW0 (s/η).
                              2πi    1/2−i∞              η               Γ 2
Applying Lemma 2.7 (with s replaced by s + δm (0 ≤ m ≤ M ) as required), and using the
values of κm , one has
                                                             
                           X                                 2730η 3 ,
                                                             
                                                                                    t = 0,
                                 |κm E2 (s + δm )| <
                         0≤m≤M                               53η 2 + 1605η 3 ,
                                                             
                                                                                    t>H

as required.



2.4     The trigonometric polynomial P

As we briefly mentioned in §2.2, we require a non-negative trigonometric polynomial
                                                    X
                                      P (x) :=                ak cos kx ≥ 0
                                                 0≤k≤K

with coefficients satisfying a0 < a1 and ak ≥ 0 (0 ≤ k ≤ K). This is a generalization of
the polynomial 2(1 + cos x)2 = 3 + 4 cos x + cos 2x used in the classical proof. The task of
finding favourable polynomials was considered in [Kon77] and [MT14]. In this chapter we
will use the following trigonometric polynomial, which is a rational approximation to the




                                                         33
CHAPTER 2. THE CLASSICAL ZERO-FREE REGION


polynomial found in [MT14, Table 5]
                                                                         2
                                               1        X
                             P (x) :=                          ck eikx
                                           14912370
                                                       0≤k≤K

where we recall that K = 16, c0 = 4 and

          c1 = −8,               c2 = 2,                c3 = 20,                 c4 = −9,

          c5 = −34,              c6 = 27,               c7 = 91,                 c8 = −27,

          c9 = −201,             c10 = 32,              c11 = 895,               c12 = 1949,

          c13 = 2389,            c14 = 1896,            c15 = 949,               c16 = 239.

One may verify via a routine computation that all assumed conditions on P (x) are satisfied.
In particular one has
                                      865534                   X             2919857
                 a0 = 1,     a1 =            ,         a :=           ak =           .         (2.34)
                                      497079                                 828465
                                                              1≤k≤K

In standard proofs of the zero-free region, the non-negativity of P (x) is used to establish
the inequality
                              X Λ(n)
                                            f (log n)P (t log n) ≥ 0.
                                      nσ
                              n≥1

In the remainder of this section, we show that it is possible to strengthen this inequality
by replacing the right side with a positive quantity. This improvement will be used later
to obtain a larger zero-free region.

Lemma 2.9. Let f (u) be defined in Definition 2.2 and P (x) be defined as above. Then
                           X Λ(n)                                1
                                       f (log n)P (t log n) ≥       f (0).
                                 nσ                              12
                           n≥1


Our main idea is that instead of applying the estimate P (t log n) ≥ 0 pointwise, we first
sum over terms of the form n = pm (m ≥ 1) where p is a fixed prime. Here we do not
attempt to obtain the sharpest possible bounds; rather, our goal is merely to show that
some improvement is possible. In particular we only consider the contributions of the
primes 2 and 3 since our method is not sharp enough to extract meaningful contributions
from the other primes; however, in principle a sufficiently careful analysis should allow
one to consider all primes of size O(exp(1/η)).


                                                  34
                                            2.4. THE TRIGONOMETRIC POLYNOMIAL P


Lemma 2.10. If sk = σ + ikt with σ ∈ [σ0 , 1) and t ∈ R, then
                                                 
                         X       ak              0.21, p = 2,
                                                 
                    Re                 >   δp :=
                               psk − 1           
                                                 0.02, p = 3.
                       0≤k≤K



Proof. Write ϕp = ϕp (σ) := (pσ + p−σ )/2 and x = t log p so that
                    X        ak      p2σ − 1 X        ak        1 X
              Re                   =                          −     ak .               (2.35)
                           psk − 1     4pσ     ϕp (σ) − cos kx 2
                   0≤k≤K                      0≤k≤K                           0≤k≤K

Viewed as a function of x, the expression on the right side has period 2π so it suffices to
lower-bound
                             X             ak
                                                           (−π ≤ x < π).
                                    ϕp (σ) − cos kx
                            0≤k≤K

If y = cos x then cos kx = Tk (y), where Tk is the kth Chebyshev polynomial of the first
kind, the first few of which are

                   T0 (y) = 1,           T1 (y) = y,              T2 (y) = 2y 2 − 1,

                      T3 (y) = 4y 3 − 3y,               T4 (y) = 8y 4 − 8y 2 + 1.

Therefore, noting that 1 < ϕp (σ) < ϕp (1) so each term in the sum is positive, one has
          X            ak         X        ak         X     ak
                                >                   +              = hp (y),
                ϕp (σ) − cos kx     ϕp (1) − Tk (y)     ϕp (1) + 1
        0≤k≤K                        0≤k≤4                       5≤k≤K

say. We compute the minimum of hp (y) for y ∈ [−1, 1]. In the case of p = 2 one has

                                                   16510991
                                        h2 (y) >
                                                   2485395

since this inequality is equivalent to

          −2894773829632y 10 + 1704044564480y 9 + 7702262585344y 8

                      − 3358453172736y 7 − 7070063534336y 6 + 1811757646464y 5

                      + 2378922223840y 4 − 172357278640y 3 − 109315955656y 2

                      + 6202997496y + 2536883955 > 0




                                                   35
CHAPTER 2. THE CLASSICAL ZERO-FREE REGION


which one may verify by numerically isolating all 10 complex roots and checking that
none exist on the interval [−1, 1]. Similarly, one may show that h3 (y) > 8592212/2485395.
Combining this with (2.35), and using (p2σ − 1)/pσ ≥ (p2σ0 − 1)/pσ0 , one obtains

                            ak    p2σ0 − 1              1 X
                       Re s     >          inf hp (y) −     ak > δp ,
                         p k −1     4pσ0 |y|≤1          2
                                                               0≤k≤K

as required.

Lemma 2.11. If f is as defined in Definition 2.2, then f (u) ≥ κf (0) for 0 ≤ u ≤ 59,
         P
where κ = 0≤m≤M κm .


Proof. Note that w′ (x) ≤ 0 for x ≥ 0, so w(ηu) ≥ w(η0 u). Meanwhile, 2σ0 − 1 ≤ δm ≤ m
so
                                        X                             X
                  η −1 f (u) = w(ηu)           κm e−δm u ≥ w(η0 u)           κm e−dm u ,
                                       0≤m≤M                         0≤m≤M

where dm = (2σ0 − 1)m if κm < 0 and dm = m otherwise. The right side is a pure function
of u, and a routine computer-assisted computation is used to verify that
                                                     X
                              −κw(0) + w(η0 u)             κm e−dm u ≥ 0
                                                   0≤m≤M

for 0 ≤ u ≤ 59. The result follows from multiplying by η.


Using the non-negativity of f (u) and P (x), we drop all terms with n > N := exp(59).
Together with Lemma 2.11, we have
      X Λ(n)                                     X Λ(n)
                 σ
                   f (log n)P (t log n) ≥ κf (0)            P (t log n)
             n                                          nσ
      n≥1                                       1≤n≤N
                                                   X         X                 X
                                        = κf (0)Re    log p        ak                      p−msk .
                                                     p       0≤k≤K     1≤m≤log N/ log p

The inner sum can be estimated using
                 X                  1        X              1     p−(R−1)σ0
        Re             p−msk = Re s     − Re   p−msk ≥ Re s     −           ,
                                 p k −1                  p k −1   1 − p−σ0
             1≤m≤R                                 m>R




                                                   36
                                                                       2.5. PROOF OF LEMMA 2.1


since σ ≥ σ0 . Consider now just the terms with p = 2 or 3. Applying the above inequality
with R = log N/ log p and using Lemma 2.10,
           X                 X                           1    pσ0       X
                  ak                      p−msk ≥ δp −                         ak > δp − 10−10 .
                                                         N 1 − p−σ0
          0≤k≤K        1≤m≤log N/ log p                                0≤k≤K

Therefore, dropping all terms with p > 3,
           X Λ(n)                                        X
                      σ
                        f (log n)P (t log n) ≥ κf (0)        (δp − 10−10 ) log p > 0.086f (0)
                  n
           n≥1                                          p=2,3

from which Lemma 2.9 follows.



2.5       Proof of Lemma 2.1

To complete the argument we require a few more results to estimate certain terms appear-
ing on the right side of Lemma 2.8; these are recorded in the next few lemmas.

Lemma 2.12. Let sk = σ + ikt with k ≥ 0 an integer, σ0 ≤ σ < 1 and t ≥ H. If κm
(0 ≤ m ≤ M ) and ak (0 ≤ k ≤ K) are as defined in (2.16) and Section 2.4 respectively,
then

                                         1 Γ′ sk + δm
                                                       
          X            X         1                            aκ
               ak            κm − log π + Re          +1    ≤    log t − 1.568
                                 2       2   Γ   2             2
       0≤k≤K        0≤m≤M
              P                                            P
where a :=       1≤k≤K ak = 2919857/828465, κ :=                0≤m≤M κm = 433/859.



Proof. For k ≥ 1, we use the following estimate, due to [Kad05, Equation (24)]: if z = x+iy
with 0 < x < |y|, then

                            Γ′                                             x2
                                                                                
                                             1 x                      1
                          Re (z) = log |y| −        + O∗                  + 2        .
                            Γ                2 |z|2                 12x|y| 2y

Applying this estimate with z = (sk + δm )/2 + 1, all error terms for k ≥ 1 are O(t−1 ) so we
can afford to estimate them roughly. In particular, for 1 ≤ x ≤ (M + 3)/2 and y ≥ H/2
one has
                                       1 x      1     x2
                                   −         +      +     < 10−10 .
                                       2 |z|2 12x|y| 2y 2



                                                    37
CHAPTER 2. THE CLASSICAL ZERO-FREE REGION


Therefore
                                                  1 Γ′ sk + δm
                                                                      
                  X           X        1
                    ak         κm − log π + Re                       +1
                                       2          2   Γ       2
              1≤k≤K    0≤m≤M
                                                                  
                         κ X                          kt       −10
                       ≤           ak − log π + log      + 10                                           (2.36)
                         2                             2
                            1≤k≤K
                         aκ          aκ                         κ X
                       =     log t +     (− log 2π + 10−10 ) +          ak log k.
                          2           2                         2
                                                                                  1≤k≤K

For the term corresponding to k = 0 we use the fact that Γ′ /Γ(x) is increasing for x > 0,
so
             Γ′                                      Γ′                         Γ′
                                                                                           
                      σ0 + (2σ0 − 1)m                         σ + δm                      m+3
                                      +1           ≤                 +1       ≤                     .
             Γ               2                       Γ           2              Γ          2
Therefore, by choosing the appropriate bound depending on the sign of κm , we find
                                    Γ′ σ + δm
                                                 
                        1 X
                                κm             + 1 ≤ −0.041.                      (2.37)
                        2           Γ      2
                                 0≤m≤M

The result follows from combining (2.36) and (2.37).

Lemma 2.13. Suppose sk = σ + ikt (0 ≤ k ≤ K) with σ0 ≤ σ < 1 and t ≥ H. Let
η ∈ (0, η0 ] and F be as defined in Definition 2.2. Then
                                 X
                         Re           ak F (sk − 1) ≤ a0 F (σ − 1) + 10−10 η.
                              0≤k≤K


Proof. Suppose that k ≥ 1. By Lemma 2.3, for Re z ≥ ν, |z| ≥ r, one has
                                          w(0)Re z
                              ReW (z) =            + O∗ (C(ν, r)|z|−3 ).
                                            |z|2
Taking z = (sk + δm − 1)/η, so that |z| ≥ t/η ≥ H/η0 > 1010 and Re z ≥ (σ − 1)/η > −1.
A computer verification gives C(−1, 1010 ) < 51 which, together with Re z/|z|2 < M η/H 2 ,
gives
                                              
                                 sk + δm − 1           w(0)M    51
                       ReW                         <      2
                                                             η + 3 η 3 < 10−12 η.
                                      η                 H       H
Therefore, for all k ≥ 1,
                                                                             
                                         X                      sk + δm − 1
                  ReF (sk − 1) ≤                   κm W                           < 10−11 η.
                                                                     η
                                       0≤m≤M

The result follows from summing over k.


                                                       38
                                                                    2.5. PROOF OF LEMMA 2.1


Lemma 2.14. Suppose σ ∈ [σ0 , 1), η ∈ (0, η0 ] and

                                   1 − σ0           1−σ
                                          − 10−10 ≤     ≤ 1.
                                     η0              η

Let F be as defined in Definition 2.2 and a0 , a1 be as defined in (2.34). Then

          a1 F (σ − 1 + η) + a1 F (σ − η) − a0 F (σ − 1)

                                         ≥ C1 ((1 − σ)/η) + 3.909η + 26η 2 − 3897η 3 ,

where C1 (x) := 0.87637 + 0.12002x + 0.01017x2 − 0.00073x3 .


Proof. For convenience, denote

                           1−σ                1 − σ0
                    µ :=       ,      µ0 :=          − 10−10 = 0.91198 . . . .            (2.38)
                            η                   η0

First we bound the expression a1 F (σ − 1 + η) − a0 F (σ − 1), which may be written as
                       X
                              κm (a1 W (δm /η − µ + 1) − a0 W (δm /η − µ)).
                      0≤m≤M

For any a, ε, with a ̸= 0 and a + ε ̸= 0 one has

                                    1   1  ε    ε2
                                       = − 2+ 2                                           (2.39)
                                   a+ε  a a  a (a + ε)

so that
                                                         
                            ηw(0)                 δm              w(0)    xw(0) 2
           W (δm /η + x) =         + W0              +x       =        η−   2
                                                                               η + E(η)
                           δm + xη                 η               δm      δm

where
                                    x2 w(0)
                                                                      
                                                                  δm
                           E(η) = 2           η 3 + W0               +x .
                                 δm (δm + xη)                      η
If |x| ≤ x0 and δm + xη ≥ x1 > 0, then by Lemma 2.3,

                                     x20 w(0)
                                                               
                                                    C(x1 , x1 )
                      |E(η)| ≤                    +               η3.
                                 (2σ0 − 1)2 m2 x1      x31

In particular, if m ≥ 1 and x = −µ then we may take x0 = µ0 and x1 = (2σ0 − 1)m − µ0 η0
so that, writing Sk := 1≤m≤M κm /mk ,
                        P

                  X                           w(0)η        µw(0)η 2
                        κm W (δm /η − µ) ≤           S1 −           S2 + 1124η 3 .
                                              2σ − 1      (2σ − 1)2
               1≤m≤M



                                                  39
CHAPTER 2. THE CLASSICAL ZERO-FREE REGION


Similarly, if x = −µ + 1 then we may take x0 = 1 − µ0 and x1 = (2σ0 − 1)m − (1 − µ0 )η0
to get
             X                                     w(0)η       (µ + 1)w(0)η 2
                      κm W (δm /η − µ + 1) ≥              S1 −                S2 − 1185η 3 .
                                                   2σ − 1        (2σ − 1)2
           1≤m≤M

We explicitly evaluate S1 = −0.689736127 . . . and S2 = −0.818779265 . . .. Furthermore,
substituting the values of a0 , a1 and w(0), and applying µ ≥ µ0 and σ ≥ σ0 , one finds
        X
              κm (a1 W (δm /η − µ + 1) − a0 W (δm /η − µ))
         1≤m≤M
                              w(0)(a1 − a0 )          w(0)((a1 − a0 )µ + a1 )                               (2.40)
                       >η                    S1 − η 2                         S2 − 3188η 3
                                 2σ − 1                     (2σ − 1)2
                       > −2.939η + 11η 2 − 3188η 3 .

Next, the term corresponding to m = 0 is
                                                      Z 2θ cot θ
                  a1 W (−µ + 1) − a0 W (−µ) =                      eµu (a1 e−u − a0 )w(u)du.
                                                         0

For all 0 ≤ u ≤ 2θ cot θ = 1.05923293 . . . one has
                                                x2 x3       1
                                       1+x+       +   − ex < x4
                                                2   6       18
so that
                                                                                           c∗
                                                                                               
                                                         c2                           c3
                  a1 W (−µ + 1) − a0 W (−µ) > c0 + c1 µ + µ2 +                           −          µ3
                                                         2                            6    18
where
                 Z 2θ cot θ                                          Z 2θ cot θ
         cn :=                un (a1 e−u − a0 )w(u)du,       c∗ :=                u3 |a1 e−u − a0 |w(u)du
                  0                                                   0

are computable constants. In particular, we find, using computer assistance,

          c0 = 0.8763706262 . . . ,        c1 = 0.1200272738 . . . ,        c2 = 0.0203537951 . . . ,

          c3 = 0.0004382722 . . . ,        c∗ = 0.0190417514 . . . ,

so that
                                    a1 W (−µ + 1) − a0 W (−µ) > C1 (µ).                                     (2.41)

It remains to bound a1 F (σ − η). One has
                                               X
                                F (σ − η) =           κm W ((σ − η + δm )/η).
                                              0≤m≤M



                                                      40
                                                                 2.5. PROOF OF LEMMA 2.1


Applying (2.39) with a = 1 + δm and ε = σ − 1 − η = −(µ + 1)η,
                                                          
                   σ + δm − η                1       1+µ
              W                 = ηw(0)          +          η + E1 (η)
                        η                 1 + δm (1 + δm )2

where
                                 w(0)(1 + µ)2
                                                                                   
                                                                       σ + δm − η
              E1 (η) =                                η 3 + W0                          .
                         (1 + δm )(1 + δm − (1 + µ)η)                       η
Recall that µ0 ≤ µ ≤ 1, η ≤ η0 and δm ≥ dm := (2σ0 − 1)m. Furthermore, for all m ≥ 0,
one has (σ + δm − η)/η ≥ σ0 /η0 − 1 so that by Lemma 2.3
                                                                               
                                 4w(0)                 C(σ0 /η0 − 1, σ0 /η0 − 1) 3
      |E1 (η)| ≤                                     +                           η .
                   (1 + dm )(1 + dm − (1 + µ0 )η0 )m          (σ0 − η0 )3

Next we write Tk (σ) := 0≤m≤M κm /(1 + δm )k , so that
                         P

                                                 X         mκm
                              Tk′ (σ) = −2k
                                                        (1 + δm )k+1
                                              1≤m≤M

is positive for k = 1, 2 and all σ0 ≤ σ ≤ 1. Therefore,

            a1 F (σ − η) = a1 ηw(0) (T1 (σ) + η(1 + µ)T2 (σ)) + a1 E1 (η)

                         ≥ a1 ηw(0)T1 (σ0 ) + a1 η 2 w(0)(1 + µ0 )T2 (σ0 ) + a1 E1 (η)

                         > 6.848η + 15η 2 − 709η 3 .

