#A34                              INTEGERS 24 (2024)
   EFFECTIVE ESTIMATES FOR SOME FUNCTIONS DEFINED
                    OVER PRIMES


                              Christian Axler
   Heinrich Heine University Düsseldorf, Faculty of Mathematics and Natural
             Sciences, Mathematical Institut, Düsseldorf, Germany
                          christian.axler@hhu.de


   Received: 5/12/22, Revised: 7/21/23, Accepted: 3/15/24, Published: 4/8/24



                                    Abstract
In this paper we give effective estimates for some classical arithmetic functions
defined over prime numbers. First we find the smallest real number x0 so that
some inequality involving Chebyshev’s ϑ-function holds for every x ≥ x0 . Then we
give some new results concerning the existence of prime numbers in short intervals.
Also we derive new upper and lower bounds for some functions defined over prime
numbers, for instance the prime counting function π(x), which improve current best
estimates of similar shape.



1. Introduction
                                                   P
First, we consider Chebyshev’s ϑ-function ϑ(x) = p≤x log p, where p runs over all
primes not exceeding x. Since there are infinitely many primes, we have ϑ(x) → ∞
as x → ∞. Hadamard [37] and de la Vallée-Poussin [23] independently proved a
result concerning the asymptotic behavior for ϑ(x), namely
                              ϑ(x) ∼ x        (x → ∞),                           (1)
which is known as the Prime Number Theorem. In a later paper [24], where the
existence of a zero-free region for the Riemann zeta function to the left of the line
Re(s) = 1 was proved, de la Vallée-Poussin also estimated the error term in the
Prime Number Theorem by showing that
                                          √
                     ϑ(x) = x + O(xe−c0       log x
                                                      )   (x → ∞),               (2)
where c0 is a positive absolute constant. The current best explicit version of this
result is due to Fiori, Kadiri, and Swidinsky [34, Corollary 14]. They found that
                                              3/2        r        !
                                         log x                log x
                 |ϑ(x) − x| ≤ 121.0961              exp −2                      (3)
                                           R                    R
   DOI: 10.5281/zenodo.10943995
INTEGERS: 24 (2024)                                                                  2

for every x ≥ 2, where R = 5.5666305. The work of Korobov [42] and Vinogradov
[71] implies the current asymptotically strongest error term in (1), namely
                                                        
         ϑ(x) = x + O x exp −c1 log3/5 x(log log x)−1/5          (x → ∞),   (4)

where c1 is a positive absolute constant. An explicit version of (4) was recently
given by Johnston and Yang [40, Theorem 1.4]. Now, (2)–(4) each imply that for
every positive integer k and every positive real number ηk there is real number
x1 = x1 (k, ηk ) > 1 so that for every x ≥ x1 , we have
                                                    ηk x
                                 |ϑ(x) − x| <             .                        (5)
                                                   logk x

In the case where k = 3 and η3 = 0.024334, Broadbent et al. [12] found that
                                      0.024334x
                       |ϑ(x) − x| <                       (x ≥ e29 ).              (6)
                                        log3 x

In our first result, we compute the smallest positive integer N so that (6) holds for
every x ≥ N .

Proposition 1. The inequality (6) holds for every x ≥ 1, 757, 126, 630, 797 =
p64,707,865,143 .

   Estimates for ϑ(x) of the form (5) can be used to specify short intervals containing
at least one prime number. Here, we find the following result.

Theorem 1. Let a = (1.42969 × 1012 − 1)−1 and b = (1.59753 × 1012 − 1)−1 .
Further, let n be a positive integer with 1 ≤ n ≤ 5. Then there is a prime number
p such that                                          
                                                 an
                               x<p≤x 1+
                                               logn x
for every x ≥ Xn , where an and Xn are given as in Table 1.

                n               an                             Xn
                                             −11
                1    43a = 3.00 . . . × 10          952, 527, 672, 606, 693
                        n                    −9
                2    46 b = 1.32 . . . × 10         684, 943, 746, 324, 434
                        n                    −8
                3    46 b = 6.09 . . . × 10         543, 684, 371, 469, 081
                       n                      −6
                4   46 b = 2.802 . . . × 10         336, 149, 866, 771, 577
                5   46n b = 1.289 . . . × 10−4      246, 782, 656, 239, 427

                      Table 1: Explicit values for an and Xn .
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Remark 1. Note that the values of Xn , where 1 ≤ n ≤ 5, are the smallest positive
integers so that there is always a prime number in the interval (x, x(1 + an / logn x)].
                                         (1)      (2)
For n ≥ 6, we are only able to find Xn and Xn , so that analogous results are
                            (1)        (2)                                           (2)
valid for all x ∈ R with Xn ≤ x ≤ Xn . To prove these results for every x > Xn
as well, we would need estimates of the form ϑ(x) > x − ηn x/ logn x with n ≥ 6.
  Let π(x) denote the number of primes not exceeding x. Chebyshev’s ϑ-function
and the prime counting function π(x) are connected by the identity
                                         Z x
                                 ϑ(x)          ϑ(t)
                         π(x) =        +           2 dt,                   (7)
                                 log x    2  t log  t
which holds for every x ≥ 2 (see [2, Theorem 4.3]). If we combine (3) and (7), we
see that                                  √
                   π(x) = li(x) + O(xe−c2 log x )    (x → ∞),                  (8)
where c2 is a positive absolute constant. Here, the integral logarithm li(x) is defined
for every x ≥ 0 as
                         Z x              Z 1−ε          Z x        
                              dt                   dt            dt
                li(x) =           = lim                 +
                          0 log t   ε→0+    0     log t    1+ε log t

and plays an important role in this paper. The current best explicit version of (8)
is due to Johnston and Yang [40, Corollary 1.3]. Again, the work of Korobov [42]
and Vinogradov [71] implies the current asymptotically strongest error term for the
difference π(x) − li(x), namely
                                                           
