                                           SOME EXPLICIT ESTIMATES FOR THE ERROR TERM IN THE
                                                         PRIME NUMBER THEOREM




arXiv:2204.01980v2 [math.NT] 20 Apr 2022
                                                                DANIEL R. JOHNSTON AND ANDREW YANG


                                                   Abstract. By combining and improving recent techniques and results, we
                                                   provide explicit estimates for the error terms |π(x) − li(x)|, |θ(x) − x| and
                                                   |ψ(x) − x| appearing in the prime number theorem. For example,   √     we show
                                                   for all x ≥ 2 that |ψ(x) − x| ≤ 9.39x(log x)1.515 exp(−0.8274 log x). Our
                                                   estimates rely heavily on explicit zero-free regions and zero-density estimates
                                                   for the Riemann zeta-function, and improve on existing bounds for prime-
                                                   counting functions for large values of x.




                                                                               1. Introduction
                                           1.1. Overview. Let
                                                             X                      X                            X
                                                      π(x) =   1,          θ(x) =         log p   and ψ(x) =            log p
                                                                 p≤x                p≤x                         pm ≤x

                                           denote the standard prime-counting functions (see e.g. [Apo13, Chapter 4]). The
                                           prime number theorem is equivalent to any of the statements
                                                                 π(x) ∼ li(x),      θ(x) ∼ x,      or ψ(x) ∼ x,
                                                          Rx
                                           where li(x) = 0 log1 t dt is the logarithmic integral. At the core of many results in
                                           analytic number theory are sharp estimates on the functions
                                                               |π(x) − li(x)|,     |θ(x) − x|     and |ψ(x) − x|.                    (1.1)
                                           As a result, many authors [RS62; FK15; Tru16; Axl18; Büt18; Dus18; PT21b;
                                           Bro+21] have obtained explicit bounds on these functions. In particular, Platt and
                                           Trudgian [PT21b, Theorem 1] recently employed an argument due to Pintz [Pin80,
                                           Theorem 1] to provide estimates of the form
                                                                                              p
                                                               |ψ(x) − x| ≤ A(log x)B exp(−C log x)                      (1.2)
                                           for some positive constants A, B and C, and x ≥ exp(1000). They also give related
                                           estimates for |θ(x) − x| and |π(x) − li(x)| [PT21b, Corollaries 1 and 2].
                                              In this paper we make refinements to Platt and Trudgian’s method, in turn giving
                                           improvements to their estimates that hold for a wider range of x. Most notably, we
                                           modify a technique recently employed by Broadbent et al. [Bro+21, Section A.2]
                                           and make use of a new explicit error term for the Riemann-von Mangoldt formula
                                           [CHJ21]. We also give improvements to other preliminary results in [PT21b] by
                                           incorporating a recent verification of the Riemann hypothesis up to height 3 · 1012
                                           [PT21a]. Some of these improvements are also discussed in [CHJ21, Section 6].

                                             Date: April 21, 2022.
                                                                                           1
2                     DANIEL R. JOHNSTON AND ANDREW YANG


1.2. Statement of results. We obtain the following bounds for |ψ(x) − x|.
Theorem 1.1. For all x ≥ 2,
                |ψ(x) − x| ≤ 9.39x(log x)1.515 exp(−0.8274 log x).
                                                          p
                                                                                (1.3)
More generally, for corresponding values of X, A, B, C and ǫ0 in Table 1,
                                                   p     
                    |ψ(x) − x| ≤ Ax(log x)B exp −C log x                        (1.4)
and
                                  |ψ(x) − x| ≤ ǫ0 x
for all log x ≥ X.
   Notably, our values of ǫ0 in Table 1 are an order of magnitude smaller than the
corresponding values in [PT21b, Table 1] or [CHJ21, Table 1].
   From Theorem 1.1, we then apply standard procedures (see Section 4) to obtain
similar results for |θ(x) − x| and |π(x) − li(x)| (cf. [PT21b, Corollaries 2 and 3]).
Corollary 1.2. For corresponding values of X, A, B and C in Table 1,
                                         p
         |θ(x) − x| ≤ A1 x(log x)B exp(−C log x) for all log x ≥ X,             (1.5)
where A1 = A + 0.01.
Corollary 1.3. For all x ≥ 2,
                                                             p
               |π(x) − li(x)| ≤ 9.59x(log x)0.515 exp(−0.8274 log x).           (1.6)
    We also we give estimates of the form
               |ψ(x) − x| ≤ A(log x)B exp(−C log3/5 x(log log x)−1/5 )
(and similarly for |θ(x) − x| and |π(x) − li(x)|) by using an explicit Vinogradov–
Korobov zero-free region for ζ(s) due to Ford [For02, Theorem 5].
Theorem 1.4. For all x ≥ 23, we have
                                                                       
   |ψ(x) − x| ≤ 0.026x(log x)1.801 exp −0.1853 log3/5 x(log log x)−1/5          (1.7)
                                                                       
    |θ(x) − x| ≤ 0.027x(log x)1.801 exp −0.1853 log3/5 x(log log x)−1/5         (1.8)
                                                                       
   |π(x) − x| ≤ 0.028x(log x)0.801 exp −0.1853 log3/5 x(log log x)−1/5 .        (1.9)

   Certainly, the estimates (1.7)–(1.9) are asymptotically superior to (1.3)–(1.6).
However, the bounds in Theorem 1.4 are worse for values of x most likely to appear
in applications. In particular, the bounds in Theorem 1.1 are better than (1.7) for
all exp(59) ≤ x ≤ exp(2.8 · 1010 ).
   In [PT21b], Platt and Trudgian also apply their results to an inequality studied
by Ramanujan. Namely, they use their estimates for |θ(x) − x| to show that
                                            ex  x 
                                 π 2 (x) <       π                            (1.10)
                                           log x   e
for all x ≥ exp(3915). By substituting (1.5) for X = 3000 into the equations on
page 879 of [PT21b], one immediately obtains the following improvement.
Theorem 1.5. Ramanujan’s inequality (1.10) holds for all x ≥ exp(3361).
  Note that (1.10) is conjectured to hold for all x ≥ 38, 358, 837, 683 (see [DP15])
and has also been shown to hold for all 38, 358, 837, 683 ≤ x ≤ exp(103) [Joh21].
       ESTIMATES FOR THE ERROR TERM IN THE PRIME NUMBER THEOREM                    3


          Table 1. Values of X, A, B, C and ǫ0 for Theorem 1.1 and
         Corollary 1.2. Here, σ and K are parameters that appear in the
                              proof of Theorem 1.1.

            X         σ       K     A      B       C            ǫ0
           log 2   0.985692   4   9.39   1.515 0.8274         23.17
           3000    0.986688   4   8.86   1.514 0.8288      3.14 · 10−14
           4000    0.988164   4   8.15   1.512 0.8309      3.43 · 10−17
           5000    0.989238   4   7.65   1.511 0.8324      8.14 · 10−20
           6000    0.990000   4   7.22   1.510 0.8335      3.35 · 10−22
           7000    0.990718   4   6.99   1.510 0.8345      2.14 · 10−24
           8000    0.991258   4   6.78   1.509 0.8353      1.89 · 10−26
           9000    0.991714   4   6.58   1.509 0.8359      2.22 · 10−28
          10000 0.992100      5   6.72   1.508 0.8369      3.27 · 10−30
           105     0.997312   1   23.13 1.503 0.8659       9.12 · 10−111
           106     0.998974   1   38.57 1.502 1.0318       3.12 · 10−438
           107     0.999662   1   42.90 1.501 1.0706      6.62 · 10−1459
           108     0.999890   1   44.41 1.501 1.0839      2.18 · 10−4694
           109     0.999964   1   44.97 1.501 1.0886 5.86 · 10−14936
           1010    0.999988   1   45.17 1.501 1.0903 3.45 · 10−47335


1.3. Outline of paper. The structure of the paper is as follows. In Section 2, we
state some preliminary lemmas. In Section 3, we prove Theorem 1.1. In Section
4 we prove Corollaries 1.2 and 1.3. In Section 5 we prove Theorem 1.4. Finally,
in Section 6 we detail potential improvements to our results. An appendix is also
included for lengthy bounding arguments required in Sections 3.3 and 5.1.