The result follows from combining this bound with (2.40) and (2.41).


We now proceed to the main inductive argument.

Lemma 2.15 (Iteration lemma). Suppose that ζ(σ + it) ̸= 0 in the region

                                                A
                                  σ >1−             ,    t≥H
                                              log t

for some 1/6 < A ≤ A0 . Then ζ(σ + it) ̸= 0 in the region

                                           A+ε
                                  σ >1−          ,        t≥H
                                           log t

with ε = 10−100 .




                                                 41
CHAPTER 2. THE CLASSICAL ZERO-FREE REGION


Proof. Suppose for a contradiction that ρ0 = β0 + it is a zero of ζ(s) with

                           A ≤ (1 − β0 ) log t < A + ε and t ≥ H.                      (2.42)

Fix sk = σ + ikt with 0 ≤ k ≤ K and σ = 1 − A/ log(Kt + T0 ), so that σ ∈ [σ0 , 1). Let

                                                η := 1 − β0

so that η ∈ (0, η0 ], and choose
                                                             X
                                  f (u) = ηw(ηu)                 κm e−δm u
                                                        0≤m≤M

as in Definition 2.2. By Lemma 2.8 with s = sk ,

                                                            1 Γ′ sk + δm
          X Λ(n)                                                              
                                        X        1
       Re           f (log n) = f (0)       κm − log π + Re                 +1
               nsk                               2          2   Γ      2
          n≥1                         0≤m≤M
                                                                X
                                            + ReF (sk − 1) − Re    F (sk − ρ) + Ek ,
                                                                             ρ

where |Ek | ≤ 2730η 3 if k = 0 and |Ek | ≤ 53η 2 + 1605η 3 if k ≥ 1. Next, let
                                                   X
                                       P (x) =               ak cos kx
                                                  0≤k≤K

be the non-negative trigonometric polynomial constructed in Section 2.4. By Lemma 2.9
and Lemma 2.8 one has
      f (0) X Λ(n)
           ≤       f (log n)P (t log n)
       12      nσ
                n≥1
                      X         X Λ(n)
            = Re           ak               f (log n)
                                      nsk
                   0≤k≤K        n≥1
                                                                                       (2.43)
                                                    1 Γ′ sk + δm
                                                                  
                 X               X         1
            =         ak f (0)       κm − log π + Re              +1
                                           2        2   Γ      2
                0≤k≤K          0≤m≤M
                                              X                
                          + ReF (sk − 1) − Re   F (sk − ρ) + Ek .
                                                             ρ

Now we invoke a series of estimates to bound the terms on the right side. First, Lemma 2.12
is used to bound the main term of order log t:

                                      1 Γ′  sk
                                                   aκ
         X        X          1
              κm       ak − log π + Re          +1   ≤   log t − 1.568,                (2.44)
                             2        2     Γ 2        2
      0≤m≤M        0≤k≤K



                                                        42
                                                                   2.5. PROOF OF LEMMA 2.1

             P
where κ :=       0≤m≤M κm = 433/859. Next, by Lemma 2.13 one has

                                X
                        Re             ak F (sk − 1) ≤ a0 F (σ − 1) + 10−10 η                  (2.45)
                               0≤k≤K

and by Lemma 2.5 one has
                    X          X
             Re           ak        F (sk − ρ) > F (σ − β0 ) + F (σ − 1 + β0 ) − 10−7 .        (2.46)
                  0≤k≤K         ρ

Finally, Lemma 2.8 gives a bound for the error term
                                      X
                                            ak Ek < 187η 2 + 8387η 3 .                         (2.47)
                                    0≤k≤K

Substituting (2.45), (2.46), (2.44) and (2.47) into (2.43), we have

             aκ
        0≤      f (0) log t + a0 F (σ − 1) − a1 (F (σ − β0 ) + F (σ − 1 + β0 ))
              2
                     + 10−7 + (−w(0)/12 − 1.568w(0) + 10−10 )η + 187η 2 + 8387η 3 .

At this point we note that by our assumption (2.42) (recalling that η = 1 − β0 , A > 1/6
and µ0 is as defined in (2.38)), we have

                                      1−σ    A        log t
                               µ :=       >                     > µ0
                                       η    A + ε log(Kt + T0 )

and also µ < 1, so that we may apply Lemma 2.14 to obtain

                                aκ
                                   f (0) log t ≥ C1 (µ) + C2 (η) − 10−7
                                 2

where C1 is defined in Lemma 2.14 and C2 (η) := 13.27η − 161η 2 − 12284η 3 . Note that C1
and C2 are positive and increasing on [µ0 , 1] and [0, η0 ] respectively. Thus we may use (in
light of A > 1/6)

                   A            log t              2.78                                 1
           µ>                                  >1−                   and        η>
                  A + ε log(K + T0 /H) + log t     log t                             6 log t

to obtain (with the aid of any symbolic algebra package)
                                                                
                  aκ                        2.78              1
                     f (0) log t > C1 1 −           + C2             − 10−7
                   2                       log t           6 log t
                                              1.8275     4.4106      56.855
                                 > 1.00583 +          −          −          .
                                               log t    (log t)2 (log t)3



                                                   43
CHAPTER 2. THE CLASSICAL ZERO-FREE REGION


The right side is decreasing for t ∈ [H, H1 ] so we may lower-bound this expression by its
value at t = H1 , which is 1.036428 . . .. Therefore

                                1.03642
                   η log t >            = 0.205677 . . . > A0 + ε ≥ A + ε,
                               aκw(0)/2

as required.


To complete the proof of Lemma 2.1 it remains to note that all zeroes β+iγ with 3 ≤ γ < H
are known to lie on the critical line thanks to the computational verification performed in
[PT21], so Lemma 2.1 follows in this region. For γ ≥ H, we use Lemma 2.15 combined
with any existing zero-free region of the form σ > 1 − c/ log t for some c > 1/6, e.g. those
proved in [Kad05], [MT14] or [MTY24].




                                              44
Chapter 3


Littlewood’s zero-free region


3.1     Introduction

In 1922, Littlewood [Lit22] showed that for sufficiently large t, ζ(σ + it) ̸= 0 in the region

                                                   log log t
                                     σ >1−A                                              (3.1)
                                                     log t

for some absolute constant A > 0, thereby enlarging the “classical” zero-free region by a
factor of log log t. The key input that facilitates this improvement is a non-trivial upper
bound on |ζ(s)| inside the critical strip, which is obtained from certain exponential sum
estimates. A number of tools may then be used to pass from estimates of ζ(s) to a zero-free
region, most notably the “local” method of Landau.

The goal of this chapter is to prove an explicit version of this result. Specifically, we will
prove Theorem 1.2, which follows from combining Lemma 2.1 from the previous chapter
with the following lemma:

Lemma 3.1. If t ≥ exp(56.5) then ζ(σ + it) ̸= 0 for

                                              1 log log t
                                   σ >1−                  .
                                            19.62 log t




                                              45
CHAPTER 3. LITTLEWOOD’S ZERO-FREE REGION


The arguments in this chapter are a refinement of those in an earlier manuscript [Yan24] by
the author, where a weaker zero-free region was obtained with a constant of (21.233)−1 .
The proof requires two main ingredients. The first ingredient is the following explicit
estimate of ζ(s) along certain vertical lines inside the critical strip.

Lemma 3.2 (Yang [Yan24] Theorem 1.1). For an integer k ≥ 4 and σk := 1 − k/(2k − 2),
one has
                                                   k
                        |ζ(σk + it)| ≤ 1.546t1/(2 −2) log t    (t ≥ 3).


The second ingredient is a method of detecting zeroes due to Ford [For02b] (Lemma 3.4
below), which replaces Landau’s method in our treatment. Landau’s approach involves
taking an integral on a small circle centered at s, while Ford takes the integral on two
vertical lines, which makes more efficient use of our knowledge of ζ(s).

The analysis in [Yan24] focuses on deriving the first ingredient; Ford’s method was then
applied with minimal changes. In contrast, in this chapter we shall leave Lemma 3.2
unchanged, and focus on refinements to the zero-detection method. In particular, a number
of ideas we encountered in the previous chapter can be directly imported into Ford’s
argument, such as evaluating the zero-detector at a point s inside the critical strip (instead
of on the line Re s = 1) and using an iterative argument to progressively enlarge the
zero-free region. Furthermore we take advantage of more precise tail estimates of the
form F0 (z) ≪ |z|−3 (compared to F0 (z) ≪ |z|−2 used in [For02b]) which provide sharper
estimates of various error terms. Our argument unifies some of the ideas and approaches
presented in [Kad05] and [For02b].



3.2       The zero detector

We begin by recalling Ford’s zero-detector, which is the main tool we use to relate upper
bounds on ζ(s) to the behaviour of zeroes of ζ(s) close to the line Re s = 1.

Lemma 3.3 (Ford [For02b] Lemma 2.2). Let f be the quotient of two entire functions of
finite order. Suppose f has no pole or zero at z = s and z = 0. For all λ > 0, except for


                                              46
                                                                  3.2. THE ZERO DETECTOR


a set of Lebesgue measure 0 (possibly depending on f and s), we have

              f ′ (s)
                                                            
                           π      X                 π(ρ − s)
         −Re          = Re                 mρ cot
              f (s)        2λ                          2λ
                              |Re(s−ρ)|≤λ
                           Z ∞
                         1       log |f (s − λ + 2λiu/π)| − log |f (s + λ + 2λiu/π)|
                      +                                                              du
                        4λ −∞                          (cosh u)2

where ρ runs through all zeroes and poles of f (with multiplicity) and mρ = 1 if ρ is a zero
of f , or −1 if ρ is a pole of f .


One could in principle use the above lemma directly, with f (s) = ζ(s), along with the
arguments in the rest of this chapter, to construct a zero-free region of the form (3.1).
However, to obtain better constants we instead work with ζ ′ (s)/ζ(s) “mollified” by a
smoothing function. This also ultimately allows us to take s inside the critical strip.
The choice of smoothing function is similar to that in Definition 2.2, except that one no
longer has access to Stechkin’s device because in Lemma 3.3 we are only summing over a
subset of all complex zeroes of ζ(s), so that the sum no longer contains both the zeroes
ρ and 1 − ρ. Nevertheless, one may still readily apply Heath-Brown’s [HB92a] family of
smoothing functions. We start by recalling the following function w(u) from Definition
2.1 of the previous chapter, reproduced here for convenience:

Definition 3.1 (The functions w, W and W0 ). For fixed θ ∈ (0, π/2), let w be as defined
as

          w(u) := sec2 θ(sec2 θ(θ cot θ − u/2) cos(u tan θ) + 2θ cot θ − u

                           + sin(2θ − u tan θ) csc 2θ − 2(1 + sin(θ − u tan θ) csc θ))

                                                                       R ∞ −zu
for 0 ≤ θ ≤ 2θ cot θ, and w(u) := 0 elsewhere. Let W (z) :=             0 e    w(u)du denote its
Laplace transform, and let W0 (z) := W (z) − w(0)/z.

Definition 3.2 (The functions f , F and F0 ). Let w(u) be defined in Definition 3.1 with
respect to some fixed 0 < θ < π/2. For fixed η > 0, we define the functions f : R → R,
F : C → C and F0 : C → C as
                                           Z ∞
                                                                                     f (0)
           f (u) := ηw(ηu),     F (z) :=         e−zu f (u)du,   F0 (z) := F (z) −         .
                                           0                                           z


                                                  47
CHAPTER 3. LITTLEWOOD’S ZERO-FREE REGION


By inspection, one has F (z) = W (z/η) and F0 (z) = W0 (z/η). Furthermore, we recall that
one has the estimate

                       |W0 (z)| ≤ C(ν, r)|z|−3             (Re z ≥ ν, |z| ≥ r),          (3.2)

where C(ν, r) is defined in Lemma 2.3.

Lemma 3.4 (Zero-detector for ζ(s)). Let s = σ + it where 3/4 < σ < 1 and ζ(s) ̸= 0.
For all 1 − σ < λ < 1/2 except a set of Lebesgue measure zero, one has
                                       X Λ(n)
                                  Re              f (log n) = I(s, λ)
                                             ns
                                       n≥1

where
                           Z ∞
                   f (0)     log |ζ(s − λ + 2λiu/π)| − log |ζ(s + λ + 2λiu/π)|
        I(s, λ) :=                                                             du
                    4λ   −∞                        (cosh u)2
                                   X                       X
               + ReF ∗ (s − 1) −         ReF ∗ (s − ρ) −       ReF0 (s − ρ) + E(s),
                                      ρ=β+iγ                     ρ=β+iγ
                                      β≥σ−λ                      β<σ−λ
                                                                            
                                                          π       πz  1
                           F ∗ (z) := F (z) + f (0)          cot       −
                                                          2λ       2λ    z
and
                                                               Γ′  z 
                                   Z 1/2+i∞
                            1
                E(s) := Re                     F0 (s − z)Re             dz + ReF0 (s).
                           2πi        1/2−i∞                   Γ 2

Proof. We apply the version of Weil’s [Wei52] explicit formula found in [Kad05, Theorem
3.1] with q = 1 and                  
                                     (f (0) − f (y))e−ys ,
                                     
                                                                 y ≥ 0,
                            ϕ(y) =
                                     
                                     0,                         otherwise.
For all complex numbers s not coinciding with a zero or pole of ζ,
                                             ζ ′ (s)
              X Λ(n)                                                        
                                                           1      X       1
           Re        f (log n) = f (0) − Re          − Re       +     Re
                  ns                         ζ(s)         s−1      ρ
                                                                         s−ρ
              n≥1
                                                      X                                  (3.3)
                                    + ReF (s − 1) −       ReF (s − ρ) + ReE(s),
                                                                ρ

where ρ runs through non-trivial zeroes of ζ. First, we check that all sums over zeroes in
(3.3) are absolutely convergent. One has
                                       1          |σ − β|
                                 Re       =
                                      s−ρ   (σ − β)2 + (t − γ)2


                                                   48
                                                                      3.3. PROOF OF LEMMA 3.1

                  −2                                                            −1 |.
        P                                                    P
and      ρ (t − γ)        converges, and hence so does           ρ |Re(s − ρ)           Furthermore, F is
holomorphic with |ReF (z)| ≪ |z|−2 uniformly for |Re z| ≤ 1 and |Imz| ≥ 1, and there are
                                                                          P
finitely many zeroes ρ for which |Im(s − ρ)| < 1, so by the same reasoning ρ |ReF (s − ρ)|
converges.

We may thus freely change the order of summation over zeroes. Applying F (z) = f (0)/z +
F0 (z), we get
             X Λ(n)                             ζ ′ (s)                  X
        Re             s
                         f (log n) = −f (0)Re           + ReF0 (s − 1) −   ReF0 (s − ρ) + E(s)
                   n                            ζ(s)                     ρ
             n≥1

for all s not coinciding with a zero or pole of zeta. We apply Lemma 3.3 with λ, s and
f = ζ. Then for all λ ∈ (1 − σ, 1/2), except for a set of Lebesgue measure zero, we have

          ζ ′ (s)
                                                                       
                         π X            π(s − ρ)         π       π(s − 1)
      −Re         = −Re           cot              + Re cot
          ζ(s)          2λ                 2λ           2λ          2λ
                           ρ=β+iγ
                           β≥σ−λ                                                     (3.4)
                       Z ∞
                     1     log |ζ(s − λ + 2λiu/π)| − log |ζ(s + λ + 2λiu/π)|
                  +                                                          du.
                    4λ −∞                       (cosh u)2

Lastly, since F ∗ (z) is holomorphic for |z| ≤ λ, one has by (3.3) and (3.4)
                                     Z ∞
        X Λ(n)               f (0)      log |ζ(s − λ + 2λiu/π)| − log |ζ(s + λ + 2λiu/π)|
  Re             f (log n) =                                                              du
              ns              4λ     −∞                     (cosh u)2
        n≥1
                                      X                     X
                   + ReF ∗ (s − 1) −       ReF ∗ (s − ρ) −       ReF0 (s − ρ) + E(s)
                                      ρ=β+iγ                     ρ=β+iγ
                                      β≥σ−λ                      β<σ−λ

for all s as in the lemma and for all λ ∈ (1 − σ, 1/2) outside a set of Lebesgue measure
zero.



3.3         Proof of Lemma 3.1

Other than the use of a different zero-detector, the proof of Lemma 3.1 is conceptually
similar to the approach taken in the previous chapter. Throughout, let

                                H1 := exp(56.5),          H2 := exp(225.4).                         (3.5)




                                                     49
CHAPTER 3. LITTLEWOOD’S ZERO-FREE REGION


As before, we use an inductive argument. For each

                          0 < A ≤ A1 := (19.62)−1 = 0.05096 . . . ,

let H(A) denote the hypothesis
                                                                 log log t
            H(A) :      ζ(σ + it) ̸= 0        for      σ ≥1−A              ,   t ≥ H1 .
                                                                   log t
In particular, H(A1 ) implies Lemma 3.1. By the Vinogradov–Korobov zero-free region,
H(A) holds for some A > 0. It therefore suffices to show

Proposition 3.1. For any 0 < A ≤ A1 and ε = 10−100 , one has

                                    H(A) =⇒ H(A + ε).


The proof of Proposition 3.1 is the object of the rest of this chapter. Suppose for sake
of contradiction that for some A ∈ (0, A1 ], H(A) holds but H(A + ε) does not. This is
equivalent to the following assumption.

Assumption 3.1. For some A ∈ (0, A1 ], the hypothesis H(A) holds, and furthermore
there exists a zero ρ0 = β0 + it satisfying
                              log log t              log log t
                1 − (A + ε)             ≤ β0 < 1 − A                 and   t ≥ H1 .
                                log t                  log t

We will show that Assumption 3.1 leads to a contradiction. As before, we take

                                          η := 1 − β0                                     (3.6)

in Definition 3.2. Meanwhile, let P (x) be the non-negative trigonometric polynomial
constructed in Section 2.4.1 For any x ∈ R, one has
                                           X
                                P (x) =             ak cos(kx) ≥ 0                        (3.7)
                                          0≤k≤K

   1 One could equally use any admissible non-negative trigonometric polynomial here,

such as the degree 40 polynomial recorded in [MTY24, Table 2]. As t increases, the cost
associated with using a polynomial of higher degree becomes less significant compared
to the savings they generate. Nevertheless, for values of t most relevant to this chapter,
polynomial (3.7) remains the most favourable (by a small margin). By contrast, in the next
chapter we will switch to a polynomial of larger degree for the proof of the Vinogradov–
Korobov zero-free region, where t is very large.