       π(x) = li(x) + O x exp −c3 (log x)3/5 (log log x)−1/5      (x → ∞),      (9)

where c3 is a positive absolute constant. Ford [36, p. 2] has found that the constant
c3 in (9) can be chosen to be equal to 0.2098. Johnston and Yang [40, Theorem
1.4] used explicit zero-free regions and zero-density estimates for the Riemann zeta-
function to show that the inequality
                                                                             
    |π(x) − li(x)| ≤ 0.028x(log x)0.801 exp −0.1853(log x)3/5 (log log x)−1/5    (10)

holds for every x ≥ 71. Panaitopol [53, p. 55] gave another completely different
asymptotic formula for the prime counting function by showing that for every pos-
itive integer m, one has
                                                                     
                              x                                  x
   π(x) =                                              +O                    (x → ∞),
                        k1
           log x − 1 − log       k2               km
                           x − log2 x − . . . − logm x       logm+2 x
                                                                                   (11)
where the positive integers k1 , . . . , km are defined by the recurrence formula

                km + 1!km−1 + 2!km−2 + . . . + (m − 1)!k1 = m · m!.
INTEGERS: 24 (2024)                                                                             4

For instance, we have k1 = 1, k2 = 3, k3 = 13, k4 = 71, k5 = 461, and k6 =
3441. The computation of the prime counting function π(x) for large values of
x is a difficult problem (the latest record is due to Baugh and Walisch and was
π(1028 ) = 157, 589, 269, 275, 973, 410, 412, 739, 598). Also the asymptotic formula
(8) (or (11)) is not very meaningful with regard to the computation of π(x) for
some fixed x. Hence we are interested in finding new effective estimates for the
prime counting function π(x) which correspond to the first terms of (11). For
instance, those estimates for the prime counting function are used to get effective
estimates for 1/π(x) (see [10]) or the nth prime number (see [7]). In this paper, we
use Proposition 1 to establish the following upper bound for π(x) which corresponds
to the first terms of the asymptotic formula (11).
Theorem 2. Let a5 = 461.364417856444 and a6 = 4331.1. Then for every x ≥ 48,
we have
                                            x
  π(x) <                                                                             . (12)
         log x − 1 − log1 x − 3.024334
                               log2 x
                                       − 12.975666
                                           log3 x
                                                   − 71.048668
                                                       log4 x
                                                               − loga55 x − loga66 x

    For all sufficiently large values of x, Theorem 2 is a consequence of (10). On the
other hand, we get the following lower bound for the π(x) which corresponds to the
first terms of (11).
Theorem 3. Let b5 = 460.634397856444 and b6 = 3444.031844143556. Then for
every x ≥ 1, 751, 189, 194, 177 = p64,497,259,289 , we have
                                                x
   π(x) >                                                                                 .   (13)
            log x − 1 − log1 x − 2.975666
                                  log2 x
                                          − 13.024334
                                              log3 x
                                                      − 70.951332
                                                          log4 x
                                                                  − logb55 x − logb66 x

  Again, for all sufficiently large values of x, Theorem 3 follows directly from (10).
The asymptotic expansion (11) implies that the slightly sharper inequality
                                                x
                           π(x) >                                                 (14)
                                    log x − 1 − log1 x − log32 x

holds for all sufficiently large values of x. In [6, Theorem 1], the present author was
able to prove that the inequality (14) holds for every x such that 65 405 887 ≤ x ≤
2.7358·1040 and every x ≥ 4.8447·1019377 . Under the assumption that the Riemann
hypothesis is true, the present author[6, Proposition 2] showed that the inequality
(14) holds for every x ≥ 65, 405, 887. In the following theorem we finally see that
the inequality (14) holds for every x ≥ 65, 405, 887 even without the assumption
that the Riemann hypothesis is true.
Theorem 4. The inequality (14) holds unconditionally for every x ≥ 65, 405, 887.
   Our next goal is to establish new explicit estimates for the functions
                               X1            X log p
                                      and            ,
                                  p                p
                                  p≤x               p≤x
INTEGERS: 24 (2024)                                                                                         5

where p runs over primes not exceeding x, respectively. Euler [32] proved that the
sum of the reciprocals of all prime numbers diverges. Mertens [49, p. 52] found that
log log x is the right order of magnitude for this sum by showing
                          X1                               
                                                        1
                                = log log x + B + O           .                 (15)
                              p                       log x
                            p≤x

Here B denotes the Mertens’ constant and is defined by
                        X            1
                                         
                                             1
                                               
                B=γ+          log 1 −      +      = 0.26149 . . . ,                                      (16)
                          p
                                       p     p

where γ = 0.577215 . . . denotes the Euler-Mascheroni constant. In Section 6, we
apply Proposition 1 to some identity obtained by Rosser and Schoenfeld [63] and
derive the following result which improves all other results of this form.
Theorem 5. For every x ≥ 1, 757, 126, 630, 797, we have

                  X1                                                               
                                                    0.024334                 15
                            − log log x − B ≤                        1+                   .              (17)
                  p≤x
                        p                            3 log3 x             4 log x

  In 1874, Mertens [49] showed that
                                     X log p
                                                = log x + O(1).                                          (18)
                                           p
                                     p≤x

Landau [45, §55] improved (18) by finding
                    X log p                                p
                                     = log x + E + O(exp(− 14 log x)),
                                p
                    p≤x

where E is a constant defined by
                                           X     log p
                        E = −γ −                        = −1.3325 . . . .                                (19)
                                           p
                                               p(p − 1)
                                                                                              P
Similar to Theorem 5, we establish the following explicit estimates for                           p≤x log(p)/p
which improve [5, Proposition 8].
Theorem 6. For every x ≥ 1, 757, 126, 630, 797, we have

                  X log p                                                        
                                                     0.024334               2
                                    − log x − E ≤                    1+               .
                  p≤x
                            p                         2 log2 x            log x

Remark 2. Note that the positive integer N0 = 1, 757, 126, 630, 797 in Theorem 6
might not be the smallest positive integer N so that the inequality given in Theorem
6 holds for every x ≥ N .
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2. Proof of Proposition 1

In the following proof of Proposition 1, we first utilize an identity investigated
by Rosser and Schoenfeld [63] to express Chebyshev’s ϑ-function in terms of the
difference π(x) − li(x). Then we apply Walisch’s primecount C++ code [74] to find
a lower bound for π(x) − li(x) in a certain restricted interval.