1.4. Remarks on recent correspondence and literature. Upon announcing
the first version of this article, H. Kadiri informed us that she was working on
similar work with A. Fiori and J. Swidinsky. Subsequently, they announced their
work [FKS22]. Compared with the results of this paper, [FKS22] contains better
bounds on |ψ(x) − x| for lower values of X (X ≤ 10000) and at present, does
not feature any corresponding results for |π(x) − li(x)|. However, as [FKS22] is
currently unpublished, and our results on |ψ(x) − x| for low X are used to establish
subsequent results, we have refrained from using the work in [FKS22] here.
   Importantly though, H. Kadiri and G. Hiary brought to our attention an un-
reliable result in [Hia16] which was previously used in this work. In particular,
in [Hia16] an error in [CG04] was used to compute bounds on |ζ(1/2 + it)|. For-
tunately, this error can be remedied to give the slightly worse result (cf. [Hia16,
4                       DANIEL R. JOHNSTON AND ANDREW YANG


Theorem 1.1])
                          |ζ(1/2 + it)| ≤ 0.77t1/6 log t,     t ≥ 3.                    (1.11)
For more detail on how to obtain this modified result, see [Pat21, Section 2.2.1].
The bound (1.11) has been incorporated throughout this paper, in particular in
Lemmas 2.6, 2.8 and 2.9 where [Hia16, Theorem 1.1] was previously employed.

                                   2. Useful lemmas
   In this section, we list a series of useful lemmas. Most of the following results
are estimates regarding the zeros of the Riemann zeta-function which follow from
existing results in the literature. Throughout, we use the notation f (x) = O∗ (g(x))
to mean |f (x)| ≤ g(x) for all x under consideration. Moreover, any sum will be
assumed to be over the non-trivial zeros ρ of the Riemann zeta-function ζ(s).
   We begin by quoting a recent rigorous verification of the Riemann hypothesis up
to height 3 · 1012 .
Lemma 2.1 (Riemann height [PT21a]). Let 0 < β < 1 and H = 3 000 175 332 800.
Then, if ζ(β + it) = 0 and |t| ≤ H we have β = 21 .
   Lemma 2.1 allows us to update some useful bounds on |ψ(x) − x|, |θ(x) − x| and
|π(x) − li(x)| due to Büthe [Büt16].
Lemma 2.2. The following estimates hold:
                       √
                        x
         |ψ(x) − x| <     log2 x,          for 59 < x ≤ 2.169 · 1025 ,
                       8π
                       √
                        x
          |θ(x) − x| <    log2 x,         for 599 < x ≤ 2.169 · 1025 ,
                       8π
                       √
                        x
      |π(x) − li(x)| <    log x,         for 2657 < x ≤ 2.169 · 1025 .
                       8π
Proof. Substitute T = H (from Lemma 2.1) into [Büt16, Theorem 2].                          
Lemma 2.3. For all x ≥ exp(2000), we have
                                 ψ(x) − x
                                          ≤ 1.570 · 10−12 .
                                    x
Proof. Repeat the computations in [Büt16, Section 6] using the new Riemann height
(Lemma 2.1) along with parameters1 α = 0, c = 33.6 and ǫ = 1.12 · 10−11 .       
   Next we give a recent estimate on the error term in the Riemann-von Mangoldt
formula due to Cully-Hugill and the first author [CHJ21]. Here we convert their
result into a specific form that is useful for our application.
Lemma 2.4. Let T and x be such that max{50, log x} < T /1.8 < (x1/35 − 2)/4.
Then,
               ψ(x) − x     X xℜ(ρ)−1        4.3128
                        ≤                 +         log0.6 x           (2.1)
                  x                |ℑ(ρ)|      T
                                    |ℑ(ρ)|≤T

for all x ≥ exp(1000).
   1As noted in [Bro+21, Theorem 16] there is a small error in [Büt16, Theorem 1]. However, in
the case when α = 0 there is no difference to the result.
        ESTIMATES FOR THE ERROR TERM IN THE PRIME NUMBER THEOREM                       5


Proof. Using the values of α, ω, M and xM given in the final row of [CHJ21, Table
5], we have by [CHJ21, Theorem 1.2] that
                ψ(x) − x       X xρ−1          
                                                 4.3128     0.6
                                                                
                                             ∗
                          =              +O              log x
                    x                 ρ             T
                              |ℑ(ρ)|≤T
                                   ∗


for some T ∈ [T /1.8, T ]. Applying the triangle inequality and the bound T ∗ ≤ T
           ∗

then gives the desired result.                                                  
  We also require the following bound for the sum over 1/ℑ(ρ) up to height T .
Lemma 2.5 ([STD15, Lemma 2.10], [BPT21, Lemma 8]). If T ≥ 4πe, then
                                                       
         1      2  T               X      1     1     2  T
            log       − 0.9321 ≤             ≤    log        .
        4π         2π                   ℑ(ρ)   4π        2π
                                         0<ℑ(ρ)≤T

  Now, let N (σ, T ) denote the number of zeros in the box σ < ℜ(s) < 1 and
0 < ℑ(s) < T . In [KLN18, Lemma 4.14], estimates are given for N (σ, T ) for
σ ∈ [0.75, 1) of the form
                N (σ, T ) ≤ C1 (σ)T 8(1−σ)/3 log5−2σ T + C2 (σ) log2 T.            (2.2)
where C1 (σ) and C2 (σ) are positive constants. As discussed in [CHJ21, Section 6],
one can improve the values of C1 (σ) and C2 (σ) given in [KLN18] by incorporating
the recent Riemann height (Lemma 2.1) and replacing (3.13) in [KLN18] with
Theorem 2 of [CHT21]. Moreover, since the results of [KLN18] rely on Hiary’s
unreliable bound for |ζ(1/2 + it)| (see Section 1.4) we use (1.11) in place of (3.2) in
[KLN18]. This amounts to setting a1 = 0.77 and a2 = 3.161 on page 27 of [KLN18].
Making these adjustments, we recalculated C1 (σ) and C2 (σ) for a variety of values
of σ close to 1.
Lemma 2.6 (Zero-density estimates). For corresponding values of σ, C1 (σ) and
C2 (σ) in Table 3 (see Appendix B), the bound (2.2) holds.
   The following lemmas are all explicit zero-free regions for the Riemann zeta-
function. As our main results hold for a large range of x, it is worthwhile to use of
a variety of zero-free regions that are better (i.e. wider) at different heights.
Lemma 2.7 (Classical zero-free region [MT15]). For |t| ≥ 2 there are no zeros of
ζ(β + it) in the region β ≥ 1 − ν1 (t) where
                                              1
                                 ν1 (t) =
                                          R0 log |t|
and2 R0 = 5.5666305.
Lemma 2.8. For |t| ≥ 5.45 · 108 there are no zeros of ζ(β + it) in the region
                                           1
                             β ≥1−                   ,
                                      R(|t|) log |t|
where
                             J(t) + 0.685 + 0.155 log log t
                    R(t) =                               ,
                                                  0.0196
                              log t 0.04962 − J(t)+1.15

   2This value of R is lower than that appearing in [MT15, Theorem 1]. However, since the
                    0
Riemann hypothesis has now been verified to a higher height (Lemma 2.1), we can take R0 =
5.5666305 as discussed in [MT15, Section 6.1].
6                        DANIEL R. JOHNSTON AND ANDREW YANG


with
                                     1
                            J(t) =     log t + log log t + log(0.77).
                                     6
Proof. Same as [For02, Theorem 3] with an improved expression for J(t). To achieve
this, we replace Ford’s (1.6) with the estimate (1.11) given in Section 1.4.    