                                                50
                                                                      3.3. PROOF OF LEMMA 3.1


where K = 16. This property guarantees that for any t ∈ R
                   ∞                                       K          ∞
                   X Λ(n)                                  X          X Λ(n)
              0≤             σ
                               f (log n)P (t log n) = Re         ak               f (log n).
                         n                                                  nsk
                   n=1                                     k=0        n=1

We apply Lemma 3.4 to all terms on the right side with k ≥ 1, and Lemma 2.6 from the
previous chapter to the term with k = 0. One obtains, for all λ ∈ (1 − σ, 1/2) except a set
of Lebesgue measure zero,
                                       X
                                 0≤           ak I(sk , λ) + a0 Q(σ),                           (3.8)
                                      1≤k≤K

where

                       1 Γ′  σ    
             
               1                                       X
 Q(σ) = f (0) − log π + Re      + 1 + ReF (σ − 1) − Re   F (σ − ρ) + E(σ) (3.9)
               2       2   Γ 2                         ρ

and E(σ) is defined in Lemma 3.4. As before we now seek to upper-bound the right
side of (3.8). This requires a more careful treatment than that of the previous chapter,
because the ratio of the error and “main” terms decay to zero rather slowly (at a rate
of (log log t)−1 ) as t → ∞. To assist in the tracking of constants, we begin by fixing the
choices of θ, σ and λ. These choices were made with the benefit of hindsight; however,
for clearer presentation we record them before specifying the arguments that justify their
choice. Hereafter we take
                                           θ := 1.1272                                         (3.10)

in the definition of w(u) and f (u), which appears close to optimal. In [HB92a], it was
shown that the asymptotically optimal choice of θ (as t → ∞) is given by the unique
solution to the equation

                                    a1
                         sin2 θ =      (1 − θ cot θ)       (0 < θ < π/2),
                                    a0

which for our trigonometric polynomial is approximately 1.136. The difference between
these choices of θ is due to the presence of a secondary error term that remains significant
for the range of t relevant to our application. This error term arises from the contribution
to I(sk , λ) of those hypothetical zeroes close to ρ0 . For large t, one can use a “local”
zero-density estimate to bound the number of such zeroes; however, such bounds are not



                                                  51
CHAPTER 3. LITTLEWOOD’S ZERO-FREE REGION


sufficiently strong when t is small. Consequently, we divide our argument into two regimes.
For t < H2 we estimate the contribution of such zeroes “trivially” using explicit bounds on
the zero-counting function N (T ). For t ≥ H2 we switch over to an argument that makes
use of our knowledge of the local density of zeroes.

This two-part argument also influences our choice of σ and λ. Specifically, we eventually
choose σ = σ ∗ and λ = λ∗ , given by:

Definition 3.3 (The parameters σ ∗ and λ∗ ). With K, H1 and H2 defined in (3.7) and
(3.5), define
                                            log log(Kt + T0 )
                   σ ∗ := σ ∗ (t) = 1 − A                     ,       T0 := 1010 .
                                              log(Kt + T0 )
Furthermore, let λ∗ = λ∗ (t) be defined as
                                             −1
                      ∗             24.2
                     λ := 8 −                            (H1 ≤ t < exp(216)),
                                  log log t
                                    67.19 −1
                                            
                      ∗
                     λ := 16 −                           (exp(216) ≤ t < H2 ),
                                   log log t
                                 (log log t)2
                     λ∗ := 2.248                         (t ≥ H2 ).
                                     log t


One may verify that the assumptions σ ∈ (3/4, 1) and λ ∈ (1 − σ, 1/2) are satisfied by our
choices for t ≥ H1 . The parameter σ ∗ is chosen so that Assumption 3.1 implies ζ(s) ̸= 0
for all Re s ∈ [σ ∗ , 1] and Im s ∈ [t, Kt]. The shape of λ∗ is different for t < H2 and t ≥ H2 ,
reflecting the two different approaches we take to bound the sum over zeroes appearing in
I(s, λ).

For convenience, we also define the constants

                       σ1 := σ ∗ (H1 ) = 0.99648 . . . ,

                       η1 := (A1 + ε) log log H1 / log H1 = 0.00363 . . . ,
                                                                                          (3.11)
                       λ1 := λ∗ (H1 ) = 0.49966 . . . ,

                       λ2 := λ∗ (H2 ) = 0.29275 . . . .




                                                    52
                                                                   3.3.1   Bounding the main terms


These constants are defined so that

                λ∗ ∈ (0, λ1 ]      (t ≥ H1 ),              λ∗ ∈ (0, λ2 ]    (t ≥ H2 ),

                σ ∗ ∈ [σ1 , 1)      (t ≥ H1 ),             η ∈ (0, η1 ]    (t ≥ H1 ).



3.3.1    Bounding the main terms


In this section we use Lemma 3.2 to bound the “main” integral term appearing in I(s, λ):
                1 ∞ log |ζ(s − λ + 2λiu/π)| − log |ζ(s + λ + 2λiu/π)|
                  Z
                                                                      du.
                2 −∞                    (cosh u)2
Due to the rapid decay of (cosh u)−2 as |u| → ∞, this integral is morally log |ζ(s − λ)| −
log |ζ(s + λ)|. Since Re(s − λ) < 1 < Re(s + λ) by assumption, we can estimate this
integral using two pointwise estimates of ζ(s), one (slightly) inside the critical strip and
one outside the critical strip. For the first estimate we use Theorem 3.2, and for the second
bound we use an estimate due to Ramaré [Ram15, Lemma 5.4] that
                                                 eγ(σ−1)
                                 |ζ(σ + it)| ≤                 (σ > 1)                      (3.12)
                                                  σ−1
where γ = 0.57721 . . . is the Euler–Mascheroni constant.

With our choices of σ and λ, this integral generates a leading term of size O(log t/ log log t).
In comparison, the main term in the previous chapter comes from ReΓ′ /Γ(s/2 + 1) ∼ log t,
and this saving of a factor of log log t explains the corresponding enlargement of the zero-
free region by the same factor.

For optimization purposes it is convenient to have an upper bound on ζ(s) that holds
uniformly inside the critical strip instead of just along a discrete set of vertical lines. To
do this we interpolate the bounds on ζ(s) along two neighbouring vertical lines using the
Phragmén–Lindelöf principle (Lemma 3.5 below).

Lemma 3.5 (Fiori [Fio25] Corollary 3). Let s = σ + it and suppose that for σ1 ≤ σ ≤ σ2 ,
f (s) is holomorphic and satisfies |f (s)| ≪ exp(c exp(c1 |t|)) for some c > 0 and 0 < c1 <
π/(σ2 − σ1 ). Furthermore suppose that for sℓ = σℓ + it (ℓ = 1, 2) we have

                 |f (sℓ )| ≤ Aℓ |Q1 + sℓ |αℓ | log(Q2 + sℓ )|β | log log(Q3 + sℓ )|γ


                                                  53
CHAPTER 3. LITTLEWOOD’S ZERO-FREE REGION


where Aℓ , αℓ (ℓ = 1, 2), Qj (1 ≤ j ≤ 3) and β, γ jointly satisfies α1 ≥ α2 , Q1 + σ1 > 0,
Q2 + σ1 > 1 and Q3 + σ1 > e. Then, for all Re s ∈ [σ1 , σ2 ],

                         w2
           |f (s)| ≤ Aw
                      1 A2 |Q1 + s|
                       1            α1 w1 +α2 w2
                                                 | log(Q2 + s)|β | log log(Q3 + s)|γ

where w1 := (σ2 − σ)/(σ2 − σ1 ) and w2 := (σ − σ1 )/(σ2 − σ1 ).


We use the above lemma to interpolate between two bounds from Lemma 3.2.

Lemma 3.6. For integers n ≥ 3, let σn = 1−n/(2n −2). If σn ≤ σ ≤ σn+1 and t ≥ H1 /2,
then
                                                                 (1 − σ)2n − 1
                       |ζ(σ + it)| ≤ Cn tµ log t,           µ=                 .
                                                                 (n − 1)2n + 2
where C3 = 0.0728 · 2.50214σ/3 and Cn = 1.549 for n ≥ 4.


Proof. Let zn := σn + it, µn = 1/(2n − 2) and suppose for j ∈ {n, n + 1},

                                  |ζ(zj )| ≤ cj tµj log t    (t ≥ 3)                     (3.13)

for some constants cj > 1/2. Note that 1/2 ≤ σn < 1. First, we verify computationally
that on {s ∈ C : 1/2 ≤ Re s ≤ 1, |Ims| ≤ 3}, one has |(s − 1)ζ(s)| < Q and | log(Q + s)| > 1
with Q = 3. Furthermore, since cj > 1/2, this implies

                  |(zj − 1)ζ(zj )| < cj |Q + zj |1+µj | log(Q + zj )|       (|t| ≤ 3).

Meanwhile by (3.13), and since | log(Q+zj )| ≥ log |Q+zj | > log t and |zj −1|, t < |Q+zj |,
one in fact has

                  |(zj − 1)ζ(zj )| ≤ cj |Q + zj |1+µj | log(Q + zj )|       (t ∈ R).

Taking j = n, n + 1 and applying Lemma 3.5 gives
                                     w
                                cw   n+1
                                 n cn+1 |Q + s|
                                  n             wn µn +wn+1 µn+1 | log(Q + s)|
                     |ζ(s)| ≤                                                            (3.14)
                                                  |s − 1|

for all σn ≤ σ ≤ σn+1 , where wn := (σn+1 − σ)/(σn+1 − σn ) and wn+1 = 1 − wn . Next, for
z = x + iy with x > 0, y > 1, we have the crude bounds

                                     |z| < x + y ≤ (1 + ε0 )y,


                                                   54
                                                                3.3.1   Bounding the main terms


            | log z| = | log |z| + i arctan(y/x)| < ((log y + log(1 + ε0 ))2 + π 2 )1/2

                                                 < ((1 + ε1 )2 + ε2 )1/2 log y,

where ε0 = x/y, ε1 = log(1 + ε0 )/ log y and ε2 = (π/ log y)2 . Taking z = Q + s and using
t ≥ H1 /2 gives ε0 , ε1 < 10−10 and ε2 < 2.5 · 10−3 . Applying these inequalities to (3.14),
along with |s − 1| ≥ t and µn ≤ 1/6, gives

                        |ζ(s)| ≤ ctwn µn +wn+1 µn+1 log t       (t ≥ H1 /2)

where
                                                                    w
                         c = (1 + ε0 )1/6 ((1 + ε1 )2 + ε2 )1/2 cwn  n+1
                                                                 n cn+1 .


If n ≥ 4, by Lemma 3.2 we may take cn = cn+1 = 1.546 so that c ≤ 1.549. If n = 3, then
we use the following estimate due to [HPY24]:

                          |ζ(1/2 + it)| ≤ 0.618t1/6 log t        (t ≥ 3)

so we may take c3 = 0.618. Combined with c4 = 1.546 from Lemma 3.2, one has

                                    c ≤ 0.0728 · 2.50214σ/3 .

This completes the proof.


The next lemma makes explicit the notion that the integral term can be bounded effectively
using a pointwise estimate of ζ(s) inside the critical strip. This lemma is largely the same
as [For02b, Lemma 3.4], the main difference being that we only require a bound on ζ(σ+it)
to hold for large t instead of for all t ≥ 3. This technical change allows us to avoid applying
the Phragmén–Lindelöf principle at very small values of t where it is poorly suited.

Lemma 3.7. Let 0 < δ ≤ 1/2, t ≥ 1000 and 1/2 ≤ σ ≤ 1 − t−1 be fixed, and suppose that

                          |ζ(σ + iy)| ≤ Xy Y (log y)Z        (y ≥ t/2)

for some constants X, Y > 0 and Z ≥ 1/2. Then

              1 ∞ log |ζ(σ + it + iuδ)|
                Z
                                        du ≤ log X + Y log t + Z log log t.
              2 −∞       (cosh u)2


                                                55
CHAPTER 3. LITTLEWOOD’S ZERO-FREE REGION


Proof. We follow essentially the same argument as [For02b, Lemma 3.4] by splitting the
integral into three parts. For convenience let us write

                                           q := t/δ

so that q ≥ 2000. Let
                 Z ∞                                  Z −3q/2       Z −q/2       Z ∞
                       log |ζ(σ + it + iuδ)|
                                             du =               +            +
                    −∞       (cosh u)2                 −∞            −3q/2       −q/2

                                                 = I1 + I2 + I3 ,

say. If u ≤ −3q/2 then −(t + uδ) ≥ t/2. By the lemma’s assumption and applying
|ζ(s)| = |ζ(s)| and log(t + x) ≤ log t + x/t,

             log |ζ(σ + i(t + uδ))| ≤ log X + Y log(−t − uδ) + Z log log(−t − uδ)
                                                                          
                                                                 Z      u
                                    ≤ log(XtY (log t)Z ) − Y +            +2 .
                                                               log t    q
Furthermore, using (cosh u)2 > eu /4 (for u > 0) and δ ≤ 1/2,
              Z −3q/2                   Z ∞
                       u/q + 2
           −                   2
                                 du < 4     (u/q + 2)e−2u du = (6 + 1/q)e−2q .
               −∞     (cosh u)           q

Therefore,
                                          Z −3q/2
                                                        du
                              Y       Z
                  I1 ≤ log(Xt (log t) )                        + (6 + 1/q)e−2q .        (3.15)
                                           −∞        (cosh u)2
If −3q/2 ≤ u ≤ −q/2 then |t + uδ| ≤ t/2. Writing s = σ + i(t + uδ), we have |s − 1| ≥
|Re(s − 1)| ≥ t−1 by assumption. Also, |s| < t/2 + 1. Since σ ≥ 1/2 and t ≥ 1000 we have,
via a classical identity (see e.g. Titchmarsh [Tit86, (2.1.4)]),
                              Z ∞
                                   ⌊x⌋ − x + 1/2                 1 |s| ∞ dx
                                                                      Z
                 1      1                                 1
    |ζ(s)| =         + +s                        dx ≤          + +           < 2t,
               s−1 2           1        xs+1           |s − 1| 2    2 1 xσ+1
so that, since t = δq ≤ q/2 and (cosh u)2 > exp(2|u|)/4,
                                 Z −q/2
                                           du
                    I2 ≤ log(2t)                 2
                                                   < 4e−q q log q < e−q/2 .             (3.16)
                                  −3q/2 (cosh u)

Finally, if u ≥ −3q/2, then t + uδ ≥ t/2. By the lemma’s assumption, and applying
log(1 + x) ≤ x and log(1 + x) ≤ x − 12 x2 + 31 x3 gives

    log |ζ(σ + i(t + uδ))| ≤ log X + Y log(t + uδ) + Z log log(t + uδ)
                                                                uδ (uδ)2 (uδ)3
                                                                              
                                   Y       Z             Z
                           ≤ log(Xt (log t) ) + Y +                −       +       .
                                                       log t     t     2t2   3t3



                                                56
                                                                 3.3.1     Bounding the main terms


Applying x + 12 x2 + 13 x3 < 10x3 (for x > 21 ), (cosh u)2 > 14 e2u (for u > 0) and q ≥ 2000
yields
                Z ∞                                     Z ∞ Z −q/2
                       u/q − (u/q)2 /2 + (u/q)3 /3
                                                   du =       −
                  −q/2         (cosh u)2                  −∞     −∞
                                       Z ∞
                               π2           u/q + (u/q)2 /2 + (u/q)3 /3
                          =−        +                                   du
                              12q 2     q/2          (cosh u)2
                               π2      40 ∞ 3 −2u               π2     10−100
                                          Z
                          ≤−        +           u e    du < −        +        ,
                              12q 2    q 3 1000                12q 2     q3
so that
                                            Z ∞
                                Y       Z         du        π2     10−100
                  I3 ≤ log(Xt (log t) )                2
                                                         −       +        .                 (3.17)
                                         −q/2 (cosh u)     12q 2       q3
                                                 R∞
Combining (3.15), (3.16) and (3.17), and since −∞ (cosh u)−2 du = 2, we have
                Z ∞
                      log |ζ(σ + i(t + uδ))|
                                     2
                                             du < 2 log(XtY (log t)Z ) + R,
                  −∞         (cosh u)
where (recalling that q ≥ 2000 and t = qδ ≤ q/2)
                                                           π2     10−100
                                                                      
                     −q/2           Z               −2q
               R=e        + Y +           (6 + 1/q)e    −       +
                                  log t                   12q 2     q3
                               1
                 < e−q/2 − 2         < 0,
                            4q log q
as required.


The next two lemmas are the main results of this section.