Proof of Proposition 1. By (6) and [12, Corollary 11.1], it suffices to check that the
inequality
                                          0.024334x
                              ϑ(x) > x −                                          (20)
                                            log3 x
holds for every x satisfying 1, 757, 126, 630, 797 ≤ x ≤ e29 . Using [63, (2.26)] with
f (x) = log x, we get
                                                             Z x
                                                                 π(t) − li(t)
         ϑ(x) = x − 2 + li(2) log 2 + (π(x) − li(x)) log x −                  dt  (21)
                                                              2       t

for every x ≥ 2. Now we can use [54, Corollary 1] to see that
                      Z x                                          Z 9
                            π(t) − li(t)                                 π(t) − li(t)
− 2 + li(2) log 2 −                      dt ≥ −2 + li(2) log 2 −                      dt ≥ 0.129
                       2         t                                  2         t
                                                                                            (22)
for every x with 9 ≤ x ≤ e29 . Applying (23) to (21), we get

                               ϑ(x) > x + (π(x) − li(x)) log x                             (23)

for every x so that 9 ≤ x ≤ e29 . Now we use Walisch’s primecount C++ code [74]
to get
                                            0.024334x
                           π(x) − li(x) ≥ −                               (24)
                                              log4 x
for every x with 1, 760, 505, 892, 241 ≤ x ≤ 2, 342, 911, 050, 819 and every x with
2, 346, 094, 807, 193 ≤ x ≤ 4 × 1012 . If we combine (24) with (23), we get (20)
for every x satisfying 1, 760, 505, 892, 241 ≤ x ≤ 2, 342, 911, 050, 819 and every
x with 2, 346, 094, 807, 193 ≤ x ≤ e29 ≤ 4 × 1012 . In order to verify the re-
quired inequality (20) in the case where x satisfies 1, 757, 126, 630, 797 ≤ x <
1, 760, 505, 892, 241, we can check with a computer that ϑ(pn ) > g(pn+1 ) for every
integer n such that π(1, 757, 126, 630, 797) ≤ n ≤ π(1, 760, 505, 892, 241). Finally,
a direct computer check shows that the inequality (20) also holds for every x such
that 2, 342, 911, 050, 819 ≤ x ≤ 2, 346, 094, 807, 193.

  The present author [5, Theorem 1, Proposition 1, and Equations (4.4) and (4.5)]
INTEGERS: 24 (2024)                                                                7

utilized [30, Table 1 and Corollary 4.5] to show that
                                  0.043x
                     |ϑ(x) − x| <                (x ≥ e40 ),                    (25)
                                   log3 x
                                  0.15x
                     |ϑ(x) − x| <               (e35 ≤ x < e5000 ),             (26)
                                  log3 x
                                  99.07x
                     |ϑ(x) − x| <                (x ≥ e25 ),                    (27)
                                   log4 x
                                   100x
                     |ϑ(x) − x| <               (x ≥ 70, 111).                  (28)
                                  log4 x

Broadbent et al. [12, p. 2299] pointed out that the main theorem of [59] is incor-
rect and thus bounds claimed in [30] are likely affected, in particular [30, Table 1]
for bounds for ψ(x), and consequently the inequalities (25)–(28). Except for the
corresponding line for the value b = 2500, all other explicit values in [30, Table
1] were confirmed and even improved by Broadbent et al. [12, Table 8] while the
corresponding line for the value b = 2500 was recently confirmed and even improved
by Fiori, Kadiri, and Swidinsky [33, Table 5]. Hence, we can recover [30, Table 1].

Proposition 2. The explicit values for ε given in [30, Table 1] are correct.

  To show that the inequalities (25)–(28) still hold, it suffices to note that [30,
Proposition 4.4] combined with [12, Proposition 4] yield the correctness of [30,
Corollary 4.5]. Hence, we get

Proposition 3. The inequalities (25)–(28) for Chebyshev’s ϑ-function are correct.

Remark 3. Note that Proposition 1 already provides the correctness of the in-
equalities (25) and (26).

Remark 4. To find other explicit estimates for ϑ(x) in the restricted interval
[2, 1020 ], one can also apply the method used by Dusart in [31]. Let
                                               (
                         π(x − ε) + π(x + ε)    π(x) − 1/2, if x is prime,
            π0 (x) = lim                     =
                     ε→0          2             π(x),        otherwise.

Riemann [60] published the formula
                                        ∞
                                        X   µ(n)
                             π0 (x) =            f (x1/n ),                     (29)
                                        n=1
                                             n

where µ(n) is the Möbius function, and f (x) is the Riemann prime counting function
                                            Z ∞
                               X                        dt
               f (x) = li(x) −    li(xρ ) +          2 − 1) log t
                                                                  − log 2.
                                ρ            x    t(t
INTEGERS: 24 (2024)                                                                   8

Here the sum means limT →∞ |ρ|≤T li(xρ ), and the ρ’s are the nontrivial zeros of
                            P

the Riemann zeta function. A first proof of (29) was given by von Mangoldt [73] in
1895. Now let
                           ∞                             ∞
                           X   µ(n)                 X    logk x
                  R(x) =            li(x1/n ) = 1 +               .                (30)
                           n=1
                                n                     k!kζ(k + 1)
                                                        k=1

The latter series for it is known as Gram series. Since log x < x for every real x > 0,
this series converges for all positive x by comparison with the series for ex . In [61],
Riesel and Göhl showed that the function
                                            1   1         π
                        g(x) = R(x) −          + arctan
                                          log x π       log x
is a quite good approximation to π0 (x). The difference between g(x) and π0 (x)
                                                   √
heuristically oscillates with an amplitude of about x/ log x. So we define
                                                             
                                          1     1         π     log x
              ∆(x) = π0 (x) − R(x) +          − arctan           √ ,       (31)
                                        log x π         log x      x
the function which represents the fluctuations of the distribution of primes. We can
use (30) and (31) to get
                                      √
                       1                x              1      1           π
        π(x) − li(x) ≤ + f2 (x) +          × ∆(x) −        + arctan           , (32)
                       2             log x           log x π            log x
where
                                          ∞
                                          X µ(n)
                               fk (x) =              li(x1/n ).
                                                n
                                          n=k