  We convert this result into a slightly smaller zero-free region that is easier to
work with and holds for all |t| ≥ 3.
Lemma 2.9. For |t| ≥ 3, there are no zeroes of ζ(β +it) in the region β ≥ 1−ν2 (t),
where                                                        
                                1            8.02 log log |t|
                  ν2 (t) =                1−                    .            (2.3)
                           3.359 log |t|         log |t|
Proof. Due to the symmetry of zeros of ζ(s) about the line ℑ(s) = 0, it suffices to
prove the lemma for positive t ≥ 3. So, first we note that for 3 ≤ t ≤ exp(91.2853)
the result holds by Lemmas 2.1 and 2.7. Now, let t ≥ exp(91.2853), R(t) and J(t)
be as defined in Lemma 2.8, and
                            0.0196              6 · 0.0196     0.526 log log t
           a1 (t) :=                         ≤               ≤                 ,
                     0.04962(J(t) + 1.15)      0.04962 log t        log t
                     6(1.155 log log t + log(0.77) + 0.685)    7.494 log log t
           a2 (t) :=                                         ≤                 .
                                       log t                       log t
Then,
                                                             
                           1         6 · 0.04962 1 − a1 (t)
                                   =
                        R(t) log t       log t     1 + a2 (t)
                                                                      
                                     6 · 0.04962       a1 (t) + a2 (t)
                                   =               1−
                                         log t           1 + a2 (t)
                                                                    
                                           1          8.02 log log t
                                   ≥              1−
                                     3.359 log t           log t
as required.                                                                          

Lemma 2.10 (Vinogradov–Korobov zero-free region [For02, Theorem 5]). For |t| ≥
3 there are no zeros of ζ(β + it) in the region β ≥ 1 − ν3 (t) where
                                                           1
                              ν3 (t) =           2/3
                                                                                   (2.4)
                                         c log         |t|(log log |t|)1/3
and c = 57.54.
    By way of comparing the zero-free regions above, let
                               ν(t) = max{ν1 (t), ν2 (t), ν3 (t)}.
Then,
                       ν(t) = ν1 (t) for 3 ≤ |t| ≤ exp(91.2),
                       ν(t) = ν2 (t) for exp(91.3) ≤ |t| ≤ exp(54563),
                       ν(t) = ν3 (t) for |t| ≥ exp(54563.1).
       ESTIMATES FOR THE ERROR TERM IN THE PRIME NUMBER THEOREM                                     7


                                3. Proof of Theorem 1.1
3.1. Estimates for small values of x (2 ≤ x ≤ exp(2488)). For 2 ≤ x ≤ 59,
Theorem 1.1 can be verified by a straight-forward computation. For 59 < x ≤
exp(50), Theorem 1.1 holds by Lemma 2.2. Then, to cover the range exp(50) <
x ≤ exp(2000) we use [Bro+21, Table 8]. Finally, for exp(2000) < x ≤ exp(2488)
we apply Lemma 2.3.
3.2. Proof of Theorem 1.1 for X ≤ 10000. Let X ≤ 10000 correspond to one
of the initial rows of Table 1. Suppose x ≥ x0 where x0 = exp(X) except in the
case X = log 2 whereby we set x0 = exp(2488) and use Section 3.1 to cover smaller
values of x. In what follows, we build on the methods from [PT21b,
                                                               p       Section 3]
and [Bro+21, Section A.2]. To begin with, we set T = exp(2 log x/R0 ) where
R0 = 5.5666305 as in Lemma 2.7. By Lemma 2.4
                    ψ(x) − x      X xℜ(ρ)−1      4.3128
                              ≤                +        log0.6 x            (3.1)
                        x               |ℑ(ρ)|     T
                                            |ℑ(ρ)|≤T

where the sum is over all non-trivial zeros ρ of the Riemann zeta-function with
|ℑ(ρ)| ≤ T . For some choice of σ ∈ [0.98, 1), we write the sum in (3.1) as
                 X xℜ(ρ)−1         X xℜ(ρ)−1            X xℜ(ρ)−1
                               =                   +                  .     (3.2)
                       |ℑ(ρ)|               |ℑ(ρ)|            |ℑ(ρ)|
                |ℑ(ρ)|≤T                     |ℑ(ρ)|≤T               |ℑ(ρ)|≤T
                                              ℜ(ρ)≤σ                 ℜ(ρ)>σ

Let H be the Riemann height from Lemma 2.1. By Lemma 2.5, the first sum on
the right-hand side of (3.2) satisfies
       X xℜ(ρ)−1          X xℜ(ρ)−1        X   xℜ(ρ)−1
                     =                   +
             |ℑ(ρ)|               |ℑ(ρ)|        |ℑ(ρ)|
    |ℑ(ρ)|≤T               |ℑ(ρ)|≤H                     H<|ℑ(ρ)|≤T
     ℜ(ρ)≤σ                 ℜ(ρ)≤σ                       ℜ(ρ)≤σ
                                1
                                                                                                !
                         x− 2 log2 ( 2π
                                     H
                                        )                    log2 ( 2π
                                                                    T
                                                                       ) log2 ( 2π
                                                                                H
                                                                                   )
                       ≤                  + xσ−1                        −            + 1.8642
                               2π                               2π          2π
                       = s1 (x, σ), say.                                       (3.3)
                                                                        p
Now, let ν1 (t) = 1/(R0 log t) be as defined in Lemma 2.7 and t0 := exp( log x/R0 ).
Note also that since x ≥ exp(2488), we have T > H. For the second sum on the
right-hand side of (3.2),
        X xℜ(ρ)−1           X x−ν1 (ℑ(ρ))
                       ≤
                |ℑ(ρ)|               |ℑ(ρ)|
     |ℑ(ρ)|≤T               |ℑ(ρ)|≤T
      ℜ(ρ)>σ                 ℜ(ρ)>σ
                                Z T
                                        x−ν1 (t)
                           =2                    dN (σ, t)
                                    H     t
                                    Z t0                           Z T                     !
                                            x−ν1 (t)                     x−ν1 (t)
                           ≤2                        dN (σ, t) +                  dN (σ, t)     (3.4)
                                        H     t                     t0     t
with the understanding that if t0 < H, we remove the first integral. Next, let K ≥ 1
and define                                   r        !
                                            k     log x
                         tk := exp     1+
                                           K       R0
8                      DANIEL R. JOHNSTON AND ANDREW YANG


for 0 ≤ k ≤ K. Noting that x−ν1 (t) /t is decreasing if and only if t ≥ t0 , the second
integral in (3.4) is bounded by
   Z T −ν1 (t)                                       K−1
       x                   x−ν1 (t0 ) t1             X x−ν1 (tk ) Z tk+1
                                     Z
               dN (σ, t) ≤               dN (σ, t) +                     dN (σ, t)
    t0     t                 t0       t0                   tk      tk
                                                               k=1
                                                                            K−1
                             x  −ν1 (t0 )                                   X      x−ν1 (tk )
                         ≤                  (N (σ, t1 ) − N (σ, t0 )) +                       N (σ, tk+1 )
                                  t0                                                 tk
                                                                             k=1
                             K−1
                             X      x−ν1 (tk )                x−ν1 (t0 )
                         =                     N (σ, tk+1 ) −            N (σ, t0 ).                   (3.5)
                                      tk                        t0
                             k=0