Lemma 3.8 (Main term estimate for small t). Let H1 ≤ t ≤ H2 and suppose σ ∗ and λ∗
are as in Definition 3.3. If s∗k = σ ∗ + ikt and 0 < δ ≤ 1/2 is a constant, then
                          Z ∞
             1 X                log |ζ(s∗k − λ∗ + iuδ)| − log |ζ(s∗k + λ∗ + iuδ)|
                       ak                                                         du
            4λ             −∞                       (cosh u)2
                 1≤k≤K
                                                               log t
                                                  ≤ 3.4431             .
                                                             log log t

Proof. Throughout, let T = kt for integers 1 ≤ k ≤ K and suppose σn := 1−n/(2n −2) for
integers n ≥ 3. For convenience, let us denote (T3 , T4 , T5 ) = (H1 , e216 , H2 ) and (E3 , E4 ) =
(24.2, 67.19), so that for n = 3, 4 one has
                                               −1
                        ∗         n     En
                      λ (t) = 2 −                        (Tn ≤ t ≤ Tn+1 ).
                                      log log t


                                                  57
CHAPTER 3. LITTLEWOOD’S ZERO-FREE REGION


The constants En are chosen so that σ ∗ (t) + λ∗ (t) ∈ [σn , σn+1 ] for all t ∈ [Tn , Tn+1 ].
Writing δn := 1/((n − 1)2n + 2) and applying Lemma 3.6 and Lemma 3.7,
                   Z ∞
               1       log |ζ(s∗k − λ∗ + iuδ)|
                                               du
              4λ    −∞        (cosh u)2
                            1
                        ≤ ∗ (δn (2n (1 − σ ∗ + λ∗ ) − 1) log T + log log T + log Cn )
                          2λ
                                   1 − σ∗         δn En log T      log log T + log Cn
                        = 2n−1 δn         log T +               +                     .
                                     λ∗             2 log log t            2λ∗

However, for n = 3, 4 and Tn ≤ t ≤ Tn+1 one may verify that

                2n−1 δn (log log t)2   2n−1 δn log log t (2n log log t − En )
                                     =                                        < 0.133
                  λ∗       log t                        log t

so that, since log T < log(Kt + T0 ) and log log(Kt + T0 ) < 1.012 log log t for t ≥ H1 , and
A ≤ A1 by assumption,

              1 − σ∗                       log log(Kt + T0 ) log T log t            log t
    2n−1 δn       ∗
                     log T < 0.133(A1 + ε)                             2
                                                                         < 0.0069           .
                λ                            log(Kt + T0 ) (log log t)            log log t

Meanwhile, using log T = log t + log k and log log T < log log t + 0.02 (t ≥ H1 ),
        Z ∞
              log |ζ(s∗k − λ∗ + iuδ)|
                                                                         
      1                                      δn En          log k               log t
                                      du ≤            1+            + 0.0069
     4λ −∞           (cosh u)2                 2           log H1             log log t
                                                log log t + 0.02 + log Cn
                                             +                            .             (3.18)
                                                            2λ∗

On the other hand, we use an estimate of Ford [For02b] (adapted to degree K polynomials)
to get
                         K     Z ∞
                      1X           log |ζ(s∗k + λ∗ + iuδ)|
                    −    ak                                du ≤ a0 log ζ(σ ∗ + λ∗ ).       (3.19)
                      2         −∞        (cosh u)2
                         k=1

Now, applying (3.12) and since ζ(σ) is decreasing for σ > 1, and σ ∗ ≥ σ1 , one has

                   ζ(σ ∗ + λ∗ )   ζ(σ1 + λ∗ )   γ(σ1 − 1 + λ∗ ) − log(σ1 − 1 + λ∗ )
                                ≤             ≤                                     .      (3.20)
                       2λ∗           2λ∗                        2λ∗

Combining (3.18), (3.19) and (3.20), and substituting the value of σ1 and the trigonometric
polynomial coefficients ak ,
                     Z ∞
        1 X                log |ζ(s∗k − λ∗ + iuδ)| − log |ζ(s∗k + λ∗ + iuδ)|
                  ak                                                         du
       4λ             −∞                       (cosh u)2
              1≤k≤K                                                                        (3.21)
                                                                    log t
                                         ≤ (1.7761δn En + 0.0244)           + E(t)
                                                                  log log t


                                                  58
                                                                   3.3.1    Bounding the main terms


where
                      3.53(log log t + log Cn ) − log(λ∗ (t) − 0.004) + 0.07 γ
             E(t) =                                                         + .
                                              2λ∗ (t)                        2
If n = 3 then by σ ∗ < 1,

                            14 ∗
               log Cn =       (σ − λ∗ ) log 2.502 + log 0.0728 < 1.66 − 4.27λ∗
                            3

so that
                            3.53 log log t − log(λ∗ (t) − 0.004) + 1.73
                 E(t) =                                                 − 7.24.
                                               2λ∗ (t)
Substituting λ∗ = (8 − E3 / log log t)−1 and computing derivatives, one finds (with the aid
of computer assistance) that for t ≥ H1

                                          log log t
                                   E(t)             ≤ 0.87153 . . .
                                            log t

with the maximum occurring near t = exp(111.12 . . .).

On the other hand if n = 4 then Cn = 1.549, so that

                             3.53 log log t − log(λ∗ (t) − 0.004) + 3.28 γ
                   E(t) =                                               + .
                                                 λ∗ (t)                  2

In this case, one finds (with the aid of computer assistance) that for t ≥ T4 , one has
E(t) log log t/ log t ≤ 1.0319 . . . with the maximum occurring near t = exp(222.75 . . .).
The result follows from substituting these bounds into (3.21).

Lemma 3.9 (Main term estimate for large t). Let t ≥ H2 and suppose σ ∗ and λ∗ are as
in Definition 3.3. If s∗k = σ ∗ + ikt and 0 < δ ≤ 1/2 is a constant, then
                    Z ∞
             1 X        log |ζ(s∗k − λ∗ + iuδ)| − log |ζ(s∗k + λ∗ + iuδ)|
                 ak                                                       du
            4λ       −∞                     (cosh u)2
                1≤k≤K
                                                                log t
                                                   ≤ 3.4431             .
                                                              log log t

Proof. Once again let σn := 1 − n/(2n − 2) (for n ≥ 3) and T = kt (for 1 ≤ k ≤ K).
Furthermore let t3 = H2 and tn (n ≥ 4) be defined implicitly via

                                     σ ∗ (tn ) − λ∗ (tn ) = σn .                             (3.22)




                                                  59
CHAPTER 3. LITTLEWOOD’S ZERO-FREE REGION


Since σ ∗ (H2 ) − λ∗ (H2 ) = 0.94475 . . . ∈ [σ3 , σ4 ] and σ ∗ (t) − λ∗ (t) is continuous and decreas-
ing for all t ≥ H2 , tn is uniquely defined and increasing in n for all n ≥ 3.

Now let n be such that t ∈ [tn , tn+1 ), so that σ ∗ − λ∗ ∈ [σn , σn+1 ). By Lemma 3.6 and
Lemma 3.7, and using σ ∗ < 1,

    1 ∞ log |ζ(s∗k − λ∗ + iuδ)|        λ ∗ 2n − 1
      Z
                                du <               log T + log log T + log 1.549.
    2 −∞          (cosh u)2          (n − 1)2n + 2

By definition,

                 λ∗ 2n − 1
                                                                                  
       log t                        1                                    log t
                       n
                            =                       2n B log log t −                   = gn (t),   (3.23)
     log log t (n − 1)2 + 2   (n − 1)2n + 2                            log log t

say, where B = 2.248. Next we show that for all n ≥ 3,

                               gn (t) ≤ 3.1018          (tn ≤ t ≤ tn+1 ).

First suppose that 3 ≤ n ≤ 7. Rearranging (3.22) gives

                             n                 ∗           (log log tn )2
                                  = 1 − σ n > λ  (tn ) = B                .
                           2n − 2                             log tn

By explicitly solving this inequality for tn (for fixed 3 ≤ n ≤ 7), one finds that log tn ≥ rn ,
where r3 = log H2 , r4 = 234, r5 = 531, r6 = 1155 and r7 = 2469. With the aid of computer
assistance, we verify that for 3 ≤ n ≤ 7,

                                 max       gn (t) ≤ max gn (t) < 3.1018.
                              tn ≤t≤tn+1           t≥rn


Now suppose that n ≥ 8. The function h(x) = 2n B log x − x/ log x is maximized on (1, ∞)
at x = x∗ , where x∗ satisfies

                                  2n B       1        1
                                       +           −       = 0.
                                   x∗    (log x∗ )2 log x∗

For n ≥ 8 one has x∗ > 1000 and so (via a computer verification)
                                                            
                              ∗           ∗              1       2
                   n
              log(2 B) = log x − log log x + log 1 −       ∗
                                                               > log x∗ .
                                                     log x       3

Therefore for all x > 1,

                                                      3
                         h(x) ≤ h(x∗ ) < 2n B log x∗ < 2n B log(2n B)
                                                      2


                                                   60
                                                                   3.3.1   Bounding the main terms


and thus, applying the above estimate with x = log t and by n ≥ 8,

                            2n B log x − x/ log x   3B 2n (n log 2 + log B)
                 gn (t) =                         <                         < 3.1,
                               (n − 1)2n + 2         2    (n − 1)2n + 2

as required.

Now combining this with

               log T       log k                     log log T            log k
                      ≤1+                  and                  ≤1+
                log t     log H1                      log log t     log H1 log log H1

one obtains
                       Z ∞
                1 X        log |ζ(s∗k − λ∗ + iuδ)|
                    ak                             du < 14.48 log log t + 1.6.              (3.24)
                2       −∞        (cosh u)2
                 1≤k≤K

For the integral outside the critical strip, we once again follow the argument of [For02b]
to get
                       K      Z ∞
                   1X             log |ζ(s∗k + λ∗ + iuδ)|
                 −    ak                          2
                                                          du ≤ a0 log ζ(σ ∗ + λ∗ ).
                   2           −∞        (cosh u)
                      k=1

For t ≥ H2 , applying (3.12) one has

                                                              (log log t)2    log log t
         log ζ(σ ∗ + λ∗ ) ≤ γh(t) − log h(t),     h(t) := B                −A           .   (3.25)
                                                                 log t          log t

Using t ≥ H2 and 0 < A ≤ A1 ,

                                           log log H2            (log log H2 )2
                   (B log log H2 − A1 )               ≤ h(t) < B                .
                                              log t                 log H2

Substituting the appropriate bound into (3.25) and evaluating constants, one finds log ζ(σ ∗ +
λ∗ ) < log log t − 4.01. Combining this with (3.24), and recalling t ≥ H2 ,
                       Z ∞
                1 X        log |ζ(s∗k − λ∗ + iuδ)| − log |ζ(s∗k + λ∗ + iuδ)|
                    ak                                                       du
               4λ       −∞                     (cosh u)2
                  1≤k≤K
                      15.48 log log t    log t                 log t
                  <                               2
                                                    < 3.4431           ,
                           2B         (log log t)            log log t

as required.




                                                   61
CHAPTER 3. LITTLEWOOD’S ZERO-FREE REGION


3.3.2     Bounding the sums over zeroes


In this section, we focus on lower-bounding
                              X                               X
                                     ReF ∗ (s∗k − ρ) +                  ReF0 (s∗k − ρ),
                          ρ=β+iγ                             ρ=β+iγ
                         β≥σ ∗ −λ∗                          β<σ ∗ −λ∗

where we recall that F ∗ and F0 were defined in Lemma 3.4 and Definition 3.1 respectively.

Although both sums could in theory contain infinitely many zeroes, only those zeroes
close to sk have a significant contribution, since both F ∗ (z) and F0 (z) decay quickly to
0 as |z| → ∞. If k = 1 then the first sum contains ρ0 , and our goal is to estimate the
contribution of ρ0 while bounding the contribution of all the other zeroes. By the definition
of F ∗ (z), and since all relevant sums are absolutely convergent, we may rearrange the sums
as
                 X                                   X
                             ReF ∗ (s∗k − ρ) +                   ReF0 (s∗k − ρ)
             ρ:|s∗k −ρ|≤λ∗                       ρ:|s∗k −ρ|>λ∗

                                                                                        π(s∗k − ρ)
                                                                                                    
                                                    π             X                                          (3.26)
                                            + f (0) ∗                      Re cot                        .
                                                   2λ                                      2λ∗
                                                               ρ=β+iγ
                                                              β≥σ ∗ −λ∗
                                                             |s∗k −ρ|>λ∗

The first sum, containing those zeroes closest to s∗k , is fortunately easy to bound since
we assumed that s∗k lies sufficiently far inside the zero-free region2 so that, for all such
zeroes, Re(s∗k − ρ) ≥ 0 and hence ReF (s∗k − ρ) ≥ 0. The majority of this section focuses
on bounding the second sum, which produces the main error term. To estimate it we first
establish a “local” zero-density estimate to bound the number of zeroes close to s∗k . We
also require an estimate of the total number of non-trivial zeroes in order to bound the
last two sums of (3.26).

Definition 3.4 (The quantities N (T ) and N (z, δ)). For any T > 0, define N (T ) as the
number of zeroes ρ of ζ, counted with multiplicity, such that 0 < Im ρ ≤ T . In addition,

     2 Paradoxically, one may have expected bounding this sum would pose the greatest

difficulty due to the proximity of zeroes to s∗k ; indeed it would be difficult to control this
sum without assuming that s∗k lies deep enough inside a known zero-free region.




                                                       62
                                                        3.3.2   Bounding the sums over zeroes


for any δ > 0 and z ∈ C, define N (z, δ) as the number of zeroes ρ of ζ, counted with
multiplicity, such that |z − ρ| ≤ δ.


The function N (T ) is commonly studied in the literature; here we record an explicit result
due to Rosser [Ros41].3

Lemma 3.10 (Rosser [Ros41] Theorem 19). For any t ≥ 2 one has

                                          T       T  7
                               N (T ) =      log    + + R(T )
                                          2π     2πe 8

where |R(T )| ≤ R0 (T ) := 0.137 log T + 0.443 log log T + 1.588.


Meanwhile, N (z, δ) may be bounded using Lemma 3.3. We follow the approach taken in
[For02b, Lemma 4.2], except we use Lemma 3.2 instead of a Vinogradov type estimate.

Recall that λ2 = 0.29275 . . . is a constant defined in (3.11).

Lemma 3.11 (Local zero density estimate). Let 0 < δ ≤ λ2 and z = σ + iT with T ≥ H2
and 1 − δ/100 ≤ σ < 1. Then

                                   7.6δ
                    N (z, δ) ≤ −         log T + 1.7(log log T − log δ) + 16.
                                   log δ

Proof. For convenience, let us write

              α := 0.333,       β := 1.529,       u := 1 + αδ + iT,       v := βδ.

The constants α and β are chosen to approximately minimize the bound on N (z, δ) when
T = H2 and δ = λ2 , which are the values most relevant to our application. Note in
particular that Re(z − λ) = 1 + (α − β)δ ≥ 1/2. By Lemma 3.3 with f = ζ, we have

                  ζ ′ (u)
                                                       
                                   X           π(u − ρ)
           −2v Re         = −πRe         cot
                  ζ(u)                            2v
                                    ρ=β+iγ
                                   β≥Re(u−v)
                         Z ∞
                     1       log |ζ(u − v + 2vix/π)| − log |ζ(u + v + 2vix/π)|
                   +                                                           dx.
                     2    −∞                     (cosh x)2

   3 See also the recent preprint [Bel24] containing a refinement of this result.




                                                63
CHAPTER 3. LITTLEWOOD’S ZERO-FREE REGION


Write ϕ := ((β − α)δ log 2)−1 for convenience. If we take
                                                         
                                        log ϕ + log log ϕ
                                  k=                                                  (3.27)
                                              log 2

then (since ϕ > e)
                                                             
                         k   k      1               log log ϕ
                       k
                           > k ≥                 1+             > (β − α)δ
                      2 −2  2    ϕ log 2              log ϕ

i.e. Re(u − v) > σk . By Lemma 3.7
             Z ∞
         1      log |ζ(u − v + 2vix/π)|        1
                                 2
                                        dx ≤ k   log t + log log t + log 1.549.       (3.28)
         2   −∞         (cosh x)            2 −2

However (3.27) implies 2k > (ϕ log ϕ)/2 > 2 so

                                  1          2                 δ
                                      <             < −4.576       .
                               2k − 2   ϕ log ϕ − 4          log δ

Next, for σ > 1 one has

                              1        X µ(n)     X
                                     =     σ+it
                                                ≤   n−σ = ζ(σ)
                         |ζ(σ + it)|     n
                                          n≥1           n≥1

where µ(n) denotes the Möbius function. By (3.12), Re(u + v) = 1 + (α + β)δ, δ ≤ λ2 and
R∞          −2
 −∞ (cosh x) dx = 2, we get
         Z ∞                                            Z ∞
     1      log |ζ(u + v + 2vix/π)|     1                      dx
   −                        2
                                    dx ≤ ζ(u + v)                   2
     2   −∞         (cosh x)            2                −∞ (cosh x)                  (3.29)
                                          ≤ γ(α + β)δ − log(α + β)δ < −0.3 − log δ.

Also, by [For02b, Lemma 3.1], for σ > 1 one has | − ζ ′ /ζ(σ + it)| ≤ (σ − 1)−1 and thus

                                 ζ ′ (u)     ζ ′ (u)      1      1
                            Re           ≤ −         ≤         =    .                 (3.30)
                                 ζ(u)        ζ(u)      Reu − 1   αδ

Next, for real x, y, one has

                                                       sin 2x
                               Re cot(x + iy) =                    .                  (3.31)
                                                  cosh 2y − cos 2x

We use this explicit formula to verify that
                        
                         πξ
               Re cot         ≥ 0.1933      (|ξ − α| ≤ 101/100, Re ξ > α).            (3.32)
                         2β




                                                 64
                                                              3.3.2    Bounding the sums over zeroes


By the maximum modulus principle applied to the function ξ 7→ e− cot(πξ/2β) , it suffices to
check that the above inequality holds on the boundary of the semicircular region {ξ ∈ C :
|ξ − α| ≤ 1.01, Re ξ ≥ α}. Meanwhile, if |z − ρ| ≤ δ then

                         u−ρ       1    u−ρ     1−σ
                             −α −     ≤     −α−     ≤ 1.
                          δ       100    δ       δ

Therefore, applying (3.32) termwise with ξ = (u − ρ)/δ,
                                              
                            X         π(u − ρ)
                       Re       cot              ≥ 0.1933N (z, δ).
                                         2v
                               ρ=β+iγ
                               |z−ρ|≤δ

On the other hand, the contribution of the other zeroes is
                                                      
                                 X            π(u − ρ)
                           Re           cot              ≥0
                                                 2v
                                     ρ=β+iγ
                                    β≥Re(u−v)
                                     |z−ρ|>δ

thanks to (3.31) and 0 < Re(u − ρ)/v ≤ 1. Combining this with (3.28), (3.29) and (3.30),

           2β                       4.576δ
       −      + 0.1933πN (z, δ) ≤ −        log T + log log T + log 1.549 − 0.3 − log δ
           α                         log δ

and the result follows upon rearranging and rounding the constants appropriately.

Lemma 3.12. If T ≥ H1 then
                                         X
                                                |γ − T |−3 < log T,
                                      ρ=β+iγ
                                     |γ−T |>1

where ρ runs through the non-trivial zeroes of the Riemann zeta-function.