Since µ(4) = 0 and f5 (x) is strictly decreasing on (1, ∞), the inequality (32) implies
that                                   √                  √
                                    li( x) li(x1/3 )        x
                π(x) − li(x) ≤ −           −          +        × ∆(x)             (33)
                                       2         3       log x
for every x ≥ 2, 000. Similarly, we see that
                                   5                          √
                                   X µ(n)                        x
                  π(x) − li(x) ≥                li(x1/n ) +         × ∆(x)         (34)
                                   n=2
                                           n                  log x

for every x ≥ 10, 326. Applying (33) and (34) to (21), we get
                                                        √
                           √               √    √    li( 5 x) log x
   ϑ(x) > x + (∆(x) − 1) x − max ∆(t) × li( x) − x −
                                                3
                                                                    + c1
                            2000≤t≤x                        5
for every x ≥ 10, 326, where c1 is a constant. Analogously, we see that the inequality
                                                                 √
                       √                         √     √      li( 5 x) log x   √
ϑ(x) < x + (∆(x) − 1) x − min ∆(t) × li( x) − 3 x +                          − 5 x + c2
                             10,236≤t≤x                              5
INTEGERS: 24 (2024)                                                                 9

holds for every x ≥ 10, 326, where c2 is a constant. Now one can use the exten-
sive table of the minimum and maximum values of ∆(x) in [44] to obtain explicit
estimates for ϑ(x) in the restricted interval [2, 1020 ].
Remark 5. Under the assumption that the Riemann hypothesis is true, von Koch
                                                     √
[72] deduced the asymptotic formula ϑ(x) = x + O( x log2 x). An explicit version
was given by Schoenfeld [66, Theorem 10]. Under the assumption that the Riemann
hypothesis is true, Schoenfeld has found that
                                            √
                                              x
                               |ϑ(x) − x| <     log2 x                         (35)
                                            8π
for every x ≥ 599. Recently, Schoenfeld’s result was slightly improved by Dusart
[31, Proposition 2.5]. In 2016, Büthe [14, Theorem 2] investigated a method to
show that the inequality (35) holds unconditionally for every x such that 599 ≤ x ≤
1.4 × 1025 . Büthe’s result was improved by Platt and Trudgian [55, Corollary 1].
They proved that the inequality (35) holds unconditionally for every x satisfying
599 ≤ x ≤ 2.169 × 1025 . Recently, Johnston [39, Corollary 3.3] extended the last
result by showing that the inequality (35) holds unconditionally for every x with
599 ≤ x ≤ 1.101 × 1026 .



3. Proof of Theorem 1

Bertrand’s postulate states that for each positive integer n there is a prime number p
with n < p ≤ 2n. It was proved, for instance, by Chebyshev [18]. In the following,
we note some improvements of Bertrand’s postulate. The first result is due to
Schoenfeld [66, Theorem 12]. He discovered that for every x ≥ 2, 010, 759.9 there
is a prime number p with x < p < x(1 + 1/16, 597). Ramaré and Saouter [58,
Theorem 3] proved that for every x ≥ 10, 726, 905, 041 there is a prime number p
so that x < p ≤ x(1 + 1/28, 313, 999). Further, they gave a table of sharper results
which hold for large x, see [58, Table 1]. Kadiri and Lumley [41, Table 2] obtained
a series of improvements. For instance, they showed that for every x ≥ 4 × 1018
there is a prime number p such that x < p < x(1 + 1/36, 082, 898). Recently, Cully-
Hugill and Lee [22, Theorem 1] improved the results of Kadiri and Lumley. In
particular, they found that for every x ≥ 4 × 1018 there is a prime number p so that
x < p ≤ x(1+1/1, 429, 689, 999, 999). Dusart [28, Théorème 1] proved that for every
x ≥ 3, 275 there exists a prime number p such that x < p ≤ x(1 + 1/(2 log2 x)) and
then reduced the interval himself [29, Proposition 6.8] by showing that for every
x ≥ 396, 738 there is a prime number p satisfying x < p ≤ x(1 + 1/(25 log2 x)).
Trudgian [70, Corollary 2] proved that for every x ≥ 2, 898, 242 there exists a prime
number p with                                          
                                                  1
                            x<p≤x 1+                      .                       (36)
                                             111 log2 x
INTEGERS: 24 (2024)                                                                  10

In [3, Theorem 1.26], it is shown that for every x ≥ 58, 833 there is a prime number
p such that                                          
                                               1.274
                               x<p≤x 1+                 .
                                               log3 x
This was improved in [4, Theorem 1.5] by showing that for every x ≥ 58, 837 there
is a prime number p such that x < p ≤ x(1 + 1.1817/ log3 x). Dusart [30, p. 243]
used (the recovered) Table 1 of [30] (cf. Proposition 2) to show the inequality
                                      0.499x
                       |ϑ(x) − x| <              (x ≥ 4 × 1018 ).                  (37)
                                      log3 x

Alternatively, the inequality (37) follows directly from Proposition 1. Then, he [30,
Proposition 5.4] utilized the inequality (37) to see that for every x ≥ 89, 693 there
exists a prime number p such that
                                                      
                                                  1
                               x<p≤x 1+                                          (38)
                                                log3 x
and concluded from this that for every x ≥ 468, 991, 632 there exists a prime number
p such that                                             
                                                 1
                          x<p≤x 1+                                               (39)
                                           5, 000 log2 x
which improves Trudgian’s result (36). In [5, Theorem 4], the present author com-
bined (39) and the (recovered) inequality (25) (cf. Proposition 3) to obtain that for
every x ≥ 6, 034, 256 there exists a prime number p such that
                                                    
                                              0.087
                              x<p≤x 1+                 .                        (40)
                                              log3 x

Further, the present author [5, Theorem 4] used the (recovered) inequality (27) (cf.
Proposition 3) and (39) to find found that for every x > 1 there is a prime number
p with                                             
                                              198.2
                             x<p≤x 1+                 .                        (41)
                                             log4 x
Now we give a proof of Theorem 1 where we give improvements of (38)–(41) by
decreasing the coefficient of the term 1/ logn x and on the other hand by increasing
the exponent of the log x term.