Meanwhile, the first integral in (3.4) is bounded by
                                  x−ν1 (t0 ) t0
        Z t0 −ν1 (t)
             x                                              x−ν1 (t0 )
                                            Z
                      dN (σ, t) ≤               dN (σ, t) =            N (σ, t0 )
         H       t                   t0      H                t0
since N (σ, H) = 0.
   Therefore, writing N0 (σ, tk ) = C1 (σ)T 8(1−σ)/3 log5−2σ tk + C2 (σ) log2 tk as the
upper bound for N (σ, tk ) from Lemma 2.6, we have
                                                    K−1
                                            X x−ν1 (tk )
                       X         xℜ(ρ)−1
                                         ≤2              N0 (σ, tk+1 )
                                  |ℑ(ρ)|        tk
                     |ℑ(ρ)|≤T                       k=0
                      ℜ(ρ)>σ

                                              = s2 (x, σ, K), say.                                     (3.6)
Substituting everything back into (3.1) gives
                       ψ(x) − x
                                ≤ s1 (x, σ) + s2 (x, σ, K) + s3 (x),                                   (3.7)
                          x
where s1 (x, σ) and s2 (x, σ, K) are as in (3.3) and (3.6), and
                                         4.3128
                                s3 (x) =        log0.6 x.                                              (3.8)
                                           T
  We now note that for fixed K, each term
                                 x−ν1 (tk )
                                       2    N0 (σ, tk+1 )
                                     tk
appearing in s2 (x, σ, K) decreases towards
                                5−2σ           B          r       !
                            k+1             log x               log x
             2C1 (σ) 1 +                              exp −Ck                                          (3.9)
                             K               R0                  R0
for sufficiently large x, where
                                                         
         5 − 2σ              K +k    K   8            k+1
      B=            and Ck =      +     − (1 − σ) 1 +       .                                        (3.10)
            2                 K     K+k  3             K
Therefore, if we let
                                       C′ =             min    Ck                                    (3.11)
                                               k∈{0,...,K−1}
and                                "                                            !#−1
                                                   B                r
                                           log x                 ′       log x
                    F (x, σ) =                          exp −C                                       (3.12)
                                             R                             R
        ESTIMATES FOR THE ERROR TERM IN THE PRIME NUMBER THEOREM                            9


then s2 (x, σ, K)F (x, σ) is decreasing for sufficiently large x. Note in particular
that3
                                     8σ − 2     8
                         C ′ ≤ C0 =         −      (1 − σ) < 2.
                                        3      3K
Thus, s1 (x, σ)F (x, σ) and s3 (x)F (x, σ) are also decreasing for sufficiently large x.
We then define
      A′ = A′ (x, σ, K) := s1 (x, σ)F (x, σ) + s2 (x, σ, K)F (x, σ) + s3 (x)F (x, σ)
so that provided A′ (x, σ, K) is decreasing for x ≥ x0 , we have
                                                B            r       !
              ψ(x) − x       ′             log x             ′   log x
                         ≤ A (x0 , σ, K)            exp −C               ,
                 x                           R                     R
for all x ≥ x0 . To compute the values of A, B and C in Table 1, we let B be as
                   ′          ′
in (3.10), A = RAB , C = √CR , and optimised over K and σ for different values of
x0 = exp(X) (or x0 = exp(2488) when X = log 2).
   Note that the values of σ are optimised to a higher precision than presented in
Table 3. To achieve this, we note that if σ1 < σ < σ2 , then one can take C1 (σ2 )
and C2 (σ1 ) in the zero-density bound (2.2). This is justified as C1 and C2 are
respectively increasing and decreasing in σ (using the computational method we
modified from [KLN18]).
   We also remark that Platt and Trudgian’s approach [PT21b, Section 3] can be
viewed as a less optimised version of the case K = 1. Namely, in our notation,
Platt and Trudgian have that B = (5 − 2σ)/2, C ′ = (16σ − 10)/3 and A′ →
2.0025 · 25−2σ · C1 (σ) as x → ∞ (see [PT21b, page 875]). Comparing this to (3.9)–
(3.12), we see an improvement in C ′ and the limiting value of A′ . In particular,
for fixed σ, large K and large x, the value of A′ obtained via our method will be
approximately 8 times smaller than that using Platt and Trudgian’s approach.
3.3. Proof of Theorem 1.1 for X ≥ 105 . For these values of X, we use the
zero-free region in Lemma 2.3 to obtain a better result. As this zero-free region
is more complicated to work with, we use a simplified argument equivalent to the
case K = 1 in Section 3.2. One could parametrise K as in the previous section, but
the improvement will only be small as we are now considering much larger X.
   Analogously to the previous section, we seek to maximise the function
                                     x−ν2 (t)
                                        t
for t ≥ H = 3 000 175 332 800. Namely, we wish to find bounds for
                                   1
                        T :=         x−ν2 (t)
                                              = min txν2 (t) .                         (3.13)
                             maxt≥H             t≥H
                                        t
This is done in Appendix A (Lemma A.2), whereby we show for all x ≥ x0 = exp(X)
that                         p                   p
                      exp(B2 log x) ≤ T ≤ exp(B3 log x),                  (3.14)
where B2 = B2 (x0 ) and B3 = B3 (x0 ) are constants given in Table 2. Moreover, in
Lemma A.1, we show that if t0 is the point where the maximum of x−ν2 (t) /t occurs,
then x−ν2 (t) /t is increasing for t ∈ [H, t0 ) and decreasing for t > t0 .

  3In fact, for all the cases we consider, C ′ = C .
                                                  0
10                         DANIEL R. JOHNSTON AND ANDREW YANG


    We now show how each error term from Section 3.2 changes in this setting. So
firstly, by (3.13), we have for all log x ≥ X,

                              1
                                                                                             !
                                     2
         X        xρ−1              H
                         x− 2 log ( 2π )                  log2 ( 2π
                                                                 T
                                                                    ) log2 ( 2π
                                                                             H
                                                                                )
                       ≤                 + xσ−1                      −            + 1.8642
                   ρ           2π                            2π          2π
       |ℑ(ρ)|≤T
        ℜ(ρ)≤σ
                              1
                                                                                                 !
                          x− 2 log2 ( 2π
                                      H
                                         )                                   2
                                                                             H
                                                          (B3 )2 log x log ( 2π )
                        ≤                  + xσ−1                     −           + 1.8642
                                2π                            2π          2π
                        = s′1 (x, σ), say.
Then,

                            Z t0                          Z T                     !
       X        xρ−1               x−ν2 (t)                     x−ν2 (t)
                     ≤2                     dN (σ, t) +                  dN (σ, t)
                  ρ           H      t                     t0     t
     |ℑ(ρ)|≤T
      ℜ(ρ)>σ

                                                                         x−ν2 (t0 )
                          −ν2 (t0 )              −ν2 (t0 )                                  
                          x                       x
                      ≤2             N (σ, T ) +             N (σ, T ) −            N (σ, t0 )
                             t0                     t0                     t0
                        2N (σ, T )
                      =            ,
                           T
and,
                                                        C2 (σ) log2 T
                                                                     
           2N (σ, T )
                      ≤ 2 C1 (σ)T (5−8σ)/3 log5−2σ T +                  ,
              T                                              T
                                                        
                                       B2 (5 − 8σ) p             p        5−2σ
                      ≤ 2 C1 (σ) exp                 log x   B2 log x
                                            3
                                                                              
                                                            p
                                        + C2 (σ) exp −B2 log x (B2 )2 log x

                      = s′2 (x, σ), say.
Finally,
       4.3128                  4.3128
              log0.6 x ≤                  log0.6 x = 4.3128 log0.6 x exp(−B2 log x)
                                                                            p
                                   √
         T               exp(B2 log x)
                       = s′3 (x, σ), say.
Then, for all x ≥ x0 ,
                                                                                     