Proof. One has
                    X                         X            1           X           1
                             |γ − T |−3 =                        +                       .
                                                       (γ − T )3                (T − γ)3
                   ρ=β+iγ                   ρ:γ>T +1                 ρ:γ<T −1
                  |γ−T |>1

We estimate the sums using bounds on the zero-counting function N (T ). By Lemma 3.10
                           Z ∞
         X         1           dN (T + x)
                         =
               (γ − T )3    1        x3
      ρ:γ>T +1
                                                                 Z ∞
                           (1 − T −2 ) log(T + 1) log 2π    1         dR(T + x)
                         =                       −       +     +
                                     4π             4π     4πT    1       x3



                                                       65
CHAPTER 3. LITTLEWOOD’S ZERO-FREE REGION


and, integrating by parts,

                                    R(T + x) ∞
                 Z ∞                             Z ∞
                      dR(T + x)                        R(T + x)
                             3
                                =       3
                                                +3              dx.
                  1        x           x      1     1     x4

Via a routine computation, for x > 0 one has R0 (T + x) < R0 (T ) + x/T . As T ≥ H1 , one
has [R(T + x)/x3 ]∞
                  1 < R0 (T ) + 10
                                   −11 . In addition,

          Z ∞                   Z ∞                    
               |R(T + x)|                R0 (T )    1         1
                      4
                          dx <               4
                                                 +    3
                                                          dx ≤ R0 (T ) + 10−11 .
           1        x            1         x       Tx         3

Combining these estimates, one has
                            X            1         1     T
                                              3
                                                <    log    + 2R0 (T ) + 10−10 .
                                     (γ − T )     4π     2π
                          ρ:γ>T +1

Via a similar computation we find (since R0 (2) > 0)
                               Z T −2
             X         1               dN (T − x)
                           3
                             =
                    (T − γ)     1           x3
         ρ:2≤γ<T −1

                                           R(T − x) T −2
                                                            Z T −2
                               log T                                 |R(T − x)|
                             ≤        +          3
                                                          +3                    du
                                 4π            x      1         1        x4
                                1       T
                             <     log     + 2R0 (T ) + 10−10 .
                               4π      2π

Also, since the locations of zeroes are symmetric about the real axis,
                          Z ∞
        X        1              dN (x)
                     3
                       =
             (T − γ)       2   (T + x)3
      ρ:γ<−2
                             Z ∞                           Z ∞
                           1      log((T + x)/(2π))             |R(x)|
                       ≤                       3
                                                    dx + 3             4
                                                                         du < 10−10 .
                          2π 2         (T + x)              2  (T +  x)

Lastly, since there are no zeroes of ζ(s) with |γ| ≤ 2, combining the last three inequalities
gives (for T ≥ H1 )
                      X                        1     T
                               |γ − T |−3 ≤      log    + 4R0 (T ) + 10−9 < log T,
                                              2π     2π
                     ρ=β+iγ
                    |γ−T |>1

as required.

Lemma 3.13. If T ≥ H1 , z = σ + iT , σ ∈ [σ1 , 1) then, for any 0 < λ < 1/2,
                 X                                                             N (z, λ)
                          |z − ρ|−3 ≤ 0.355(λ−3 + (2σ1 − 1 − λ)−3 ) log T −             .
                                                                                 λ3
                ρ=β+iγ
               |γ−T |≤1
               |z−ρ|>λ




                                                     66
                                                             3.3.2     Bounding the sums over zeroes


Proof. Partition S := {ρ : |Im ρ − T | ≤ 1, |z − ρ| > λ} into

                             S1 := {ρ ∈ S : |1 − z − ρ| > λ},

                             S2 := S \ S1 = {ρ ∈ S : |1 − z − ρ| ≤ λ}.

First, suppose ρ ∈ S1 . If ρ lies on the critical line, then |z − ρ| ≥ σ1 − 1/2 so

                                       |z − ρ|−3 ≤ (σ1 − 1/2)−3 .

On the other hand, if ρ lies off the critical line, then 1 − ρ is another zero in S1 . Since
Re z ≥ σ1 , the total contribution of these zeroes is bounded by

                |z − ρ|−3 + |z − (1 − ρ)|−3 ≤ λ−3 + (2σ1 − 1 − λ)−3 = 2J(λ),

say. Since x−3 is convex for x > 0, by Jensen’s inequality one has J(λ) ≥ J(σ1 − 1/2) =
(σ1 − 1/2)−3 . Thus, in either case one has
                                       X
                                              |z − ρ|−3 < J(λ)|S1 |.
                                       ρ∈S1

The contribution of each zero in S2 is at most (2σ1 − 1 − λ)−3 , so that since N (1 − z, λ) =
N (z, λ),
                             X
                                    |z − ρ|−3 ≤ (2σ1 − 1 − λ)−3 N (z, λ).
                             ρ∈S2

It follows from |S| = |S1 | + 2N (z, λ) that
                     X                   λ−3 + (2σ1 − 1 − λ)−3       N (z, λ)
                           |z − ρ|−3 ≤                         |S| −          .
                                                   2                   λ3
                     ρ∈S

It remains to bound |S|. By Lemma 3.10,

             |S| < N (T + 1) − N (T − 1)
                    T +1     T +1 T −1     T −1
                ≤        log      −    log      + Q0 (T + 1) + Q0 (T − 1).
                     2π       2πe   2π      2πe

If f (x) := x log(x/e) then f ′ (x) = log x is increasing, so

            T +1     T +1 T −1     T −1  1    T +1  1    T
                 log     −     log      ≤ log      < log    + 10−10 .
             2π       2π   2π       2π   π     2π   π    2π




                                                     67
CHAPTER 3. LITTLEWOOD’S ZERO-FREE REGION


Furthermore, since log x and log log x are both concave, we have log(T − 1) + log(T + 1) ≤
2 log T and log log(T − 1) + log log(T + 1) ≤ 2 log log T by Jensen’s inequality. Therefore,
for T ≥ H1
                                  1     T
                         |S| ≤      log    + 2Q0 (T ) + 10−10 < 0.71 log T.                   (3.33)
                                  π     2π
The result follows.

Lemma 3.14. Let t ≥ H2 and suppose z = σ + iT with T ≥ t and σ1 ≤ σ < 1. Then, for
any 0 < λ ≤ λ2 one has
                                                                            
               X
                              −3             17.09                  log λ
                        |z − ρ|    ≤ − 140 + 2        log T − 1.7         + 48
                                            λ log λ                  λ3
              ρ=β+iγ
             |γ−T |≤1
             |z−ρ|>λ
                                                                                 N (z, λ)
                                        + (λ−3 − 39)(1.7 log log T + 16) −                .
                                                                                   λ3

Note that for the given ranges of λ and T , the right side is always positive.


Proof. We essentially follow the approach taken by Ford [For02b, Lemma 4.3]. Let S :=
{ρ : |Imρ − T | ≤ 1, |z − ρ| > λ} be partitioned into the following sets:

                          S1 := {ρ ∈ S : |z − ρ| > λ2 , |1 − z − ρ| > λ2 },
                                                                                              (3.34)
                          S2 := S \ S1 .

We estimate the sum over S1 in the same way as done in the proof of Lemma 3.13. If
ρ ∈ S1 then 1 − ρ ∈ S1 so that since Re z ≥ σ1 the total contribution of these zeroes is at
most λ−3                 −3
      2 + (2σ1 − 1 − λ2 ) . Thus

                                      X
                                             |z − ρ|−3 ≤ J(λ2 )|S1 |,                         (3.35)
                                      ρ∈S1

where as before J(x) = x−3 + (2σ1 − 1 − x)−3 . Now suppose that ρ ∈ S2 . Then either
|1 − z − ρ| ≤ λ2 or |z − ρ| ≤ λ2 , and (by the symmetry of zeroes about the critical line)
there are N (z, λ2 ) zeroes in each case. The zeroes of the former category contribute
                            X
                                      |z − ρ|−3 ≤ (2σ1 − 1 − λ2 )−3 N (z, λ2 )                (3.36)
                         |1−z−ρ|≤λ2




                                                     68
                                                                   3.3.2    Bounding the sums over zeroes


and, by integrating by parts, the zeroes of the latter category contribute

                                                  N (z, u) λ2
                              Z λ2                             Z λ2
            X
                         −3        dN (z, u)                          N (z, u)
                  |z − ρ| =             3
                                             =        3
                                                              +3               du.
                               λ      u             u       λ     λ     u4
                 |z−ρ|≤λ2

Summing the above equation with (3.35), (3.36) and noting that |S1 | + 2N (z, λ2 ) = |S|
gives
                                                                           Z λ2
                          X
                                          −3                 N (z, λ)             N (z, u)
                                |z − ρ|        ≤ J(λ2 )|S| −          +3                   du.         (3.37)
                                                               λ3           λ       u4
                          ρ∈S

Since λ/(1 − σ) > 100 for t ≥ H2 , we may apply Lemma 3.11 to obtain
            Z λ2                     Z λ2                                             
                   N (z, u)                   7.6 log T    1.7(log log T − log u) + 16
        3                   du ≤ 3          − 3          +                               du
             λ       u4               λ        u log u                  u4
                                                Z λ2                             
                                                        du             log λ                           (3.38)
                                  < −22.8 log T                + 1.7         + 48
                                                 λ    u3 log u          λ3
                                                + (λ−3 − 39)(1.7 log log T + 16).

To estimate the remaining integral, first we show
                                                   Z ∞
                                                      e−t      1.5ex
                                  Ei(x) := −              dt <               (x ≥ 2)
                                                    −x t         x

where the integral is interpreted in terms of its Cauchy principal value. The inequality
holds for x ≥ 10 since Ei(x) < − 21 ex log(1 − 2/x) [AS72, (5.1.20)]4 , and for 2 ≤ x < 10 we
verify that it holds computationally (with room to spare). Therefore
                    Z λ2
                                dx                                                        0.75
                                       = Ei(−2 log λ2 ) − Ei(−2 log λ) > 6.86 +                    .   (3.39)
                      λ     x3 log x                                                    λ2 log λ

Lastly, we established in the proof of Lemma 3.13 that |S| < 0.71 log T so that J(λ2 )|S| <
15.4 log T . Combining this with (3.37), (3.38) and (3.39), one obtains the desired result.



Lemma 3.15. Suppose T ≥ H1 and 1 − σ ≤ η ≤ λ ≤ (log T )−1/2 . Furthermore suppose
that ζ(s) ̸= 0 in the rectangle

                                     {s ∈ C : σ ≤ Re s ≤ 1, |Ims − T | ≤ T0 }.

   4 In [AS72] this inequality is stated in terms of the function E (z) =
                                                                                           R ∞ −1 −t
                                                                   1                        z t e dt, which
is related to the exponential integral via E1 (−x) = −Ei(x) for x > 0.




                                                           69
CHAPTER 3. LITTLEWOOD’S ZERO-FREE REGION


If z = σ + iT then
                                                                    
                                     X         π          π(z − ρ)
                           f (0)             Re cot                      ≥ −10−10 η.
                                               2λ            2λ
                                   ρ=β+iγ
                                    β≥σ−λ
                                   |z−ρ|>λ



Proof. Recall (3.31), which implies Re cot z ≥ 0 for 0 < Re z ≤ π/2, so
                                                       
                              X                π(z − ρ)
                                     Re cot               ≥ 0.                                         (3.40)
                                                  2λ
                                      ρ=β+iγ
                                     σ−λ≤β<σ
                                      |z−ρ|>λ

Now suppose ρ = β + iγ is a zero with β ≥ σ. By assumption, one has |γ − T | > T0 .
Meanwhile by (3.31), cosh 2y > e2y /2 and | sin 2x| ≤ |2x|, one has

                        |Re cot(x + iy)| < 10−11 cy −3            (|x| ≤ c, |y| ≥ 1010 ).

Since |γ − T | ≥ T0 and |σ − β| < η, we may apply the above bound with x + iy =
π(z − ρ)/(2λ) and c = πη/(2λ) so
                                          
                  X       π       π(z − ρ)                                    X
            f (0)      Re cot                < 10−10 η 2 λ                              |γ − T |−3 .
                         2λ          2λ
                   ρ=β+iγ                                                    ρ=β+iγ
                     β≥σ                                                    |γ−T |>T0
                   |z−ρ|>λ

By Lemma 3.12 the sum over zeroes is bounded by log T (with plenty of room to spare),
so                                                         
                         X          π            π(z − ρ)
                f (0)             Re cot                        < 10−10 η 2 λ log T ≤ 10−10 η.
                                    2λ              2λ
                        ρ=β+iγ
                          β≥σ
                        |z−ρ|>λ

The desired result follows from combining this with (3.40).

Lemma 3.16. Let η be as defined in (3.6). Let λ, ν > 0 with λ/η ≥ ν > 0 and suppose
z = σ + iT with T ≥ H1 and 1/2 < σ < 1. Suppose further that ζ(s) ̸= 0 in the rectangle

                                  {s ∈ C : σ ≤ Re s ≤ 1, |Ims − T | ≤ λ}.                              (3.41)

If T = t then
                            X                                             η3
                   Re                F ∗ (z − ρ) ≥ F ∗ (σ − β0 ) − C(0, ν) 3 N (z, λ)
                                                                          λ
                         ρ:|z−ρ|≤λ




                                                       70
                                                                3.3.2   Bounding the sums over zeroes


and if T > t then
                                X                               η3
                         Re               F ∗ (z − ρ) ≥ −C(0, ν) 3 N (z, λ).
                                                                λ
                              ρ:|z−ρ|≤λ



Proof. We will show that for |z| ≤ λ and Re z ≥ 0 one has
                                              πz  2λ             η3
                                        
            ∗                      π
        ReF (z) = ReF (z) + f (0) Re cot           −      ≥ −C(0, ν) 3 .                       (3.42)
                                   2λ          2λ     πz            λ

Applying the maximum modulus principle to exp(−F ∗ (z)), it suffices to verify the in-
equality on the boundary of the region {z ∈ C : |z| ≤ λ, Re z ≥ 0}. On Re z = 0, one has
Re(cot z − 1/z) = 0 and, by construction, ReF (z) ≥ 0 so

                                   ReF ∗ (z) ≥ 0            (Re z = 0).

Meanwhile, on |z| = λ and Re z ≥ 0 one has

                                    f (0)             f (0)
                    ReF (z) = Re          + ReF0 (z) = 2 Re z + ReF0 (z),                      (3.43)
                                      z                λ

and, by a result due to Ford [For02b, Lemma 4.4] that Re(cot z − 1/z + 4z/π 2 ) ≥ 0, one
has                              
                              π        πz  2λ     f (0)
                         f (0) Re cot       −     ≥ − 2 Re z.                                  (3.44)
                              2λ        2λ    πz      λ
Since |z/η| = λ/η ≥ ν and Re(z/η) ≥ 0, one has |F0 (z)| = |W0 (z/η)| ≤ C(0, ν)η 3 /λ3 so
that, summing (3.43) and (3.44),

                              ReF ∗ (z) ≥ ReF0 (z) ≥ −C(0, ν)η 3 /λ3 .

Therefore, in either case, (3.42) holds.

By assumption (3.41), all zeroes in the sum
                                                X
                                          Re             F ∗ (z − ρ)
                                               |z−ρ|≤λ

satisfy Re(z − ρ) ≥ 0. There are N (z, λ) zeroes in the sum, so if T ̸= t then we apply
(3.42) termwise to get
                                X                             η3
                         Re             F ∗ (z − ρ) ≥ −C(0, ν) 3 N (z, λ).
                                                              λ
                              |z−ρ|≤λ




                                                     71
CHAPTER 3. LITTLEWOOD’S ZERO-FREE REGION


If T = t we apply this bound to all zeroes except for ρ = ρ0 , to get
                        X                                             η3
                  Re             F ∗ (z − ρ) > F ∗ (σ − β0 ) − C(0, ν) 3 N (z, λ),
                                                                      λ
                       |z−ρ|≤λ

as required.


To summarize, the main conclusion of this section is:

Lemma 3.17. Suppose Assumption 3.1 holds. If t ≥ H1 and s∗k = σ ∗ + ikt then
             X       X                         X                  
                                    ∗                    ∗ ∗
                 ak          ReF0 (sk − ρ) +         ReF (sk − ρ)
                1≤k≤K          ρ=β+iγ                       ρ=β+iγ
                              β≤σ ∗ −λ∗                    β>σ ∗ −λ∗
                                                                                log t
                                             ≥ a1 F ∗ (σ ∗ − β0 ) − 0.75η                 .
                                                                             (log log t)2

Proof. For simplicity, write z = σ ∗ + iT with t ≤ T ≤ Kt. Rearranging sums, we seek a
lower bound on
                                                                                                         
     X                            X                                X            π              π(z − ρ)
          ReF ∗ (z − ρ) +                 ReF0 (z − ρ) + f (0)               Re ∗ cot                         .
       ρ                                                                       2λ                2λ∗
                                ρ=β+iγ                            ρ=β+iγ
   |z−ρ|≤λ∗                    |z−ρ|>λ∗                          β≥σ ∗ −λ∗
                                                                 |z−ρ|>λ∗

                                                                                                          (3.45)

Consider the terms in the second sum. By Assumption 3.1 one has σ ∗ > 1 − η so that
Re(z − ρ)/η > −1. Furthermore, since |z − ρ| > λ∗ and λ∗ /η > λ∗ /((A + ε) log log t/ log t)
is increasing in t, one has

                               |(z − ρ)/η| ≥ λ1 /η1 = ν = 76.36 . . . ,

say, so that
                                                                  C(−1, ν)η 3
                         |F0 (z − ρ)| = |W0 ((z − ρ)/η)| ≤                    .
                                                                   |z − ρ|3
Therefore
                       X                                           X
                               ReF0 (z − ρ) ≤ C(−1, ν)η 3                    |z − ρ|−3 .                  (3.46)
                     ρ=β+iγ                                       ρ=β+iγ
                    |z−ρ|>λ∗                                     |z−ρ|>λ∗

Note that by T ≤ Kt, the definition of σ ∗ and Assumption 3.1, there are no zeroes in the
region |Ims| ≤ T + T0 and Re s ≥ σ ∗ so that the zero-free region assumptions of Lemma
3.15 and Lemma 3.16 are satisfied with σ = σ ∗ .


                                                   72
                                                              3.3.2     Bounding the sums over zeroes


First, suppose that H1 ≤ t < H2 . Applying Lemma 3.12, Lemma 3.14, Lemma 3.15 and
Lemma 3.16, the right side of (3.45) is lower-bounded by

                                            N (z, λ∗ )
            E1 + (C(−1, ν) − C(0, ν)) η 3              − 10−10 η − C(−1, ν)η 3 C1 (T )
                                             (λ∗ )3

where E1 = F ∗ (σ − β0 ) if T = t and zero otherwise, and

                    C1 (T ) := (0.355(λ∗ −3 + (2σ1 − 1 − λ∗ )−3 ) + 1) log T.

From Lemma 2.3 one has C(−1, ν) > C(0, ν), so the term involving N (z, λ∗ ) may be
dropped. Meanwhile, as t < H2 one has by definition λ∗ ≥ 0.27789 . . ., so C1 (T ) ≤
18.6 log T . Furthermore, since L1 (t) > log T ≥ log t and L2 (t) < 1.012 log log t for t ≥ H1 ,

                                                              L2 (t)2              (log log t)2
        C(−1, ν)η 2 C1 (T ) ≤ 18.6C(−1, ν)(A1 + ε)2                   log T < 2.56              .
                                                              L1 (t)2                 log t

Hence, taking T = kt and summing over k, and since t ∈ [H1 , H2 ],
             X       X                       X                    
                                   ∗                     ∗ ∗
                 ak         ReF0 (sk − ρ) +         ReF (sk − ρ)
            1≤k≤K        ρ=β+iγ                       ρ=β+iγ
                        β≤σ ∗ −λ∗                    β>σ ∗ −λ∗
                                                                       (log log t)2
                                               X                                             
                                ∗
                       > a1 F (σ − β0 ) − η              ak       2.56              + 10−10
                                                                          log t
                                              1≤k≤K
                                                       log t
                       > a1 F (σ ∗ − β0 ) − 0.75η                ,
                                                    (log log t)2

so that the desired result holds for H1 ≤ t ≤ H2 .