Proof of Theorem 1. In order to prove that there is a prime number p with x < p ≤
x(1 + a1 / log x) for every x ≥ X1 = 952, 527, 672, 606, 693, we first consider the case
where x ≥ exp(4, 000). Here, we can use [12, Table 15] to get that
                                                                
                           a1                 x                 a1 ε
              ϑ x 1+               − ϑ(x) >         a1 − 2ε −           ≥ 0,
                          log x             log x              log x
INTEGERS: 24 (2024)                                                                  11

where ε = 5.741 × 10−13 , which implies that for every x ≥ exp(4, 000) there is a
prime number p satisfying x < p ≤ x(1 + a1 / log x). For every x with 4 × 1018 ≤
x < exp(4, 000), the claim follows directly from [22, Theorem 1]. So it suffices to
consider the case where 952, 527, 672, 606, 693 ≤ x < 4 × 1018 . Let n be an integer
so that 1, 721, 649, 982, 233, 847 ≤ pn ≤ π(4 × 1018 ) and let x be a real number
satisfying pn ≤ x < pn+1 . Then, we can utilize [51, Table 8] to see that
                                                a1 pn    a1 x
                        pn+1 − pn ≤ 1, 476 <          ≤       .
                                               log pn   log x

This implies that for every x with 1, 721, 649, 982, 233, 847 ≤ x < 4 × 1018 there is a
prime number in the interval (x, x(1 + a1 / log x)]. Similar, we can see that for every
x satisfying 1, 041, 648, 882, 338, 903 ≤ x < 1, 721, 649, 982, 233, 847 there is always
a prime in the interval (x, x(1 + a1 / log x)]. Next, we can use Walisch’s primesieve
program [75] to obtain that
                                                a1 pn    a1 x
                         pn+1 − pn ≤ 860 <            ≤
                                               log pn   log x

for every integer n satisfying 9.88 × 1014 ≤ pn < 1.042 × 1015 and every x with
pn ≤ x < pn+1 . So there exists a prime number p with x < p ≤ x(1 + a1 / log x) for
every x so that 9.88 × 1014 ≤ x ≤ 1, 041, 648, 882, 338, 903. If n is an integer with
9.53 × 1014 ≤ pn < 9.88 × 1014 and x satisfies pn ≤ x < pn+1 , we can use Walisch’s
primesieve program [75] to see that pn+1 − pn ≤ 802 < a1 pn / log pn ≤ a1 x/ log x.
This provides that for every x with 9.53 × 1014 ≤ x < 9.88 × 1014 there is always
a prime number in the interval (x, x(1 + a1 / log x)]. For every integer n satisfying
952, 527, 672, 607, 523 ≤ pn < 9.53 × 1014 and every x with pn ≤ x < pn+1 , we apply
Walisch’s primesieve program [75] to obtain that pn+1 − pn ≤ 708 < a1 pn / log pn ≤
a1 x/ log x and it turns out that for every x with 952, 527, 672, 607, 523 ≤ x <
9.53 × 1014 there is a prime number in the interval (x, x(1 + a1 / log x)]. Finally,
it suffices to consider the case where x belongs to the interval [a, b) where a =
952, 527, 672, 606, 693 and b = 952, 527, 672, 607, 523. In this situation, we have
π(x(1 + a1 / log x)) − π(x) ≥ 1 as desired. The proof of the remaining assertions is
similar to the above proof and we leave the details to the reader.

Remark 6. Beginning with Hoheisel [38], many authors have found shorter inter-
vals of the form [x − xδ , x] that must contain a prime number for all sufficiently
large values of x. The most recent result is due to Baker, Harman, and Pintz [8].
They found the value δ = 0.525. Under the assumption that the Riemann hypoth-
esis is true, much better results are known. For more details, see, for instance,
Ramaré and Saouter [58], Dudek [26], Dudek, Grenié, and Molteni [27], Carneiro,
Milinovich, and Soundararajan [16], Cully-Hugill and Dudek [19], Cully-Hugill and
Johnston [20], and Cully-Hugill and Dudek [21].
INTEGERS: 24 (2024)                                                               12

4. Proof of Theorem 2

First, we note some well known estimates for the prime counting function π(x).
A classic method of finding explicit estimates for π(x) is the following. Let k be
a positive integer and ηk a positive real number. By (5), there is a real number
x1 = x1 (k, ηk ) > 1 so that
                                                   ηk x
                                   |ϑ(x) − x| <
                                                 logk x
for every x ≥ x1 . In order to prove Theorem 2, we define the auxiliary function
                                                        Z x                   
                          ϑ(x1 )     x        ηk x             1       ηk
 Jk;ηk ;x1 (x) = π(x1 ) −        +       +            +            +         dt (42)
                          log x1 log x logk+1 x          x1  log2 t logk+2 t
and note the following both inequalities involving the prime counting function π(x).
Lemma 1. For every x ≥ x1 , we have Jk;−ηk ;x1 (x) ≤ π(x) ≤ Jk;ηk ;x1 (x).

Proof. The claim follows directly form (7) and (5).

  One of the first estimates for π(x) is due to Gauss. In 1793, he computed that

                                    π(x) ≤ li(x)                                (43)

holds for every x with 2 ≤ x ≤ 3, 000, 000 and conjectured that the inequality (43)
holds for every x ≥ 2. This conjecture was disproven by Littlewood [48]. More
precisely, he proved that the function π(x) − li(x) changes the sign infinitely many
times. Unfortunetely, Littlewood’s proof is nonconstructive and there is still no
example of x such that π(x) > li(x). Skewes [67] proved the existence of a number
x0 with x0 < exp(exp(exp(exp(7.705)))) such that π(x0 ) > li(x0 ). Lehman [47]
improved this last upper bound considerably by showing that exists a number x0
with x0 < 1.65 × 101165 such that π(x0 ) > li(x0 ). After some further improvements
(see, for instance, te Riele [69], Bays and Hudson [9], Chao and Plymen [17], Saouter
and Demichel [65], Stoll and Demichel [68]), the current best upper bound was
found by Saouter, Trudgian, and Demichel [64]. They proved that there exists a
number x0 with x0 < exp(727.951335621) such that π(x0 ) > li(x0 ). All these upper
bounds have been proved by using computer calculations of zeros of the Riemann
zeta function. The first lower bound for a number x0 with π(x0 ) > li(x0 ) was
given by the calculation of Gauss, namely x0 > 3, 000, 000. This lower bound was
improved in a series of papers. For details, see Rosser and Schoenfeld [63], Brent
[11], Kotnik [43], Platt and Trudgian [54], and Stoll and Demichel [68]. For our
further inverstigation, we use the following improvement.
Lemma 2 (Büthe [15]). For every x with 2 ≤ x ≤ 1019 , we have π(x) ≤ li(x).
Remark 7. Recently. Dusart [31, Lemma 2.2] showed that π(x) ≤ li(x) for every
x with 2 ≤ x ≤ 1020 .
INTEGERS: 24 (2024)                                                                  13

     Now we use Proposition 1 and the Lemmata 1 and 2 to give a proof of Theorem
2.