           ψ(x) − x                5−2σ                    B2 (x0 )(5 − 8σ) p
                    ≤ A(x0 , σ) log 2 (x) exp                                log x               (3.15)
              x                                                    3
provided
                                        s1 (x, σ) + s2 (x, σ) + s3 (x, σ)
                      A(x, σ) :=         5−2σ
                                                                   √       
                                    log    2(x) exp B2 (x0 )(5−8σ)
                                                             3         log x

is decreasing for x ≥ x0 = exp(X). Optimising over σ, we used (3.15) to compute
the last 6 rows of Table 1.
        ESTIMATES FOR THE ERROR TERM IN THE PRIME NUMBER THEOREM                         11


                     4. Proofs of Corollaries 1.2 and 1.3
4.1. Bounds on |θ(x) − x|. For 2 ≤ x ≤ 599, the bound on |θ(x) − x| implied
by the first row in Table 1 can be verified by a straight-forward computation. For
599 < x ≤ exp(58), the same bound follows from Lemma 2.2. For x > exp(58) we
have [Bro+21, Corollary 5.1]
                                                  1        1
                              ψ(x) − θ(x) < a1 x 2 + a2 x 3 ,                         (4.1)
                              −8
where a1 = 1 + 1.93378 · 10        and a2 = 1.01718. Thus, noting that
                         |θ(x) − x| ≤ ψ(x) − θ(x) + |ψ(x) − x|
the bounds in Corollary 1.2 for x > exp(58) follow from (4.1) and the bounds for
|ψ(x) − x| in Theorem 1.1.
4.2. Bounds on |π(x) − li(x)|. For 2 ≤ x ≤ 2657, the desired bound (1.6) can
be verified by a straight-forward computation. For 2657 < x ≤ exp(58), we have
that (1.6) holds by Lemma 2.2. For x > exp(58), we follow a similar procedure to
Platt and Trudgian [PT21b, Section 4] and Dusart [Dus98, Section 1.7]. By partial
summation and integration by parts,
                                                           Z x
                                   θ(x) − x         2          |θ(t) − t|
                |π(x) − li(x)| ≤               +         +                dt.
                                      log x       log 2      2   t log2 t
We write                    Z x
                                 |θ(t) − t|
                                            dt = I1 + I2 + I3 ,
                             2    t log2 t
where
       Z 599                        Z exp(58)                         Z x
             |θ(t) − t|                       |θ(t) − t|                      |θ(t) − t|
  I1 =              2   dt,  I 2 =                   2   dt,   I3  =                2    dt.
         2    t log   t              599       t log   t               exp(58) t log t
By direct computation, we see that
                                        I1 ≤ 5.43.
Then, by Lemma 2.2, we have
                          Z exp(58)
                                       1
                     I2 ≤              √ dt ≤ 7.87 · 1012 .
                              599    8π t
Now, let A1 = 9.40, B = 1.515 and C = 0.8274 be the parameters corresponding
to X = log 2 in Corollary 1.2. To bound I3 we first define a function
                         h(t) = A1 t log−α t exp (−Cu(t)) ,
               √
where u(t) =     log t and α is a parameter to be optimised later. We have
                                                           
             A 1 exp(−Cu(t))       log t − α
    h′ (t) =                                   −   Ctu ′
                                                         (t)  ≥ A1 logB−2 t exp(−Cu(t))
                   logα t             log t
provided
                          log t − α − Ct log tu′ (t) ≥ logB+α−1 t.                    (4.2)
In particular, (4.2) is satisfied when α = 0.45 and t ≥ exp(58). Hence
                                               A1 logB t exp(−Cu(t))
           Z x                       Z x
                    |θ(t) − t|
                          2    dt ≤                                    dt
            exp(58) t log t            exp(58)             log2 t
                                     Z x
                                  ≤            h′ (t)dt
                                     exp(58)
12                     DANIEL R. JOHNSTON AND ANDREW YANG


                                  ≤ A1 x log−α x exp(−Cu(x))
                                  = log1−B−α x · A1 x logB−1 x exp(−Cu(x))
                                  ≤ log1−B−α x0 · A1 x logB−1 x exp(−Cu(x)),

where x0 = exp(58) since log1−B−α x is decreasing. Putting everything together,
we have for all x ≥ x0 = exp(58),
                                                            
                  |π(x) − li(x)| ≤ A2 x logB−1 x exp −C log x ,
                                                       p


where A2 is given by
                                                                                        !
                                                                log1−B x0
                                                           
                                 2                                            p       
A1   1 + log1−B−α x0 +               + 5.43 + 7.87 · 1012                 exp C log x0
                               log 2                              A1 x0
 ≤ 9.59,

with A1 = 9.40, B = 1.515, C = 0.8274 and α = 0.45. This completes the proof of
Corollary 1.3.
   One could also produce a table of bounds for |π(x) − li(x)| similar to those
in Theorem 1.1 and Corollary 1.2. However, since it is much more common in
applications to use bounds on |ψ(x) − x| or |θ(x) − x| rather than |π(x) − li(x)|, we
have refrained from performing such calculations here.


                                5. Proof of Theorem 1.4
5.1. Proof of (1.7). The following argument is essentially identical to that in
Section 3.3 except we use the Vinogradov–Korobov zero-free region in Lemma 2.10
rather than the zero-free region in Lemma 2.3. We thus argue tersely as to simply
highlight the outcome of using a different zero-free region.
   Now, it suffices to show (1.7) holds for 23 ≤ x ≤ exp(59) and x ≥ exp(2.8·1010 ) as
the bounds in Theorem 1.1 are sharper than (1.7) for exp(59) ≤ x ≤ exp(2.8 · 1010 ).
By direct computation, (1.7) holds for 23 ≤ x < 59. For 59 ≤ x < exp(59), the
result follows from Lemma 2.2 and [Bro+21, Table 8]. Hence, from here onwards
we assume x ≥ x0 = exp(2.8 · 1010 ).
   Let t0 , T , B2 and B3 be as defined in Lemmas A.3 and A.4 in Appendix A.
Moreover, let
                                                 log3/5 x
                                 r = r(x) =                  .
                                              (log log x)1/5
Similar to before,

                            1
                                                                                        !
        X       xρ−1              H2
                       x− 2 log ( 2π )               log2 ( 2π
                                                            T
                                                               ) log2 ( 2π
                                                                        H
                                                                           )
                     ≤                 + xσ−1                   −            + 1.8642
                 ρ           2π                         2π          2π
     |ℑ(ρ)|≤T
      ℜ(ρ)≤σ
                            1
                                                                                       !
                                                                      2
                        x− 2 log2 ( 2π
                                    H
                                       )             (B3 )2 r2         H
                                                                 log ( 2π )
                      ≤                  + xσ−1                −            + 1.8642
                              2π                       2π           2π
                      = s′′1 (x, σ), say.
           ESTIMATES FOR THE ERROR TERM IN THE PRIME NUMBER THEOREM                            13


Meanwhile,

                X        xρ−1   2N (σ, T )
                              ≤
                           ρ       T
              |ℑ(ρ)|≤T
               ℜ(ρ)>σ
                                                                  
                                                       B2 (5 − 8σ)          5−2σ
                                      ≤ 2 C1 (σ) exp               r (B2 r)
                                                            3
                                                                                           
                                                                                     2 2
                                                            + C2 (σ) exp (−B2 r) (B2 ) r