Now consider the case t ≥ H2 , where from definitions one has 0 < λ∗ ≤ λ2 . This time we
apply the same argument, except with Lemma 3.14 in place of Lemma 3.13. The right
side of (3.45) is then lower-bounded by

                             E1 − 10−10 η − C(−1, ν)η 3 C2 (λ∗ , T )                                (3.47)

where
                                                                                       
                                      17.09                                  log λ
                 C2 (λ, T ) = − 140 + 2                  log T − 1.7               + 48
                                     λ log λ                                  λ3
                                     + (λ−3 − 39)(1.7 log log T + 16).




                                                73
CHAPTER 3. LITTLEWOOD’S ZERO-FREE REGION


Since λ∗ = λ∗ (t) ≥ λ∗ (T ) and C2 (λ, T ) is decreasing in λ, the right side is at most
C2 (λ∗ (T ), T ), which is a pure function of T . We verify, with the aid of Mathematica, that

                                       (log log T )4
                    C2 (λ∗ (T ), T )                 ≤ 1.2532 . . .         (T ≥ H2 )
                                         (log T )3

with the maximum occurring near log T = 912.82 . . .. By Assumption 3.1 and log T <
1.013 log t (for t ≥ H2 and T ≤ Kt),

                                                                      (log log t)2 (log T )3
           C(−1, ν)η 2 C2 (λ∗ , T ) < 1.26C(−1, ν)(A1 + ε)2
                                                                        (log t)2 (log log T )4
                                                log t
                                   < 0.2                  .
                                             (log log t)2

Taking T = kt and summing over k, one obtains
                X       X                                      X                           
                    ak         ReF0 (s∗k − ρ) +                           ReF   ∗
                                                                                    (s∗k − ρ)
                1≤k≤K         ρ=β+iγ                           ρ=β+iγ
                             β≤σ ∗ −λ∗                        β>σ ∗ −λ∗
                                                                              log t
                                          > a1 F ∗ (σ ∗ − β0 ) − 0.71η                  ,
                                                                           (log log t)2

so the result follows in this case too.



3.3.3    Other estimates


In this section we record some lemmas used to estimate the remaining terms of (3.8).

Lemma 3.18. For any η ∈ (0, η1 ] and 0 < λ < 1/2, let F ∗ (z) be defined with respect to η
and λ as in Lemma 3.4. If sk = σ + ikt with η/2 < 1 − σ < min(λ, η) and t ≥ H1 , then
                                          X
                                  Re             ak F ∗ (sk − 1) ≤ 0.
                                         1≤k≤K


Proof. If z = sk − 1 with k ≥ 1 then by assumption |z| ≥ H1 and −η ≤ Re z ≤ −η/2.
Then by (3.2)

            ReF (z) = ReW (z/η) = Reηw(0)/z + ReW0 (z/η)

                                        ≤ (C(−1, H1 /η1 )η/|z| − w(0)/2) η 2 |z|−2 < 0.




                                                    74
                                                                            3.3.3   Other estimates


In addition, if z = x + iy with −π/2 ≤ x < 0 and y ̸= 0, then
                                      
                                     1          sin 2x
                          Re cot z −     =                  ≤ 0.
                                     z     cosh 2y − cos 2x

It follows that since −λ ≤ Re z < 0 by assumption,

                                       πf (0)
                                                     πz  2λ 
                       ∗
                   ReF (z) = ReF (z) +        Re cot       −     ≤ 0.
                                         2λ            2λ    πz

The result follows from ak > 0.

Lemma 3.19. Suppose Assumption 3.1 holds. If sk = σ + ikt with σ ≥ σ1 and t ≥ H1 ,
then
                                X                          log log t
                                      ak |E(sk )| ≤ 2.9η             .
                                                             log t
                              1≤k≤K


Proof. We apply Lemma 2.7 from the previous chapter, noting that t ≥ H1 > H, σ ≥
σ1 ≥ σ0 and (by assumption) η ≤ η1 < η0 so that all conditions are satisfied. One has
                     X                                        X
                           ak |E(sk )| ≤ (14η 2 + 425η 3 )           ak < 55η 2 .
                   1≤k≤K                                     1≤k≤K

However Assumption 3.1 implies η ≤ (A1 + ε) log log t/ log t so the result follows.

Lemma 3.20. Suppose t ≥ H1 , σ ∈ [σ1 , 1) and 0 < η ≤ (A1 + ε) log log t/ log t. If Q(σ)
is defined in (3.9), then Q(σ) < F (σ − 1) − 2.9η.


Proof. Recall that

                         1 Γ′  σ
                                    
                   1                                     X
   Q(σ) = f (0) − log π + Re      +1    + F (σ − 1) − Re   F (σ − ρ) + E(σ),
                   2     2   Γ 2                         ρ

where E(σ) is defined in Lemma 3.4. Since F0 (z) = W0 (z/η), by Lemma 2.7 (in the
previous chapter) one has E(σ) ≤ 723η 3 < 0.01η. In addition, since Γ′ /Γ(x) is increasing
for real x > 1, and σ < 1, one has

                           Γ′  σ     Γ′
                                  + 1 < (3/2) = 2 − γ − log 4.
                           Γ 2         Γ

Lastly, all zeroes ρ = β + iγ with |γ| ≤ H = 3 · 1012 are known to lie on the critical
line [PT21], so Re(σ − ρ) > 0 and ReF (σ − ρ) ≥ 0. On the other hand, if |γ| > H then


                                               75
CHAPTER 3. LITTLEWOOD’S ZERO-FREE REGION


|σ − ρ|/η ≥ H/η1 and Re(σ − ρ)/η ≥ −1, so by (3.2) the total contribution of such zeroes
is
             X                                                 X                            X
        −            ReF (σ − ρ) ≤ C(−1, H/η1 )η 3                     |σ − ρ|−3 < 46η 3            |γ|−3 .
            ρ=β+iγ                                            ρ=β+iγ                       ρ=β+iγ
             |γ|>H                                             |γ|>H                        |γ|>H

By Lehman [Leh66, Lemma 2] with n = 3, we have
                                             X                  2 log H
                                                     |γ|−3 <
                                                                   H2
                                            ρ=β+iγ
                                             |γ|>H

and, together with η ≤ (A1 + ε) log log H1 / log H1 , gives (with room to spare)
                                              X
                                        −Re        F (σ − ρ) < 10−10 η.
                                               ρ

Recalling that f (0) = ηw(0), the result follows from combining these estimates.



3.3.4    Putting it all together


We are now ready to prove Proposition 3.1. Recall that for any λ ∈ (1 − σ, 1/2) except a
set of measure zero, one has (3.8), i.e.
                                             X
                                      0≤             ak I(sk , λ) + a0 Q(σ).                                  (3.48)
                                            1≤k≤K

First suppose that λ∗ is not in the exceptional set.                   Then, we take λ = λ∗ , σ = σ ∗ and
sk = s∗k . Under Assumption 3.1 and applying Lemma 3.8, Lemma 3.9, Lemma 3.17,
Lemma 3.18 and Lemma 3.19, one has
        X
              ak I(s∗k , λ∗ )
     1≤k≤K
                        Z ∞
         f (0) X            log |ζ(s∗k − λ∗ + 2λ∗ iu/π)| − log |ζ(sk + λ∗ + 2λ∗ iu/π)|
     ≤              a k                                                                du
         4λ∗             −∞                          (cosh u)2
              1≤k≤K
               X       X                           X                   
                                      ∗ ∗                          ∗
           −       ak            ReF (sk − ρ) +            ReF0 (sk − ρ)
              1≤k≤K              ρ=β+iγ                         ρ=β+iγ
                                β>σ ∗ −λ∗                      β≤σ ∗ −λ∗
                     X                             X
            + Re            ak F ∗ (s∗k − 1) +           ak |E(s∗k )|
                   1≤k≤K                         1≤k≤K
                         log t                                   log t            log log t
     ≤ 3.4431f (0)               − a1 F ∗ (σ ∗ − β0 ) + 0.75η              + 2.9η           .
                       log log t                              (log log t)2          log t



                                                         76
                                                                        3.3.4   Putting it all together


If instead (3.48) does not hold with respect to λ∗ , then we simply apply the same argument
with a sequence of admissible λj that tends to λ∗ from below.

Meanwhile, by Lemma 3.20,

                                     Q(σ ∗ ) ≤ F (σ ∗ − 1) − 2.9η.

Summing these estimates, one obtains

                                log t
            0 ≤ 3.4431f (0)             + a0 F (σ ∗ − 1) − a1 F ∗ (σ ∗ − β0 ) + C3 (η, t)       (3.49)
                              log log t

where                                                                        
                                             log t           log log t
                  C3 (η, t) :=       0.75              + 2.9           − 2.9a0 η.
                                          (log log t)2         log t
Since the factor in the parentheses is positive for all t ≥ H1 , we may apply η ≤ (A1 +
ε) log log t/ log t (from Assumption 3.1) to obtain, after a routine computation, that

                                        C3 (η, t) ≤ 0.00452 . . .                               (3.50)

where the maximum occurs at t = exp(1144.7 . . .).

Next we estimate the quantity F ∗ (σ − β0 ). Under Assumption 3.1, for all t ≥ H1 we have

              π(σ ∗ − β0 )
                                                       
                             π A1      L2 (t) log log t
                           <                  −           < ε0 := 7 · 10−5 .
                  2λ∗        2 λ∗ (t) L1 (t)      log t

However for 0 < x ≤ ε0 one has 1/x − cot x ≤ ε1 := 3 · 10−5 . Therefore

                                                   ε1 w(0)η
               F ∗ (σ ∗ − β0 ) ≥ F (σ ∗ − β0 ) −            > F (σ ∗ − β0 ) − 10−6 .            (3.51)
                                                      λ∗

It remains to estimate the quantity
                                                    Z 2θ cot θ
             a1 F (σ ∗ − β0 ) − a0 F (σ ∗ − 1) =                 eµu (a1 e−u − a0 )w(u)du
                                                     0

where µ := (1 − σ ∗ )/η. Repeating the steps of Lemma 2.14 of the previous chapter (with
our new value of θ), the right side is bounded from below by

                                                           c∗
                                                             
                                             c2 2     c3
                        C4 (µ) := c0 + c1 µ + µ +        −      µ3
                                             2        6    18




                                                   77
CHAPTER 3. LITTLEWOOD’S ZERO-FREE REGION


where
                Z 2θ cot θ                                           Z 2θ cot θ
        cn :=                 n
                             u (a1 e   −u
                                            − a0 )w(u)du,     ∗
                                                              c :=                u3 |a1 e−u − a0 |w(u)du
                 0                                                    0

are computable constants taking the values

          c0 = 0.8219901343 . . . ,           c1 = 0.1118643577 . . . ,     c2 = 0.0182350945 . . . ,

          c3 = −0.0003829770 . . . ,            c∗ = 0.0188374573 . . . .

Since C4 (µ) is increasing on [µ0 , 1] where µ0 = (1 − σ1 )/η1 = 0.96454 . . .,

                             a1 F (σ ∗ − β0 ) − a0 F (σ ∗ − 1) ≥ C4 (µ0 ) > 0.9376.                         (3.52)

Therefore, combining (3.49), (3.50), (3.51) and (3.52), one finally has

                      log t      η log t    0.9376 − 10−6 a1 − C3 (η, t)
        (1 − β0 )             =           >                              ≥ 0.05098 > A + ε,
                    log log t   log log t          3.4431w(0)

for all t ≥ H1 , so Proposition 3.1 (and therefore Lemma 3.1) is proved.




                                                         78
Chapter 4


The Vinogradov–Korobov
zero-free region


4.1    Introduction

As t → ∞, the largest known zero-free region for ζ(s) takes the shape

                                                    A
                              σ >1−                               ,
                                      (log t)2/3 (log log t)1/3

for some constant A > 0, proved independently by Korobov [Kor58] and Vinogradov
[Vin58] in 1958. As in the previous chapter, this zero-free region can be derived by
combining Landau’s method with certain non-trivial bounds of ζ(s) close to the line σ = 1.
In modern treatments, such bounds take the form

                                        3/2
                    ζ(σ + it) ≪ tB(1−σ)       (log t)2/3   (1/2 ≤ σ ≤ 1)              (4.1)

as t → ∞, where B is an absolute positive constant. These bounds are in turn derived
from the deep analysis of certain exponential sums using Vinogradov’s mean value theorem,
which are beyond the scope of this chapter. Estimate (4.1) was proved by Richert [Ric67] in
1967 with B = 100, and currently the sharpest known estimates of ζ(s) close to σ = 1 still
takes this form (although substantial progress have been made in reducing the constant


                                               79
CHAPTER 4. THE VINOGRADOV–KOROBOV ZERO-FREE REGION


B). Table 4.1 records the historical progression of the constant B and the corresponding
improvements in the size of the zero-free region.

                        Reference                            C            B              A
                  Richert (1967) [Ric67]                     –          100 *            –
              Heath-Brown (1992) [HB92b]                     –             –      ≈ 0.0269B −2/3 *
                  Popov (1994) [Pop94]                       –             –       0.00006888 *
                  Kulas (1999) [Kul99]                       –         18.497 *          –
              Cheng (2000) [Che99; Che00]                   175           46          (952)−1
                  Ford (2002) [For02b]                      76.2         4.45        (57.54)−1
       Mossinghoff–Trudigan–Yang (2022) [MTY24]              –             –        (55.241)−1
                  Bellotti (2024) [Bel24]                   70.7        4.438       (53.989)−1
                                                           3/2
Table 4.1: Bounds of the form |ζ(σ+it)| ≤ CtB(1−σ) (log t)2/3 along with their associated
zero-free regions of the form σ > 1−A(log t)−2/3 (log log t)−1/3 and t ≥ 3 (entries containing
an asterisk (*) indicate that the respective constant only holds for t sufficiently large).

Various methods have been developed to pass from such upper bounds on zeta to a zero-
free region, the most successful currently being the zero-detector due to Ford [For02b] that
we saw in the previous chapter. Following Ford, all explicit treatments of the Vinogradov–
Korobov zero-free region employ this zero-detector, and furthermore each evaluates the
zero-detector on the line Re s = 1. As such, an idea explored in the last two chapters,
namely, iteratively moving s deeper inside the critical strip, provides a way of refining
existing approaches. This is the first idea we shall explore in this chapter.

The second idea we explore is based on the observation that one does not require the full
strength of Richert’s bound to prove the Vinogradov–Korobov zero-free region. Instead,
we only require (4.1) for σ satisfying
                                                          2/3
                                               log log t
                                  1−σ ≫                            .
                                                 log t

It turns out that for such σ, one can slightly sharpen (4.1) by a factor of (log log t)−1/6 ,
i.e.
                                           3/2
                       ζ(σ + it) ≪ tB(1−σ)       (log t)2/3 (log log t)−1/6 ,                    (4.2)

where importantly B is the same constant appearing in (4.1). Unfortunately, this estimate
generates no improvement in the asymptotic size of the zero-free region as t → ∞, which
depends primarily on the size of B. Nevertheless, we can use this estimate to obtain a


                                               80
                                                      4.2. AN EXPLICIT ESTIMATE OF ZETA


larger zero-free region in certain finite ranges of t, in particular when t is “small”. Our
main result of this chapter is

Lemma 4.1. If t ≥ exp(544 000) then ζ(σ + it) ̸= 0 for
                                           1              1
                               σ ≥1−                                  .
                                         51.34 (log t) (log log t)1/3
                                                      2/3



Theorem 1.3 then follows by combining the above lemma with Lemma 2.1 and Lemma 3.1
from the previous chapters (applied in appropriate ranges).



4.2     An explicit estimate of ζ(s)

In this section we derive two explicit estimates of ζ(s) within certain regions inside the crit-
ical strip that are particularly relevant to our application. The first estimate (Lemma 4.2
below) is of the form (4.2) and will be used for large t, and the second estimate (Lemma 4.3)
is used for small t.

Throughout we make use of certain explicit exponential sum estimates obtained via the
analysis of Vinogradov’s integral, due to Ford [For02a] and Bellotti [Bel24]. We emphasize
that the derivations of such bounds is where the main difficulty lies, which are outside
the scope of this chapter. Our goal is this section is merely to apply these bounds more
efficiently. Let us define
                                                                 X
                       S(N, t) := max            max                    (n + u)−it
                                      u∈(0,1] N ′ ∈(N,2N ]
                                                             N <n≤N ′

and λ := log t/ log N . Ford’s article [For02a] implicitly contains an estimate of the form
                                                                 2
                             S(N, t) ≤ c1 ec2 λ N 1−1/(D1 λ )           (N ≥ 1)           (4.3)

for λ ≥ 1, where c1 = 2.459, c2 = 1/64.5625 and D1 = 133.66. This follows from
Lemma 6.2 of [For02a] in the range 1 ≤ λ ≤ 2.6, from Lemma 6.8 and Table 6.1 for
2.6 ≤ λ ≤ 87, and from Theorem 2 for λ > 87. Meanwhile, Bellotti [Bel24, Theorem 1.5]
has shown that for all λ ≥ 1,
                                                             2
                               S(N, t) ≤ c3 N 1−1/(D2 λ )            (N ≥ 1)              (4.4)


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CHAPTER 4. THE VINOGRADOV–KOROBOV ZERO-FREE REGION


with c3 = 8.7979 and D2 = 132.944.

We apply these bounds to derive two estimates of ζ(σ + it) inside the critical strip
(Lemma 4.2 and 4.3 below) which will eventually be used for large and small t respectively.
Throughout, let
                       H3 := exp(544 000),             H4 := exp(4.5 · 108 ).

Lemma 4.2 (Estimate for large t). Let t ≥ 10100 and 1/2 ≤ σ ≤ 1 − 12 (log log t/ log t)2/3 .
Then
                                                 3/2
                      |ζ(σ + it)| ≤ C1 tB(1−σ)         (log t)2/3 (log log t)−1/6

with C1 = 85, B = 4.438.