Proof of Theorem 2. First, we combine Lemma 1 with Proposition 1 to see that

                     J3;−0.024334;x1 (x) ≤ π(x) ≤ J3;0.024334;x1 (x)               (44)

for every x ≥ x1 , where x1 ≥ 1, 757, 126, 630, 797. Now, let x2 = 1018 and let f (x)
be given by the right-hand side of (12). We consider the function g(x) = f (x) −
J3,0.024334,x2 (x). By [25], we have ϑ(x2 ) ≥ 999, 999, 999, 144, 115, 634. Further,
π(x2 ) = 24, 739, 954, 287, 740, 860 and so we compute g(x2 ) ≥ 2 × 108 . Since the
derivative of g is positive for every x ≥ x2 , we get f (x) − J3,0.024334,x2 (x) > 0 for
every x ≥ x1 , and we conclude from (44) that the inequality (12) holds for every
x ≥ x1 . Comparing f (x) with the integral logarithm li(x), we see that f (x) > li(x)
for every x ≥ 121, 141, 948. Now we can utilize Lemma 2 to see that the desired
inequality also holds for every x such that 121, 141, 948 ≤ x < 1018 . A computer
check for smaller values of x completes the proof.

   Under the assumption that the Riemann hypothesis is true, von Koch [72] de-
                              √
duced that π(x) = li(x) + O( x log x) as x → ∞. Actually, one can show that the
                                    √
asymptotic formula π(x) = li(x)+O( x log x) as x → ∞ is even a sufficient criterion
for the truth of the Riemann hypothesis. An explicit version of von Koch’s result
is due to Schoenfeld [66, Corollary 1]. Under the assumption that the Riemann
hypothesis is true, Schoenfeld found that the inequality
                                                 1 √
                             |π(x) − li(x)| <        x log x                       (45)
                                                8π
holds for every x ≥ 2, 657. In 2014, Büthe [14, p. 2,495] proved that the inequality
(45) holds unconditionally for every x such that 2, 657 ≤ x ≤ 1.4 × 1025 . Platt and
Trudgian [55, Corollary 1] improved Büthe’s result by showing that the inequality
(45) holds unconditionally for every x satisfying 2, 657 ≤ x ≤ 2.169×1025 . Johnston
[39, Corollary 3.3] extended the last result by showing the following

Lemma 3 (Johnston). The inequality (45) holds unconditionally for every x sat-
isfying 2, 657 ≤ x ≤ 1.101 × 1026 .

   Now we can use Theorem 2 and the Lemmata 2 and (3) to find the following
weaker but more compact upper bounds for the prime counting function π(x) of
the form
                                    x
             π(x) <                                     (x ≥ x0 ),
                    log x − a0 − log x − · · · − logam
                                  a1
                                                     mx


where m is a integer with 0 ≤ m ≤ 5 and a0 , . . . , am are suitable positive real
numbers.
INTEGERS: 24 (2024)                                                                       14

Corollary 1. We have
                                                x
                           π(x) <                 a1       a2
                                    log x − a0 − log x − log2 x

for every x ≥ x0 , where a0 , a1 , a2 , and x0 are given as in Table 2.

    a0            1.0343                        1                          1
    a1               0                        1.109                        1
    a2               0                          0                         3.48
    x0    106, 640, 139, 304, 611    81, 250, 795, 096, 339    145, 413, 088, 724, 077

                   Table 2: Explicit values for a0 , a1 , a2 , and x0 .


Proof. Theorem 2 implies that the inequality
                                                 x
                                 π(x) <                                                  (46)
                                          log x − 1.0343
holds for every x ≥ 108, 943, 258, 198, 427. If we compare the right-hand side of
(46) with li(x), we can use Lemma 2 to see that the required inequality (46) holds
for every x with 106, 910, 668, 441, 596 ≤ x ≤ 108, 943, 258, 198, 427. Finally, we use
Walisch’s primecount program [74] to obtain that the inequality (46) is also valid
for every x satisfying 106, 640, 139, 304, 611 ≤ x ≤ 106, 910, 668, 441, 596. The proof
of each of the next three inequalities is similar to the proof of (46) and we leave the
details to the reader. Next, we show that the inequality
                                                  x
               π(x) <                                                                    (47)
                         log x − 1 − log1 x − 3.024334
                                               log2 x
                                                       − 12.975666
                                                           log3 x
                                                                   − 79.962
                                                                     log4 x

holds for every x ≥ 22. First, we can use Theorem 2 to obtain the inequality (47)
for every x ≥ 1.101 × 1026 . Let f (x) denote the right-hand side of (46). We get
                    √
that f (x) ≥ li(x) + x log(x)/(8π) for every x with 22, 066, 689, 219, 741, 110 ≤ x ≤
1.101 × 1026 . Now we can apply Lemma 3 to see that the required inequality (47)
also holds for every x satisfying 22, 066, 689, 219, 741, 110 ≤ x ≤ 1.101 × 1026 . A
comparison with li(x) shows that f (x) > li(x) for every x ≥ 259, 576, 712, 645 and
Lemma 2 yields the desired inequality (47) for every x with 259, 576, 712, 645 ≤
x ≤ 22, 066, 689, 219, 741, 110. Finally, it suffices to apply Walisch’s primecount
program [74] to see that the inequality (47) also holds for every x satisfying 22 ≤
x ≤ 259, 576, 712, 645. Again, the proof of the remaining inequality is similar to the
proof of (47) and we leave the details to the reader.