                                      = s′′2 (x, σ), say.
Finally,
                   4.3128             4.3128
                          log0.6 x ≤          log0.6 x = s′′3 (x, σ), say.
                     T                  B2 r
Therefore, for all x ≥ x0 = exp(2.8 · 1010 ),
                                                          
 ψ(x) − x                        5−2σ        B2 (5 − 8σ)
            ≤ A(x0 , σ) · (B2 r)      exp                r
    x                                             3
                                              !5−2σ                                       !
                                   log3/5 x                   B2 (5 − 8σ) log3/5 x
            = A(x0 , σ) · B2                          exp                                   .
                               (log log x)1/5                       3      (log log x)1/5
so long as
                                                                                    −1
                                                              B2 (5 − 8σ)
   A(x, σ) := [s′′1 (x, σ) + s′′2 (x, σ) + s′′3 (x, σ)] exp               r (B2 r)5−2σ
                                                                   3
is decreasing for x ≥ x0 . Letting σ = 0.9999932 we obtain (1.7). We also use
                                        1                      1
                                                   ≤
                               (log log x)(5−2σ)/5   (log log x0 )(5−2σ)/5
so that the final result is of the desired form. A factor of (log log x)−(5−2σ)/5 could be
included in the estimates (1.7)–(1.9) (with a different leading constant). However,
the authors refrained from doing so to prevent the bounds from being overly messy.
5.2. Proof of (1.8). Same as the proof of Corollary 1.5, using 1.7 in place of 1.3.
5.3. Proof of (1.9). We proceed similarly to the proof of Corollary 1.3. Namely,
we use a simple computation and Lemma 2.2 to prove (1.9) up to exp(58). For
x > exp(58) we bound I1 , I2 and I3 as defined in Section 4.2. As before, I1 ≤ 5.43
and I2 ≤ 7.87 · 1012 . To bound I3 we set u(t) = log3/5 t(log log t)−1/5 . Now,
                           3 log log t − 1
           tu′ (t) =           5/2
                                                       ≤ 1.63 · 10−5     for all t ≥ exp(58)
                       5 log         t(log log t)6/5
Thus, for α = 0.19, B = 1.801, C = 0.1853 and t ≥ exp(58)
   log t − Ct log tu′ (t) ≥ (1 − 4 · 10−6 ) log t ≥ log0.991 t + 0.19 = logB+α−1 t + α.
Repeating the argument in Section 4.2, we then have for x ≥ x0 = exp(58),
                                                                   !
                                       B−1             log3/5 x
             |π(x) − li(x)| ≤ A2 x log     x exp −C                  ,    (5.1)
                                                    (log log x)1/5
14                    DANIEL R. JOHNSTON AND ANDREW YANG


where
                                                                                            !
                                                                      (u(x0 ))C log1−B x0
                                                                 
                                       2
 A2 = A1     1 + log1−B−α x0 +             + 5.43 + 7.87 · 1012
                                     log 2                                    A1 x0
     ≤ 0.028,
when A1 = 0.027, B = 1.801, C = 0.1853 and α = 0.19. This gives (1.9) as desired.

                      6. Further possible improvements
   There are many estimates from recent literature that go into our results, several
of which are frequently updated as new techniques and computational power become
available. We bring particular attention to the zero-free regions in Lemmas 2.7–
2.10. For suppose one uses a classical zero-free region of the form
                                              1
                                 β ≥1−
                                          R log |t|
such as in Lemma 2.7. Then applying our method (or any variation on the work of
Pintz [Pin80]) one obtains an estimate
                                             p       
                      |ψ(x) − x| = O x exp(−C1 log x) ,
where C1 is any real number less than
                                         2
                                        √ .
                                           R
Similarly, using a Vinogradov–Korobov zero-free region
                                             1
                          β ≥1−        2/3
                                  c log |t|(log log |t|)1/3
such as in Lemma 2.10, one can prove the estimate
                                                                 
                |ψ(x) − x| = O x exp(−C2 log3/5 x(log log x)−1/5 ) ,
where C2 is any real number less than
                             1/5 " 2/5  3/5 #
                           5          3     2
                                          +         .
                          3c3         2     3
Thus, in either case, improving on current zero-free regions will have a large impact
on our results. In this direction, we note that the zero-free region due to Mossinghoff
and Trudgian [MT15] has yet to be updated with the most recent Riemann height
(Lemma 2.1). Moreover, Mossinghoff and Trudgian’s zero-free region is stated for
all |t| ≥ 2. It would thus be very useful if their result could be improved/generalised
for the larger values of t required in our application.
    The other zero-free regions we employ are due to Ford [For02], which have not
been updated (since 2002) with recent estimates involving ζ(s).
    We also remark that the main inefficiency in our method is bounding (see (3.5))
                        Z tk+1
                               dN (σ, t) = N (σ, tk+1 ) − N (σ, tk ).             (6.1)
                        tk
Namely, we use the trivial bound N (σ, tk+1 ) − N (σ, tk ) ≤ N (σ, tk+1 ). It would thus
be interesting to further the work in [KLN18] to produce zero-density estimates in
different intervals. That is, one could bound the number of zeros of ζ(s) in the box
       ESTIMATES FOR THE ERROR TERM IN THE PRIME NUMBER THEOREM                  15


σ < ℜ(s) < 1 and T0 < ℑ(s) < T for some choice of T0 (not necessarily 0). This
would lead to an improved upper bound for (6.1).

                            7. Acknowledgements
   Thanks to our supervisor Timothy Trudgian for his ongoing support and ad-
vice, and to Michaela Cully-Hugill for her assistance in computing the zero-density
estimates in Table 3. We also thank H. Kadiri and G. Hiary for their comments
regarding the initial version of this article, as discussed in Section 1.4.
16                     DANIEL R. JOHNSTON AND ANDREW YANG


                        Appendix A. Bounds on t0 and T
   In this appendix we prove results on the functions t0 = t0 (x) and T = T (x)
as required in Sections 3.3 and 5.1. In what follows H = 3 000 175 332 800 as in
Lemma 2.1.
Lemma A.1. Let ν2 (t) be as defined in Lemma 2.3, and t0 be the value of t
such that x−ν2 (t) /t is maximised for t ≥ H. Then, x−ν2 (t) /t is increasing for all
t ∈ [H, t0 ) and decreasing for all t > t0 . Moreover, for all x ≥ x0 ≥ exp(105 ),
                                p                       p
                         exp(B0 log x) ≤ t0 ≤ exp(B1 log x)                       (A.1)
where B1 = (3.359)−1/2 and values of B0 = B0 (x0 ) are given in Table 2.
Proof. Let D = 8.02 and R1 = 3.359. We have
        d x−ν2 (t)     x−ν2 (t) log x D(1 − 2 log log t) + log t
                                                                      
                                                                    R1
                     =                                           −         .
        dt     t           R1 t2               log3 t              log x
Thus, if t0 is such that
                           D(1 − 2 log log t0 ) + log t0    R1
                                                         =
                                     log3 t0               log x
then x−ν2 (t) /t is increasing for t ∈ [H, t0 ) decreasing for t > t0 as
                                                                       √ required.
   We now prove (A.1). So, let x ≥ x0 . Then for all t ≥ exp(B1 log x),
                                 log x log t   log x
       log3 t ≥ B12 log x log t =            ≥       (log t + D(1 − 2 log log t)) .
                                     R1         R1
      d
          −η(t) 
          x
                                         √                              √
Hence dt    t       ≤ 0 for t ≥ exp(B1 log x). That is, t0 ≤ exp(B1 log x).
                              √
  Now suppose t = exp(B0 log x) where B0 = B0 (x0 ) is as in Table 2. Then,
                                               √
                 2 log log t − 1      2 log(B0 log x0 ) − 1
              D                  ≤D            √             = Cx0 , say.
                      log t                 B0 log x0
As a result,
                    D(1 − 2 log log t) + log t   (1 − Cx0 )    R1
                                 3             ≥      2     ≥
                             log t                 log t      log x
provided
                                             1 − Cx0
                                     B02 ≤           ,                                  (A.2)
                                               R1
                                                               −ν(t) 
                                                 d              x
which is true for each B0 in Table 2. Therefore, dt                 t     ≥ 0 so that t0 ≥ t =
       √
exp(B0 log x) as required.                                                                  