Proof. We begin the same way as Ford [For02a]. By the approximate functional equation1
(see e.g. Ford [For02a]), for σ ≥ 1/2 and t ≥ 10100 one has
                                          X
                           ζ(σ + it) =           n−σ−it + O∗ (10−80 ).
                                         1≤n≤t

Let J = ⌈log t/ log 2⌉ − 1. Dividing the sum into dyadic intervals, applying partial sum-
mation and taking u → 0+ gives

          X                     J
                                X          X                                J
                                                                            X
                 n−σ−it ≤ 1 +                             n−σ−it ≤ 1 +        (2j )−σ S(2j , t).
         1≤n≤t                  j=0 2j <n≤min(2j+1 ,t)                      j=0


Applying the estimate (4.4) to each term in the sum gives

                           X                           J
                                                       X
                                 n−σ−it ≤ 1 + c3             exp (h(j log 2)) ,                    (4.5)
                         1≤n≤t                         j=0

where
                                                              x3
                                h(x) := (1 − σ)x −                    .                            (4.6)
                                                          D2 (log t)2

   1 One could also use the Riemann–Siegel formula instead of Euler–Maclaurin summation

here, to obtain an approximate functional equation for ζ(s) involving shorter Dirichlet
polynomials of length O(t1/2 ). However, in our application the dominant terms in Dirichlet
polynomial occur when n is rather small (specifically n ≍ (log t)1/3+o(1) ) and the savings
from shortening the polynomial are negligible.




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                                                                            4.2. AN EXPLICIT ESTIMATE OF ZETA


For x ≥ 0, the function h is unimodal, concave down and is maximized at x = x0 :=
p
  D2 (1 − σ)/3 log t, so
                                X                                                            Z ∞
                                         exp(h(j log 2)) ≤ e               h(x0 log 2)
                                                                                         +                eh(u log 2) du.                    (4.7)
                              0≤j≤J                                                              0

                                      √           3/2
The first term on the right side is t2 D2 /27(1−σ) . To estimate the integral we use an
explicit one-dimensional saddle-point approximation. Applying the substitution
                      r
                         1 − σ 1/6                       log 2
                  y=          D2 (log t)1/3 ,   α=     1/3
                                                                   u,
                           3                         yD (log t)2/3                                   2

one obtains
                              Z ∞                                    1/3                 Z ∞
                                         h(u log 2)        yD2 (log t)2/3                                       3   3
                                    e                 du =                                           e(3α−α )y dα.                           (4.8)
                                0                              log 2                         0

Since 3α − α3 ≤ 2 − 2(α − 1)2 for all α ≥ 0, one has
                 Z ∞                              Z ∞                                    Z ∞                                   r
         −2y 3             (3α−α3 )y 3                          −2y 3 (α−1)2                             −2y 3 (α−1)2               π
     e                 e                 dα ≤               e                   dα <                 e                  dα =            .    (4.9)
                 0                                    0                                  −∞                                        2y 3

Since σ ≤ 1 − 12 (log log t/ log t)2/3 by assumption, one has
                                                                     1/6
                                                              D
                                                          y ≥ √2 (log log t)1/3 .                                                           (4.10)
                                                                6

Writing B = 2(D2 /27)1/2 and combining (4.5), (4.7), (4.8), (4.9) and (4.10), for 1/2 ≤
σ ≤ 1 − 12 (log log t/ log t)2/3 one has
                                                                                                                         3/2
                                                                                         π (6D2 )1/4 tB(1−σ) (log t)2/3
                                                                                    r
                                            −80                 B(1−σ)3/2
           |ζ(σ + it)| ≤ 1 + 10                   + c3 t                     + c3
                                                                                         2 log 2        (log log t)1/6
                                                          3/2
                              ≤ 85 t4.438(1−σ)                  (log t)2/3 (log log t)−1/6                     (t ≥ 10100 ),

as required.

Lemma 4.3 (Estimate for small t). Suppose H3 /2 ≤ t ≤ H4 and 1/2 ≤ σ ≤ 1 −
1                    2/3 . Then
2 (log log t/ log t)

                                                                             3/2
                               |ζ(σ + it)| ≤ C2 tB2 (ϕ)(1−σ)                       (log t)2/3 (log log t)−1/6

where C2 = 25, B2 (ϕ) = 4.45(1 + 3ϕ/2 + ϕ2 /2) and ϕ = (2876(1 − σ)2 log t)−1 .




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CHAPTER 4. THE VINOGRADOV–KOROBOV ZERO-FREE REGION


Proof. As before we seek to bound S(N, t) for 1 ≤ N ≤ t. Throughout let
                       r
                         1 − σ 1/6                         log N
                  y=          D1 (log t)1/3 ,    α=      1/3
                           3                         yD (log t)2/3     1

and recall that λ = log t/ log N . Suppose first that
                                          s
                                                   3
                                     λ<2
                                              D1 (1 − σ)

so that α > α0 = 1/2. Via (4.3) and a routine calculation one obtains

                                                                           ϕ
                 N −σ S(N, t) ≤ c1 exp(g(α)y 3 ),         g(α) = 3           + 3α − α3 .
                                                                           α

By explicitly computing derivatives, we find that g(α) is locally maximized at the point
                                                 √         1/2
                                             1+    1 − 4ϕ
                                  α∗ =                             .
                                                   2

Taking a Taylor expansion at α∗ , for each α ≥ α0 there exists ξ = ξ(α) ≥ α0 such that

                                             1                      1
         g(α) = g(α∗ ) + g ′ (α∗ )(α − α∗ ) + g ′′ (α∗ )(α − α∗ )2 + g ′′′ (ξ)(α − α∗ )3 .
                                             2                      6

Note that g ′ (α∗ ) = 0. Additionally, a routine computation reveals that for relevant σ and
t one has 0 < ϕ ≤ ϕ0 = 0.027, so that
                             √           √
                        ∗      2(1 + 4ϕ + 1 − 4ϕ)
                     g(α ) =          √           < 2 + 3ϕ + ϕ2 ,
                                (1 + 1 − 4ϕ)1/2
                                                            1/2
                                   √
                                          
                         ′′   ∗                  1 − 4ϕ
                        g (α ) = −6 2             √                    < −5.81.
                                              1 + 1 − 4ϕ
Furthermore, g ′′′ (ξ) = −6 − 18ϕ/ξ 4 < 0. If α ≥ α∗ then g ′′′ (ξ)(α − α∗ )3 ≤ 0 and if
α0 ≤ α ≤ α∗ then |α − α∗ | < 1 − α0 , so in either case

             1 ′′′
               g (ξ)(α − α∗ )3 < (1 + 3ϕ0 α0−4 )(1 − α0 )(α − α∗ )2 < (α − α∗ )2 .
             6

We conclude that

      N −σ S(N, t) ≤ c1 exp(g(α)y 3 ) ≤ c1 exp((2 + 3ϕ + ϕ2 − 1.9(α − α∗ )2 )y 3 ).          (4.11)

Now suppose that                              s
                                                       3
                                      λ≥2                    ,
                                                  D1 (1 − σ)


                                                  84
                                                       4.2. AN EXPLICIT ESTIMATE OF ZETA


so that α ≤ α0 . In this case, we apply (4.3) to obtain
                                                           
                      −σ                       c3       D2 3 3
                    N S(N, t) ≤ c1 exp log + 3α −          α y .                                                  (4.12)
                                               c1       D1
For any 0 < δ ≤ 1 and α∗ < 1, the function 3α − δα3 + 1.9(α − α∗ )2 is increasing for
α ∈ [0, 1]. Therefore, for 0 ≤ α < α0 < α∗ < 1 one has
                       D2 3                         D2 3
                3α −      α + 1.9(α − α∗ )2 < 3α0 −   α + 1.9(α0 − 1)2 < 1.87.                                    (4.13)
                       D1                           D1 0
In addition, since σ ≤ 1 − 12 (log log t/ log t)2/3 by assumption, we have
                                                1/6
                                            D
                                        y ≥ √1 (log log t)1/3                                                     (4.14)
                                              6
so that for t ≥ H3
                                                      c3
                                           y −3 log      < 0.13.                                                  (4.15)
                                                      c1
Thus by (4.12), (4.13) and (4.15), we have

                           N −σ S(N, t) ≤ c1 exp((2 − 1.9(α − α∗ )2 )y 3 )                                        (4.16)

in this case.

Combining (4.11) and (4.16), one has for any 1 ≤ N ≤ t that
                                                            2          ∗ 2       3
                             N −σ S(N, t) ≤ c1 e(2+3ϕ+ϕ −1.9(α−α ) )y .

Applying this for N = j log 2 for 0 ≤ j ≤ J = ⌈log t/ log 2⌉ − 1,
                               J                                                         J
                                                                             2       3                    ∗ 2 3
        X                      X                                                         X
                n−σ−it ≤ 1 +         (2j )−σ S(2j , t) ≤ 1 + c1 e(2+3ϕ+ϕ )y                    e−1.9(αj −α ) y
       1≤n≤t                   j=0                                                       j=0

where αj is α evaluated at N = j log 2. Proceeding as in the previous lemma, we have
               J                          1/3
                                        yD1 (log t)2/3 J −1.9(α−α∗ )2 y3
                                                      Z
                             ∗ 2 3
              X
                  e−1.9(αj −α ) y ≤ 1 +                   e              dα
                                            log 2       0
                 j=0
                                                 1/3
                                          D (log t)2/3
                                                                r
                                                                     π
                                       <1+ 1                             .
                                             log 2                  1.9y
Substituting and applying (4.14), we have
                                                                                               !
                                              π 1/2 (6D )1/4 (log t)2/3                                  2     3
                                                          1
     |ζ(σ + it)| < 1 + 10−80 + c1         1+                                                       e(2+3ϕ+ϕ )y
                                              1.9       log 2 (log log t)1/6
                               2
                                 √           3/2
                  ≤ 25 t(2+3ϕ+ϕ ) D1 /27(1−σ) (log t)2/3 (log log t)−1/6 .

The result follows.


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CHAPTER 4. THE VINOGRADOV–KOROBOV ZERO-FREE REGION


4.3      Proof of Lemma 4.1

In this section we prove Lemma 4.1. The approach is similar to that of Lemma 3.1 from
the previous chapter, so we shall reuse results where possible. Throughout, let

                                A2 := (51.34)−1 = 0.019477 . . . .

In the spirit of the inductive argument used in the previous chapters, let us define, for any
A ∈ (0, A2 ], the hypothesis

                                                                     A
         H1 (A) :       ζ(σ + it) ̸= 0   for σ > 1 −                               ,   t ≥ H3 .
                                                         (log t)2/3 (log log t)1/3
Vinogradov [Vin58] and Korobov [Kor58] showed that H1 (A) holds for some A > 0. It
then suffices to show that H1 (A) =⇒ H1 (A + ε) for all 0 < A ≤ A2 where as before
ε = 10−100 is a sufficiently small constant. We show this via a contradiction to the
following assumption.

Assumption 4.1. For some 0 < A ≤ A2 , hypothesis H1 (A) holds. Furthermore, hypoth-
esis H1 (A + ε) does not hold, i.e. there exists a zero ρ0 = β0 + it with t ≥ H3 and

                          A ≤ (1 − β0 )(log t)2/3 (log log t)1/3 < A + ε.


Throughout, we find it once again convenient to write

                                           η := 1 − β0 .

We start by specifying our choices of a smoothing function f and a non-negative trigono-
metric polynomial P . Compared to previous chapters, here we choose a trigonometric
polynomial of larger degree, found in [MTY24]:
                                                X
                                     P (x) =           bk cos kx                                  (4.17)
                                               0≤k≤K

where K = 40 and the coefficients bk are recorded in [MTY24, Table 2]. In particular,
bk > 0 and
                                                         X
      b0 = 1,       b1 = 1.74600190914994,       b=             bk = 3.56453965437134.            (4.18)
                                                        1≤k≤K



                                                86
                                                                           4.3. PROOF OF LEMMA 4.1


On the other hand we retain the same choice of a smoothing function f as the previous
chapter (in Definition 3.2), albeit with a different choice of θ. Here, we take θ to be the
unique solution of the equation

                                    b1
                        sin2 θ =       (1 − θ cot θ)            (0 < θ < π/2).               (4.19)
                                    b0

Following the discussion in [HB92a], one expects this choice to be optimal as t → ∞, and
the range of t we encounter in this chapter are large enough to support this choice. For
the values of b0 , b1 in (4.18) we may explicitly compute

                                    θ = 1.13269369969232 . . . .

Equipped with these choices, we follow the argument of the previous chapter. For each
fixed σ and all 1 − σ < λ < 1/2 except for a set of Lebesgue measure zero, one has
                  X Λ(n)                                  X
             0≤             σ
                              f (log n)P (t log n) ≤              bk I(σ + ikt, λ) + Q(σ),   (4.20)
                        n
                  n≥1                                    1≤k≤K

where I and Q are defined in Lemma 3.4 and (3.9) respectively. We seek an upper bound
on the terms appearing on the right side; this is our objective throughout the rest of this
section. Lemma 4.2 and Lemma 4.3 play prominent roles in this estimate, and we recall
for convenience the constants

                              B = 4.438,        C1 = 85,               C2 = 25.

We choose λ relative to the constant B; specifically, we take λ = λ∗ where2
                                                               2/3
                    ∗         ∗          −2/3       log log t
                   λ := λ (t) = EB                                     ,       E = 1.63.     (4.21)
                                                      log t

In addition, we choose σ = σ ∗ , defined as

                                                                 A
                                   σ ∗ := σ ∗ (t) = 1 −     2/3 1/3
                                                                           ,                 (4.22)
                                                           L1 L2

   2 The constant E = 1.63 replaces (4/3(1 + b /b))2/3 = 1.428 . . . appearing in Ford’s
                                              0
treatment, and determines how we weight the main and secondary terms appearing in the
estimate in Lemma 4.4. As t → ∞, Ford’s choice is optimal; however for finite t it is more
favourable to take E slightly larger.




                                                    87
CHAPTER 4. THE VINOGRADOV–KOROBOV ZERO-FREE REGION


where
             L1 (t) := log(Kt + T0 ),          L2 := log log(Kt + T0 ),            T0 := 1010 .

We also define the constants

                                       log log H3 2/3
                                                
                                  −2/3
                       λ3 := EB                       ≈ 5.06 · 10−4 ,
                                         log H3
                                           A2
                       σ3 := 1 −         2/3      1/3
                                                      ≈ 1 − 1.24 · 10−6 ,
                                 L1 (H3 ) L2 (H3 )

so that λ∗ ∈ (0, λ3 ] and σ ∗ ∈ [σ3 , 1) for all t ≥ H3 .

In the next few lemmas we repeat the arguments of the previous chapter to derive estimates
of I and Q in our new setting. First, Lemma 4.2 gives

Lemma 4.4 (Main term estimate). Let t ≥ H3 and let σ ∗ (t), λ∗ (t) be as defined above.
If s∗k := σ ∗ + ikt, then for any fixed 0 < δ ≤ 1/2,
                     Z ∞
              1 X        log |ζ(s∗k − λ∗ + iuδ)| − log |ζ(s∗k + λ∗ + iuδ)|
                  ak                                                       du
             4λ∗      −∞                     (cosh u)2
                   1≤k≤K

                                                                        (log t)2/3
                        ≤ 9.108(log t)2/3 (log log t)1/3 + 1.263                     .
                                                                      (log log t)2/3

Proof. Suppose throughout that t ≥ H3 and y ≥ t/2. By inspection, one has
                                                           2/3                          2/3
               ∗        ∗         1          log log(t/3)              1       log log y
              σ (t) − λ (t) < 1 −                                  ≤1−                            .   (4.23)
                                  2            log(t/3)                2         log y

Suppose first that t ≥ H4X where X := 1/10. In light of (4.23) we may apply Lemma 4.2
to obtain
                                           2            1
     log |ζ(σ ∗ − λ∗ + iy)| < B(λ∗ )3/2 log y +
                                             log log y − log log log y + log C1
                                           3            6                                             (4.24)
                                 ∗ 3/2     2
                            ≤ B(λ ) log y + log log y + 3.965
                                           3
since y ≥ H4X /2. Now let T ∈ [t, Kt]. By Lemma 3.7 and the inequality log T ≤ (1 +
log K/ log H4 ) log t, one has

    1 ∞ log |ζ(σ ∗ − λ∗ + i(uδ + T ))|
     Z
                                                             2
                               2
                                       du < B(λ∗ )3/2 log T + log log T + 3.965
    2 −∞               (cosh u)                              3                                        (4.25)
                                                   < 2.897 log log T + 1           (t ≥ H4X ).



                                                    88
                                                                       4.3. PROOF OF LEMMA 4.1


Now suppose that H3 ≤ t < H4X . If t/2 ≤ y ≤ t1/X then by Lemma 4.3,
                                                                           2
                      log |ζ(σ ∗ − λ∗ + iy)| < B2 (ϕ∗ )(λ∗ )3/2 log y +      log log y
                                                                           3
                                                              1
                                                          −     log log log y + log C2
                                                              6
where ϕ∗ := (R(λ∗ )2 log t)−1 and R = 2876. However the same bound also holds for
y > t1/X thanks to Lemma 4.2, since in this range
                                         E 3/2 (4.45 − B)              3B 4/3    (log t)1/3
        (B2 (ϕ∗ ) − B)(λ∗ )3/2 log y >                    log log t +
                                                XB                    2XRE 1/2 (log log t)1/3
                                                       B 8/3      (log t)2/3
                                                 +
                                                     2XR2 E 5/2 (log log t)5/3
                                      > log(C1 /C2 ),

where the last inequality was verified with computer assistance.

Therefore, for all T ∈ [t, Kt] and by Lemma 3.7,
    1 ∞ log |ζ(σ ∗ − λ∗ + i(uδ + T ))|
      Z
                                                                 2
                               2
                                        du < 4.45(λ∗ )3/2 log T + log log T + E(t, T )
    2 −∞              (cosh u)                                   3
where
                                                 1
         E(t, T ) := (B2 (ϕ∗ ) − 4.45) log T −     log log log T + log C2
                                                 6
                             (log T )1/3        −6 (log T )
                                                             2/3     1
                   < 0.004             1/3
                                           + 10                5/3
                                                                   − log log log T + log C2 .
                           (log log T )            (log log T )      6
Substituting the definition of λ∗ and applying H3 ≤ t ≤ T ≤ Kt < KH4X yields
  1 ∞ log |ζ(σ ∗ − λ∗ + i(uδ + T ))|
    Z
                                      du < 2.897 log log T + 1 (H3 ≤ t < H4X ). (4.26)
  2 −∞            (cosh u)2
Combining the cases (4.25) and (4.26), and taking T = kt (1 ≤ k ≤ K), one obtains
               Z ∞
   1 X              log |ζ(s∗k − λ∗ + iuδ)|                                       b
     ∗
            bk                      2
                                            du < 8.555(log t)2/3 (log log t)1/3 + ∗ (4.27)
  4λ            −∞         (cosh u)                                              2λ
         1≤k≤K

for all t ≥ H3 .