Remark 8. In the appendix at the end of this paper, we give lots of other weaker
upper bounds in the case where m ∈ {0, 1, 2}.
INTEGERS: 24 (2024)                                                                     15

Corollary 2. We have
                                                     x
             π(x) <
                       log x − 1 − log1 x − 3.024334
                                             log2 x
                                                     − loga33 x − loga44 x − loga55 x
for every x ≥ x0 , where a3 , a4 , a5 , and x0 are given as in Table 3.

                a3               14.893              12.975666       12.975666
                a4                  0                    79.962      71.048668
                a5                  0                      0           533.594
                x0      142, 464, 507, 937, 911           22              32

                     Table 3: Explicit values for a3 , a4 , a5 , and x0 .


Proof. Since the proof is similar to the proof of Corollary 1, we leave the details to
the reader.

   Using an estimate for Chebyshev’s ϑ-function found by Broadbent et al. [12,
Table 15], we get the following upper bound for π(x) which improves the inequality
(12) for all sufficiently large values of x.
Proposition 4. For every x ≥ 29.53, we have
                                            x
                  π(x) <                                        .
                          log x − 1 − log1 x − log32 x − 70.935
                                                         log3 x

Proof. The proof is similar to the proof of Theorem 2 and we leave the details to the
reader. We denote the right-hand side of (4) by f (x) and let x1 = 1018 . We combine
Lemma 1 with [12, Table 15] to see that π(x) ≤ J4,57.184,x1 (x) for every x ≥ x1 .
So it suffices to compare f (x) with J4,57.184,x1 (x) to get that f (x) > π(x) for every
x ≥ 1018 . Since f (x) > li(x) for every x such that 1, 098 ≤ x < 1018 , we can apply
Lemma 2 to obtain that (4) also holds for every x such that 1, 098 ≤ x < 1018 . A
direct computation for smaller values of x completes the proof.

   Integration by parts in (8) implies that for every positive integer m, one has
                                                                                  
          x       x       2x       6x       24x          (m − 1)!x            x
π(x) =       +        +         +       +         +...+             +O
        log x log2 x log3 x log4 x log5 x                  logm x         logm+1 x
                                                                                (48)
as x → ∞. In this direction, we get the following upper bound for π(x).
Proposition 5. For every x > 1, we have
                        x      x       2x   6.024334x 24.024334x
            π(x) <         +       +      +            +
                      log x log2 x log3 x     log4 x         log5 x
                             120.12167x 720.73002x 6098x
                           +            +            +        .
                               log6 x      log7 x      log8 x
INTEGERS: 24 (2024)                                                                            16

Proof. We set x1 = 1018 . Further, let f (x) be the right-hand side of the required
inequality. We have f (x) > J3,0.024334,x1 (x) for every x ≥ x1 . So, we can use (44)
to get f (x) > π(x) for every x ≥ x1 . Since f (x) > li(x) for every x ≥ 204, 182, 829,
we can apply Lemma 2 to obtain f (x) > π(x) for every x such that 204, 182, 829 ≤
x ≤ x1 . A direct computation for smaller values of x completes the proof.

   Proposition 5 yields the following weaker but more compact upper bounds for
the prime counting function π(x).
Corollary 3. For every x ≥ x0 , we have
                                           x     x     (2 + ε)x
                                π(x) <        +   2  +          ,
                                         log x log x    log3 x
where ε and x0 are given as in Table 4.

        ε                0.21                      0.215                      0.22
        x0     160, 930, 932, 942, 272     83, 016, 503, 500, 865    43, 999, 690, 220, 699
        ε               0.225                      0.23                       0.24
        x0     23, 824, 649, 646, 672      13, 279, 102, 022, 111    4, 511, 700, 549, 332
        ε                0.25                      0.26                     0.2651
        x0      1, 615, 202, 653, 795       643, 809, 266, 445        406, 742, 886, 708
        ε                0.27                      0.28                       0.29
        x0       265, 248, 130, 170         117, 997, 473, 286         57, 720, 805, 589

                             Table 4: Explicit values for ε and x0 .

Proof. Let x0 = 160, 930, 932, 942, 272 and f (x) = x/ log x+x/ log2 x+2.21x/ log3 x.
Proposition 5 implies that π(x) < f (x) for every x ≥ 180, 250, 881, 352, 396. If we
compare f (x) with the integral logarithm li(x), we get by Lemma 2 that π(x) <
f (x) for every x ≥ 162, 791, 795, 110, 834. Next, we use a computer to verify the
inequality π(x) < f (x) for every x with x0 ≤ x ≤ 162, 791, 795, 110, 834. The
remaining inequalities can be proved in the same way.




5. Proof of Theorem 3

In order to give a proof of Theorem 3, we use (44) and a numerical calculation that
verifies the desired inequality for smaller values of x.
       1 This inequality was already known to be true for every x ≥ 8 × 1011 (see [50, Proposition

3.3]
INTEGERS: 24 (2024)                                                                             17

Proof of Theorem 3. Let x1 = 1, 757, 126, 630, 797. Further, let g(x) be the right-
hand side of (13). We can compute that J3,−0.024334,x1 (x1 ) − g(x1 ) > 6 × 103 . In
                     0
addition we have J3,−0.024334,x  1
                                   (x) > g 0 (x) for every x ≥ 44.42. Therefore, we get
J3,−0.024334,x1 (x) > g(x) for every x ≥ x1 . Using (44), we get the required inequality
for every x ≥ x1 . For smaller values of x we use a computer.

Remark 9. Let x1 = 1, 751, 189, 194, 177. Then the inequality (13) does not hold
for x = x1 − 0.1.
Remark 10. Theorem 3 improves the lower bound for π(x) obtained in [5, Theorem
3].

   In the next corollary, we establish some weaker lower bounds for the prime count-
ing function.
Corollary 4. We have
                                                  x
             π(x) >
                      log x − 1 − log1 x − 2.975666
                                            log2 x
                                                    − loga33 x − loga44 x − loga55 x

for every x ≥ x0 , where a3 , a4 , a5 , and x0 are given as in Table 5.

   a3         13.024334                13.024334              13.024334                0
   a4         70.951332                70.951332                    0                  0
   a5     460.634397856444                   0                      0                  0
   x0    1, 035, 745, 443, 241     153, 887, 581, 621      7, 713, 187, 213      54, 941, 209

                   Table 5: Explicit values for a3 , a4 , a5 , and x0 .