Lemma A.2. Let ν2 (t) be as defined in Lemma 2.3 and
                                       1
                           T =              −ν2 (t)
                                                      = min txν2 (t) .
                                 maxt>H x     t
                                                        t>H


Then, for all x ≥ x0 ≥ exp(105 ),
                            p                   p
                      exp(B2 log x) ≤ T ≤ exp(B3 log x),
where B2 = B2 (x0 ) and B3 = B3 (x0 ) are given in Table 2.
        ESTIMATES FOR THE ERROR TERM IN THE PRIME NUMBER THEOREM                      17

                                                                         √
Proof. By Lemma A.1, the maximum of x−ν(t) /t occurs at t0 = exp(θ log x) for
some θ ∈ [B0 , B1 ], with B1 = (3.359)−1/2 and B0 = B0 (x0 ) as in Table 2. Thus,
        T = t0 xν2 (t0 )
                                                               √       
                    p                   log x          8.02 log(θ log x)
          = exp(θ log x) · exp            √        1−         √
                                    3.359θ log x            θ log x
                                                     √       
                    p                1        8.02 log(θ log x)
          = exp         log x θ +          1−        √               .
                                  3.359θ           θ log x
Let                                                     √         
                                    1         8.02 log(B0 log x0 )
                            α=             1−         √              .
                                  3.359            B0 log x0
Then,
                           p             α           √ p        
                 T ≥ exp       log x θ +          ≥ exp 2 α · log x
                                           θ
                  α                                  √                             √
since f (θ) = θ + θ attains a minimum at θ = α. Therefore, setting B2 := 2 α,
                      √
we have T ≥ exp(B2 log x). Meanwhile,
                                                               
                                  p                      1
                      T ≤ exp       log x B1 +
                                                    3.359B0(x0 )
                                      1
                                                                 √
so that setting B3 (x0 ) := B1 + 3.359B0 (x0 ) gives T ≤ exp(B3 log x) as required. 


         Table 2. Values of X, B0 , B2 , B3 rounded to 7 decimal places,
                    which appear in Lemma A.1 and A.2.

                       log x0         B0            B2            B3
                             5
                        10         0.3253505 0.8721857 1.4606625
                        106        0.4923764 1.0346912 1.1502603
                        107        0.5271511 1.0716004 1.1103741
                        108        0.5390163 1.0842539 1.0979426
                        109        0.5432643 1.0887652 1.0936237
                             10
                        10         0.5447895 1.0903755 1.0920896

  We now prove analogous results for the zero-free region in Lemma 2.10.
Lemma A.3. Let x ≥ x0 = exp(2.8 · 1010 ), ν3 (t) be as defined in Lemma 2.3, and
t0 be the value of t such that x−ν3 (t) /t is maximised for t ≥ H. Then, x−ν3 (t) /t is
increasing for all t ∈ [H, t0 ) and decreasing for all t > t0 . Moreover,
                               log3/5 x                       log3/5 x
                       B0                  ≤ log t 0 ≤ B 1                ,        (A.3)
                            (log log x)1/5                 (log log x)1/5
where
                           3/5  1/5
                       2          5
            B0 =                        = 0.07633 . . .     and    B1 = 0.08228.   (A.4)
                       3c         3
18                     DANIEL R. JOHNSTON AND ANDREW YANG


Proof. Let c = 57.54 as in Lemma 2.10. We have
                                                                         !
           d x−ν3 (t)      x−ν3 (t) log x
                     
                                             2 log log t + 1         3c
                         =                                       −         .
          dt      t           3ct2        log5/3 t(log log t)4/3   log x
Thus, if t0 is such that
                                    2 log log t0 + 1             3c
                                                           =
                              log5/3 t0 (log log t0 )4/3       log x
then x−ν3 (t) /t is increasing for t ∈ [H, t0 ) and decreasing for t ≥ t0 as required.
   We now prove (A.3). So firstly, if log t = B0 log3/5 x(log log x)−1/5 , then
                           log5/3 t(log log x)1/3 (1 + 2 log log t)
log x(1 + 2 log log t) =                       5/3
                                             B0
                                                                      3            1/3
                                                                           log log x
                                                    
                                                              1
                       = 3c log5/3 t(log log t)4/3 · 1 +               · 5
                                                         2 log log t      log log t
                                                    3           1/3
                                                       log log x
                       ≥ 3c log5/3 t(log log t)4/3 · 5                ,
                                                      log log t
and,
                                 3            1              3
           log log t = log B0 + log log x − log log log x ≤ log log x.
               −ν (t)          5            5              5
           d    x 3                        3/5          −1/5
Therefore, dt     t      ≥ 0 so that B0 log x(log log x)     ≤ log t0 as desired.
  On the other hand, we set
                                     3/5  1/5
                                ′     2         1
                              B1 =                  ,
                                      3c        β
where β is a constant to be chosen later. For log t ≥ B1′ log3/5 x(log log x)−1/5 ,
                                                                              1/3
                               5/3          4/3           1          β log log x
log x(1 + 2 log log t) ≤ 3c log t(log log t) · 1 +                ·                   .
                                                     2 log log t      log log t
Moreover,
                                 1
                           1+           ≤ 1.04425 = γ, say.
                            2 log log t
Setting β = 0.4125 (so that B1′ = 0.08227 . . .), we have for all x ≥ x0
                                3             1
            log log t ≥ log B1′ +  log log x − log log log x ≥ γ 3 β log log x.
                                5             5
                                                                      −ν (t) 
                                      5/3                          d   x 3
Thus, log x(1 + 2 log log t) < 3c log t(log log t)4/3 . That is, dt       t     ≤ 0 and

                                      log3/5 x             log3/5 x
                        t0 ≤ B1′                  ≤ B 1
                                   (log log x)1/5       (log log x)1/5
as required.                                                                             
Lemma A.4. Let ν3 (t) be as defined in Lemma 2.3 and
                                 1                        
                T :=              −ν (t)  = min xν3 (t) t .
                        maxt≥H x t3           t≥H
         ESTIMATES FOR THE ERROR TERM IN THE PRIME NUMBER THEOREM                                  19


Then for all x ≥ x0 = exp(2.8 · 1010 ), we have
                        log3/5 x                     log3/5 x
                        B2          ≤ log T ≤ B 3                ,                              (A.5)
                     (log log x)1/5               (log log x)1/5
where B2 = 0.18525 and B3 = 0.20680.
Proof. By Lemma A.3, the maximum of x−ν3 (t) /t occurs at the point
                                                   !
                                       log3/5 x
                         t0 = exp θ
                                    (log log x)1/5
for some θ satisfying B0 < θ < B1 , where B0 and B1 are as in (A.4). Now,
                                          log x                                 log3/5 x
log xν3 (t0 ) t0 =                       2/3
                                                                           + θ
                               3/5
                                                              3/5
                                                                              (log log x)1/5
                  c θ (logloglog x)x1/5       log1/3 θ (logloglog x)x1/5

                                     log3/5 x(log log x)2/15                       log3/5 x
                  ≥                                                   1/3
                                                                          + B 0
                        2/3
                            log B1 + 53 log log x − 15 log log log x            (log log x)1/5
                                                                    
                      cB1
                      log3/5 x(log log x)2/15          log3/5 x
                  ≥                           + B 0
                         2/3 3           1/3       (log log x)1/5
                      cB1     5 log log x
                            log3/5 x
                  ≥ B2
                         (log log x)1/5
and
                                   log3/5 x(log log x)2/15                      log3/5 x
log xν3 (t0 ) t0 ≤                                                    1/3
                                                                          + B1
                        2/3
                      cB0   log B0 + 53 log log x − 15 log log log x
                                                                              (log log x)1/5

                        log3/5 x(log log x)2/15          log3/5 x
                  ≤                             + B 1
                        2/3
                      cB0 (0.4666 log log x)
                                            1/3       (log log x)1/5
                            log3/5 x
                  ≤ B3
                         (log log x)1/5
as required.                                                                                       
20                    DANIEL R. JOHNSTON AND ANDREW YANG


                      Appendix B. Zero-density estimates

         Table 3. Some values of C1 (σ) and C2 (σ) for Lemma 2.6. In
        terms of the notation in [KLN18], we set H0 = 3 000 175 332 800,
          k = 1, µ = 1.23623 and optimise over parameters d, α and δ.