Meanwhile, as in the previous chapter, since σ ∗ + λ∗ > 1 we apply [For02b, Lemma 5.1]
(generalized in the usual way to degree K polynomials) to obtain
                             Z ∞
                1 X              log |ζ(s∗k + λ∗ + iuδ)|         log ζ(σ ∗ + λ∗ )
             − ∗          bk                             du ≤ b0                  .             (4.28)
               4λ             −∞        (cosh u)2                      2λ∗
                       1≤k≤K



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CHAPTER 4. THE VINOGRADOV–KOROBOV ZERO-FREE REGION


Applying the estimate (3.12),

                     log ζ(σ ∗ + λ∗ ) ≤ γ(σ ∗ + λ∗ − 1) − log(σ ∗ + λ∗ − 1)
                                                                  1 − σ∗
                                                                        
                                          ∗         ∗
                                      < γλ − log λ − log 1 −               .
                                                                    λ∗

Since (1 − σ ∗ )/λ∗ < 0.003 for t ≥ H3 , and substituting the definition of λ∗ , one has
                                                                                !
        log ζ(σ ∗ + λ∗ )    1     2            2    B 2/3 log log H3                 γ
                 ∗
                         < ∗        log log t − log                  − log 0.997 +
              2λ          2λ      3            3            E                        2
                               B 2/3                            1.02 γ
                           <         (log t)2/3 (log log t)1/3 − ∗ + .
                               3E                                λ   2

Combining this with (4.27) and (4.28) gives
                        Z ∞
            1 X              log |ζ(s∗k − λ∗ + iuδ)| + log |ζ(s∗k + λ∗ + iuδ)|
              ∗
                     bk                                                        du
           4λ            −∞                      (cosh u)2
                 1≤k≤K
                                                              b/2 − 1.02 γ
                           < 9.108(log t)2/3 (log log t)1/3 +              +
                                                                  λ∗           2
                                                                      (log t) 2/3
                           < 9.108(log t)2/3 (log log t)1/3 + 1.263                ,
                                                                    (log log t)2/3

for all t ≥ H3 , as required.


Next we establish a bound on N (z, δ), which we recall is the number of zeroes ρ satisfying
|z − ρ| ≤ δ. This lemma serves a similar purpose as Lemma 3.10 (of the previous chapter)
but is sharper for values of δ, T most relevant to our application.

Lemma 4.5 (Local zero density estimate). Let 0 < δ ≤ 1/4 and z = σ + iT with T ≥ H3
and σ ∈ [1 − δ/100, 1). Then

                  N (z, δ) ≤ 6.1 δ 3/2 log T + 0.36 log log T − 0.53 log δ + 6.


Proof. Applying Lemma 3.3 with s = u and λ = v and f = ζ, one has

                ζ ′ (u)
                                                    
                                X           π(u − ρ)
         −2v Re         = −πRe        cot
                 ζ(u)                          2v
                                       ρ=β+iγ
                                      β≥Re(u−v)
                           Z ∞
                       1      log |ζ(u − v + 2vix/π)| − log |ζ(u + v + 2vix/π)|
                   +                                                            dx.
                       2   −∞                     (cosh x)2



                                                  90
                                                                            4.3. PROOF OF LEMMA 4.1


We take v = 1 + α1 δ + iT , u = α2 δ where α1 = 0.652, α2 = 2.541 (these constants are
chosen to optimize the bound on N (z, δ) when δ = λ3 and T = H3 ). By Lemma 3.7 and
[Bel24, Theorem 1.1],
        Z ∞
    1      log |ζ(u − v + 2vix/π)|                               2
                           2
                                   dx ≤ B((α2 − α1 )δ)3/2 log T + log log T + log C0
    2   −∞         (cosh x)                                      3

where C0 = 70.7. In addition, following the same argument leading up to (3.29), (3.30)
and (3.32) respectively in the previous chapter, one has the series of estimates
                       Z ∞
                   1      log |ζ(u + v + 2vix/π)|
               −                                  dx ≤ γ(α1 + α2 )δ − log(α1 + α2 )δ,
                   2   −∞         (cosh x)2

                                                       ζ ′ (u)   2α2
                                                2vRe           ≤     ,
                                                       ζ(u)      α1
                                      
                                 πξ
                   Re cot                  ≥ 0.6041     (|ξ − α1 | ≤ 101/100, Re ξ > α1 ).
                                 2α2
Combining these estimates gives

               2α2                                                 2
           −       + 0.6041πN (z, δ) ≤ B(α2 − α1 )3/2 δ 3/2 log T + log log T + log C0
               α1                                                  3
                                                  + γ(α1 + α2 )δ − log(α1 + α2 ) − log δ.

The desired result follows from substituting the values of α1 , α2 , B, C0 and applying
δ ≤ 1/4.


The following two lemmas make use of this density bound to estimate the contribution of
various sums over zeroes appearing in I(sk , λ).

Lemma 4.6. If t ≥ H3 and z = σ + iT with σ ∈ [σ3 , 1) and T ∈ [t, Kt] then
                                  X                            (log T )2   N (z, λ∗ )
                                           |z − ρ|−3 ≤ 28.3              −            .
                                                               log log T    (λ∗ )3
                              ρ=β+iγ
                             |z−ρ|>λ∗


Proof. This result is conceptually similar to Lemma 3.14 of the previous chapter. We once




                                                          91
CHAPTER 4. THE VINOGRADOV–KOROBOV ZERO-FREE REGION


again follow the argument in Ford [For02b], by dividing the zeroes in the sum into the sets

                   S1 = {ρ : |Imρ − T | > 1},

                   S2 = {ρ : |z − ρ| > δ, |1 − z − ρ| > δ, |t − Imρ| ≤ 1},

                   S3 = {ρ : |1 − z − ρ| ≤ δ, |t − Imρ| ≤ 1},

                   S4 = {ρ : λ∗ < |z − ρ| ≤ δ : |t − Imρ| ≤ 1}

for some constant δ ∈ (λ∗ , 1/2) to be chosen later. From Lemma 3.12 (of the previous
chapter), the zeroes in S1 contribute
                                     X
                                            |z − ρ|−3 < log T.
                                     ρ∈S1

Meanwhile, each pair of zeroes ρ, 1 − ρ in S2 contributes

                    |z − ρ|−3 + |z − (1 − ρ)|−3 ≤ δ −3 + (2σ − 1 − δ)−3

so that
                       X                 1
                              |z − ρ|−3 ≤ (δ −3 + (2σ − 1 − δ)−3 )|S2 |.
                                         2
                       ρ∈S2

Next, each zero in S3 contributes at most (2σ−1−δ)−3 , and there are N (1−z, δ) = N (z, δ)
of them, so
                          X
                                 |z − ρ|−3 ≤ (2σ − 1 − δ)−3 N (z, δ).
                          ρ∈S3

The main contribution comes from those zeroes in S4 . We have
                        Z δ                                   Z δ
         X
                    −3      dN (z, u)   N (z, δ) N (z, λ)         N (z, u)
             |z − ρ| =          3
                                      =     3
                                                −       3
                                                          +3               du.
                         λ     u          δ           λ        λ    u4
          ρ∈S4

By Lemma 4.5, and since δ ≤ 1/4, we have
         Z δ                 Z δ
             N (z, u)            6.1u3/2 log T + 0.36 log log T − 0.53 log u + 6
       3              du ≤ 3                                                     du
          λ    u4             λ∗                         u4
                                                                    log λ∗ log δ
                                                                                 
                         = 12.2((λ∗ )−3/2 − δ −3/2 ) log T + 0.53          −
                                                                    (λ∗ )3    δ3
                                      + ((λ∗ )−3 − δ −3 )(0.36 log log T + 5.83).

Collecting these estimates, and recalling that by (3.33) one has

                   |S2 | + 2N (z, δ) = N (T + 1) − N (T − 1) < 0.71 log T,


                                                 92
                                                                                        4.3. PROOF OF LEMMA 4.1


we obtain
                                   X                                                   N (z, λ∗ )
                                             |z − ρ|−3 ≤ C3 (λ∗ , δ, T ) −                                                 (4.29)
                                                                                        (λ∗ )3
                                ρ:|z−ρ|>λ

where
                                                                                               
                                 −3/2          −3/2                           log λ log δ
    C3 (λ, δ, T ) := 12.2(λ               −δ          ) log T + 0.53               − 3
                                                                               λ3    δ
        + (λ−3 − δ −3 )(0.36 log log T + 5.83) + (0.355(δ −3 + (2σ3 − 1 − δ)−3 ) + 1) log T.

By inspection C3 (λ, δ, T ) is decreasing in λ for fixed δ and T ≥ H3 . Also, since t ≤ T ,
λ∗ = λ∗ (t) ≥ λ∗ (T ), so that C3 (λ∗ , δ, T ) ≤ C3 (λ∗ (T ), δ, T ). At this point we take δ = 1/4,
which is permitted since λ∗ < 1/4. Using a routine computation, we verify that the
function
                                                                               log log T
                                         h(T ) = C3 (λ∗ (T ), 1/4, T )
                                                                               (log T )2
is decreasing for T ≥ H3 , so that in particular
                             X                                  (log T )2        (log T )2
                                       |z − ρ|−3 ≤ h(H3 )                 < 28.3           ,                               (4.30)
                                                                log log T        log log T
                           ρ:|z−ρ|>λ

as required.

Lemma 4.7. Suppose Assumption 4.1 holds with respect to a complex zero ρ0 = β0 + it.
Let F0 (z) and F ∗ (z) be as defined in Definition 3.2 and Lemma 3.4 with respect to η =
1 − β0 . If λ∗ = λ∗ (t) and σ ∗ = σ ∗ (t) are as defined in (4.21) and (4.22) respectively, and
s∗k = σ ∗ + ikt, then
            X                                                                    
  X                                                   X                                                               0.08
        bk           ReF0 (s∗k − ρ) +                         ReF   ∗
                                                                        (s∗k − ρ)       ≥ b1 F ∗ (σ ∗ − β0 ) −                  .
                                                                                                                   (log log t)2
1≤k≤K         ρ=β+iγ                               ρ=β+iγ
             β≤σ ∗ −λ∗                            β>σ ∗ −λ∗


Proof. As before we rewrite the inner sum as
             X                                   X                                        X
                       ReF0 (s∗k − ρ) +                    ReF ∗ (s∗k − ρ) =                          ReF ∗ (s∗k − ρ)
            ρ=β+iγ                              ρ=β+iγ                                ρ:|s∗k −ρ|≤λ∗
           β≤σ ∗ −λ∗                           β>σ ∗ −λ∗
                                                                                                          π(s∗k − ρ)
                                                                                                                      
                              X                                            X             π
                       +                 ReF0 (s∗k − ρ) + f (0)                       Re ∗ cot
                                                                                        2λ                   2λ∗
                             ρ=β+iγ                                       ρ=β+iγ
                           |s∗k −ρ|>λ∗                                   β≥σ ∗ −λ∗
                                                                        |s∗k −ρ|>λ∗

                       = q1 + q2 + q3 ,



                                                              93
CHAPTER 4. THE VINOGRADOV–KOROBOV ZERO-FREE REGION


say.

First consider q1 . Under hypothesis H1 (A), and by the definition of σ ∗ , one has for
1≤k≤K
                                 ζ(s) ̸= 0         (Re s ≥ σ ∗ , |Im s − kt| ≤ T0 ).                         (4.31)

Furthermore, for t ≥ H3

                                                  EB −2/3
                                     λ∗ /η >              log log t > 400 = ν,                               (4.32)
                                                  A2 + ε

say. Thus, by Lemma 3.16 of the previous chapter,

                                                                             N (s∗k , λ∗ )
                               q1 ≥ F ∗ (s∗k − ρ)1(k=1) − C(0, ν)η 3                       .                 (4.33)
                                                                               (λ∗ )3

Now suppose ρ is a zero counted by q2 . By hypothesis H1 (A), σ ∗ > 1 − η and hence
Re(s∗k − ρ)/η > −1. Furthermore, by (4.32) one has |(s∗k − ρ)/η| > λ∗ /η > ν. Thus

                                                                           C(−1, ν)η 3
                           |F0 (s∗k − ρ)| = |W0 ((s∗k − ρ)/η)| ≤                       ,
                                                                            |s∗k − ρ|3

where we recall that W0 and C are defined in Definition 3.1 and Lemma 2.3 respectively.
Therefore, applying Lemma 4.6,

                                                                                  (log kt)2    N (s∗k , λ∗ )
                                  X                                                                         
       |q2 | ≤ C(−1, ν)η   3
                                             |s∗k − ρ|−3 ≤ C(−1, ν)η 3       28.3            −                .
                                                                                  log log kt     (λ∗ )3
                                 ρ=β+iγ
                               |s∗k −ρ|>λ∗


Since k ≤ K, C(−1, ν) = 50.884 . . . and η < (A2 + ε)(log t)−2/3 (log log t)−1/3 by Assump-
tion 4.1, one has
                                                                            ∗    ∗
                                                0.02                  3 N (sk , λ )
                                 |q2 | ≤                  − C(−1, ν)η               .                        (4.34)
                                             (log log t)2                 (λ∗ )3
Lastly, by (4.31) and Lemma 3.15 (of the previous chapter) one has

                                                                     10−10
                                           q3 ≥ −10−10 η > −                   .                             (4.35)
                                                                  (log log t)2

Combining (4.33), (4.34) and (4.35) and summing over k, one obtains
         X                                              
 X                                  X                                                                    0.08
     bk           ReF0 (s∗k − ρ) +        ReF ∗ (s∗k − ρ) > b1 F ∗ (σ ∗ − β0 ) −                                   ,
                                                                                                      (log log t)2
1≤k≤K         ρ=β+iγ                              ρ=β+iγ
             β≤σ ∗ −λ∗                           β>σ ∗ −λ∗

so the result follows.


                                                             94
                                                                                4.3. PROOF OF LEMMA 4.1


Recall that for any fixed σ and λ ∈ (1 − σ, 1/2), one has (3.8), i.e.
                                                X
                                       0≤              bk I(sk , λ) + a0 Q(σ).
                                               1≤k≤K

At this point we take σ = σ ∗ and λ = λ∗ . Applying Lemma 4.4, Lemma 4.7 and Lemma
3.18 with this choice of parameters, for all t ≥ H3 one has
    X
            bk I(s∗k , λ∗ )
   1≤k≤K
                            Z ∞
             f (0) X            log |ζ(s∗k − λ∗ + 2λ∗ iu/π)| − log |ζ(sk + λ∗ + 2λ∗ iu/π)|
        ≤        ∗
                         bk                                                                du
             4λ              −∞                          (cosh u)2
                   1≤k≤K
                    X       X                          X                   
                                          ∗ ∗                          ∗
               −        bk           ReF (sk − ρ) +            ReF0 (sk − ρ)
                   1≤k≤K            ρ=β+iγ                           ρ=β+iγ
                                   β>σ ∗ −λ∗                        β≤σ ∗ −λ∗
                         X                              X
               + Re               bk F ∗ (s∗k − 1) +            bk |E(s∗k )|
                       1≤k≤K                           1≤k≤K

                                                                          (log t)2/3
        ≤ 9.108f (0)(log t)2/3 (log log t)1/3 + 1.263f (0)                             − b1 F ∗ (σ ∗ − β0 )
                                                                        (log log t)2/3
                      0.08        X
               +                +   bk |E(s∗k )|.
                   (log log t)2
                                      1≤k≤K

By Lemma 2.7, one has
                              X                                           X
                                    bk |E(sk )| ≤ (14η 2 + 425η 3 )              bk < 55η 2 ,
                         1≤k≤K                                          1≤k≤K

and by Lemma 3.20, one has

                                          Q(σ ∗ ) ≤ F (σ ∗ − 1) − 2.9η.

Summing these estimates, one obtains

   0 ≤ 9.108f (0)(log t)2/3 (log log t)1/3 + a0 F (σ ∗ − 1) − a1 F ∗ (σ ∗ − β0 ) + C3 (η, t)                  (4.36)

where
                                                 (log t)2/3        0.08
               C3 (η, t) := 1.263ηw(0)                        +              + 55η 2 − 2.9a0 η.
                                               (log log t)2/3   (log log t)2
For each fixed t ≥ H3 , C3 is increasing in η. By Assumption 4.1 one has η ≤ (A2 +
ε)(log t)−2/3 (log log t)−1/3 ; substituting this bound gives (for any x ≥ log H3 )
                                   0.139     0.08          0.05           0.03
                 C3 (η, ex ) ≤           +         2
                                                     − 2/3      1/3
                                                                    + 4/3         .
                                   log x   (log x)    x (log x)      x (log x)2/3


                                                           95
CHAPTER 4. THE VINOGRADOV–KOROBOV ZERO-FREE REGION


However the function on the right side is decreasing for x ≥ log H3 , so

                    C3 (η, t) ≤ C3 (η, H3 ) = 0.0109804538 . . .             (t ≥ H3 ).

Next, to estimate F ∗ (σ ∗ − β0 ), we use Assumption 4.1, the definition of σ ∗ , A ≤ A2 and
t ≥ H3 to get

              π(σ ∗ − β0 )
                                                                                 
                             π                      A+ε               A
                     ∗
                           < ∗                    2/3         1/3
                                                                  − 2/3 1/3           < 10−7 .
                 2λ         2λ             (log t) (log log t)     L L   1    1

Since 1/x − cot x < 10−7 for 0 < x < 10−7 , we have

                                                     10−7 w(0)η
                F ∗ (σ ∗ − β0 ) ≥ F (σ ∗ − β0 ) −               > F (σ ∗ − β0 ) − 10−8 .           (4.37)
                                                        λ∗

Lastly, we estimate the quantity a1 F (σ ∗ −β0 )−a0 F (σ ∗ −1) by repeating the analysis of the
previous chapter (with our new value of θ and using µ := (1 − σ ∗ )/η ≥ 0.9999953403 . . .)
to obtain
                                                Z 2θ cot θ
                ∗                  ∗
        a1 F (σ − β0 ) − a0 F (σ − 1) =                      eµu (a1 e−u − a0 )w(u)du > 1.01267.
                                                 0

Combining the above estimate with (4.36) and (4.37), one finally has

                                                             1.01267 − 10−8 a1 − C3 (η, t)
                 (1 − β0 )(log t)2/3 (log log t)1/3 >
                                                                     9.108w(0)
                                                      > 0.01945 > A2 + ε ≥ A + ε,

so we have reached the desired contradiction to Assumption 4.1. Thus, Lemma 4.1 (and
Theorem 1.3) is proved.




                                                     96
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