                  σ        d      α        δ     C1 (σ)   C2 (σ)
                0.980 0.3333 0.0633 0.3101 16.281         2.231
                0.981 0.3333 0.0628 0.3101 16.337         2.223
                0.982 0.3332 0.0624 0.3101 16.394         2.215
                0.983 0.3332 0.0619 0.3102 16.450         2.207
                0.984 0.3331 0.0614 0.3102 16.507         2.199
                0.985 0.3331 0.0610 0.3102 16.564         2.191
                0.986 0.3331 0.0605 0.3102 16.621         2.182
                0.987 0.3330 0.0600 0.3102 16.678         2.175
                0.988 0.3330 0.0595 0.3103 16.734         2.166
                0.989 0.3329 0.0591 0.3103 16.791         2.159
                0.990 0.3329 0.0586 0.3103 16.848         2.150
                0.991 0.3329 0.0582 0.3103 16.905         2.142
                0.992 0.3329 0.0577 0.3103 16.962         2.134
                0.993 0.3328 0.0572 0.3103 17.019         2.126
                0.994 0.3328 0.0568 0.3103 17.077         2.118
                0.995 0.3327 0.0563 0.3104 17.134         2.110
                0.996 0.3327 0.0559 0.3104 17.191         2.102
                0.997 0.3326 0.0554 0.3104 17.248         2.094
                0.998 0.3326 0.0550 0.3104 17.305         2.086
                0.999 0.3326 0.0545 0.3104 17.362         2.077
                  1      0.3325 0.0539 0.3105 17.418      2.069


Remark. More precisely, the entry for σ = 1 gives values for C1 (σ) and C2 (σ) in
the limit σ → 1.
                                 REFERENCES                                 21


                                References
[Apo13]  T. M. Apostol. Introduction to Analytic Number Theory. New York:
         Springer Science & Business Media, 2013.
[Axl18]  C. Axler. “New estimates for some functions defined over primes”. In-
         tegers 18 (2018).
[BPT21] R. P. Brent, D. J. Platt, and T. S. Trudgian. “The mean square of the
         error term in the prime number theorem”. J. Number Theory (2021).
[Bro+21] S. Broadbent, H. Kadiri, A. Lumley, N. Ng, and K. Wilk. “Sharper
         bounds for the Chebyshev function θ(x)”. Math. Comp. 90.331 (2021),
         pp. 2281–2315.
[Büt16] J. Büthe. “Estimating π(x) and related functions under partial RH as-
         sumptions”. Math. Comp. 85.301 (2016), pp. 2483–2498.
[Büt18] J. Büthe. “An analytic method for bounding ψ(x)”. Math. Comp. 87.312
         (2018), pp. 1991–2009.
[CG04]   Y. F. Cheng and S. W. Graham. “Explicit estimates for the Riemann
         zeta function”. The Rocky Mountain Journal of Mathematics (2004),
         pp. 1261–1280.
[CHJ21] M. Cully-Hugill and D. R. Johnston. “On the error term in the explicit
         formula of Riemann–von Mangoldt”. preprint available at arXiv:2111.10001
         (2021). Submitted.
[CHT21] M. Cully-Hugill and T. Trudgian. “Two explicit divisor sums”. Ramanu-
         jan J. 56 (2021), pp. 141–149.
[DP15]   A. W. Dudek and D. J. Platt. “On solving a curious inequality of Ra-
         manujan”. Exp. Math. 24.3 (2015), pp. 289–294.
[Dus98]  P. Dusart. “Autour de la fonction qui compte le nombre de nombres
         premiers”. PhD Thesis. Université de Limoges, 1998.
[Dus18]  P. Dusart. “Explicit estimates of some functions over primes”. Ramanu-
         jan J. 45.1 (2018), pp. 227–251.
[FK15]   L. Faber and H. Kadiri. “New bounds for ψ(x)”. Math. Comp. 84.293
         (2015), pp. 1339–1357.
[FKS22] A. Fiori, H. Kadiri, and J. Swidinsky. “Density results for the zeros of
         zeta applied to the error term in the prime number theorem”. preprint
         available at arXiv:2204.02588 (2022).
[For02]  K. Ford. “Zero-free regions for the Riemann zeta function”. Number
         theory for the millennium, II (Urbana, IL, 2000). A K Peters, Natick,
         MA, 2002, pp. 25–56.
[Hia16]  G. A. Hiary. “An explicit van der Corput estimate for ζ (1/2+ it)”.
         Indag. Math. 27.2 (2016), pp. 524–533.
[Joh21]  D. R. Johnston. “Improving bounds on prime counting functions by
         partial verification of the Riemann hypothesis”. preprint available at
         arXiv:2109.02249 (2021). Submitted.
[KLN18] H. Kadiri, A. Lumley, and N. Ng. “Explicit zero density for the Riemann
         zeta function”. J. Math. Anal. Appl. 465.1 (2018), pp. 22–46.
[MT15]   M. J. Mossinghoff and T. S. Trudgian. “Nonnegative trigonometric poly-
         nomials and a zero-free region for the Riemann zeta-function”. J. Num-
         ber Theory 157 (2015), pp. 329–349.
[Pat21]  D. Patel. “Explicit sub-Weyl bound for the Riemann zeta function”.
         PhD thesis. The Ohio State University, 2021.
22                                  REFERENCES


[Pin80]      J. Pintz. “On the remainder term of the prime number formula. II. On
             a theorem of Ingham”. Acta Arith. 37 (1980), pp. 209–220.
[PT21a]      D. Platt and T. Trudgian. “The Riemann hypothesis is true up to 3 ·
             1012 ”. Bull. Lond. Math. Soc. 53.3 (2021), pp. 792–797.
[PT21b]      D. J. Platt and T. S. Trudgian. “The error term in the prime number
             theorem”. Math. Comp. 90.328 (2021), pp. 871–881.
[RS62]       J. B. Rosser and L. Schoenfeld. “Approximate formulas for some func-
             tions of prime numbers”. Illinois J. Math. 6.1 (1962), pp. 64–94.
[STD15]      Y. Saouter, T. Trudgian, and P. Demichel. “A still sharper region where
             π(x) − li(x) is positive”. Math. Comp. 84.295 (2015), pp. 2433–2446.
[Tru16]      T. Trudgian. “Updating the error term in the prime number theorem”.
             Ramanujan J. 39.2 (2016), pp. 225–234.
     School of Science, The University of New South Wales, Canberra, Australia
     Email address: daniel.johnston@adfa.edu.au

     School of Science, The University of New South Wales, Canberra, Australia
     Email address: andrew.yang1@adfa.edu.au