Proof. From Theorem 3, it follows that each required inequality holds for every
x ≥ 1, 751, 189, 194, 177. For smaller values of x we use a computer.

  Let n be a positive integer. Then (48) provides the inequality
                   x     x      2x     6x    24x           (n − 1)!x
        π(x) >        +      +      +      +       + ... +
                 log x log2 x log3 x log4 x log5 x          logn x
for all sufficiently large values of x. In the following proposition, we describe a
method to find lower bounds for π(x) in the direction of (5) by using lower bounds
for π(x) in the direction of (11).
Proposition 6. Let n be a positive integer and let a0 > 0 and a1 , . . . , an be neg-
ative real numbers. Suppose that there is a positive real number x0 such that the
inequalities
                                     a2              an
                    a0 log x + a1 +       + ... +          >0                    (49)
                                    log x         logn−1 x
INTEGERS: 24 (2024)                                                                              18

and
                                                        x
                       π(x) >                        a2                an                       (50)
                                    a0 log x + a1 + log x + . . . + logn−1 x
hold simultaneously for every x ≥ x0 . Then we have
                                       b0 x   b1 x           bn x
                         π(x) >             +      + ... +
                                      log x log2 x         logn+1 x
for every x ≥ x0 , where b0 , . . . , bn are real numbers recursively defined by
                                                            k
                                                       1 X
               b0 = 1/a0 ,          and       bk = −          ai bk−1    (1 ≤ k ≤ n).           (51)
                                                       a0 i=1
                                              Pn          i
                                                                           Pn             i
Proof. For y > 0, we define R(y) =              k=0 ai /y and S(y) =         i=0 bi /y . For
i ∈ {1, . . . , 2n}, we set
                 (                                          (
          0       ai , if i ∈ {1, . . . , n},           0     bi , if i ∈ {1, . . . , n},
        ai =                                  and      bi =
                  0, otherwise                                0, otherwise.
Using (51) together with b0n+1 = 0, we can see that
                                                      2n
                                                      X   k
                                                         X a0i b0k−i
                                  R(y)S(y) = 1 +                     .
                                                         i=1
                                                              yk
                                                   k=n+1

Since a0i b0k−i ≤ 0 for every i with 1 ≤ i ≤ 2n and every k satisfying n + 1 ≤ k ≤ 2n,
we get R(y)S(y) ≤ 1. By (49), we have R(log x) > 0 for every x ≥ x0 . Now we can
use (50) to get π(x) > x/(R(x) log x) ≥ xS(log x)/ log x for every x ≥ x0 .

  The best explicit result in the direction of (50) was found in [5, Proposition 5].
The following refinements of it are a consequence of Proposition 6, Theorem 3, and
Corollary 4.
Corollary 5. We have
            x     x      2x     5.975666x    b5 x   b6 x   b7 x   b8 x
 π(x) >        +      +       +           +       +      +      +
          log x log2 x log3 x        4
                                  log x     log x log x log x log8 x
                                               5      6      7

for every x ≥ x0 , where b5 , b6 , b7 , b8 , and x0 are given as in Table 6.

          b5                 b6              b7                 b8               x0
      23.975666       119.87833           719.26998    5034.88986       1, 681, 111, 802, 141
      23.975666       119.87833           719.26998             0        721, 733, 241, 667
      23.975666       119.87833              0                  0        110, 838, 719, 141
      23.975666              0               0                  0         1, 331, 691, 853
           0                 0               0                  0           10, 383, 799

                         Table 6: Explicit values for a2 and x2 .
INTEGERS: 24 (2024)                                                                19

Proof. In order to prove the first inequality, we combine Proposition 6 and Theorem
3 to see that this inequality holds for every x ≥ 1, 751, 189, 194, 177. For smaller
values of x, we use a computer. Further, we use Proposition 6, Corollary 4, and a
direct computation for smaller values of x to verify the remaining inequalities.



6. Proof of Theorem 4

In order to prove Theorem 4, we set R = 5.5666305 and, similar to [56, p. 879], we
define the function a : R>0 → R by
              2 − log 2
            
            
                                                      if 2 ≤ x < 599,
            
            
                2
                  2
             log x
                                                       if 599 ≤ x < 1.101 × 1026 ,
            
            
                √
              8π   x
            
            
   a(x)      r               1/4       r        !
         =
                     
     6            8     log x               log x
  log x                          exp −               if 1.101 × 1026 ≤ x < e673 ,
            
            
               17π 6.455                  6.455
            
                              3/2        r        !
                         log x                log x
                                                       if x ≥ e673 .
            
            121.0961
            
                                    exp −2
                            R                    R

Then we get the following result concerning Chebyshev’s ϑ-function.

Lemma 4. For every x ≥ 2, we have

                                                a(x)x
                                 |ϑ(x) − x| ≤          .
                                                log6 x
Proof. In the case where x satisfies 2 ≤ x < 599, then the given bound is trivial.
For second bound, see Johnston [39, Corollary 3.3]. The third bound was given
by Trudgian [70, Theorem 1] and the last bound was recently established by Fiori,
Kadiri, and Swidinsky [12, Corollary 14] (cf. (3)).

     We also need the following result on our function a.

Lemma 5. Let x1 be a real number with x1 ≥ e673 . Then an (x) ≤ an (x1 ) for every
x ≥ x1 .

Proof. By a straightforward calculation of the derivative, we see that a0 (x) < 0 for
every x ≥ e673 .

     Now we use Theorem 3 and Lemmata 4-5 to give the following proof of Theorem
4.

Proof of Theorem 4. In [6, Theorem 1], the inequality was already proved for every
x with 65, 405, 887 ≤ x ≤ 2.7358×1040 . If we utilize Theorem 3, it turns out that the
INTEGERS: 24 (2024)                                                                           20

inequality (14) holds unconditionally for every x such that 65, 405, 887 ≤ x ≤ e540 .
Now, let f (x) denote the right-hand side of (14). In order to verify the required
inequality for every x with e540 ≤ x ≤ e1680 , we set c0 = 1 − 1.6341 × 10−12 . By
[34, Table 3], we have ϑ(x) ≥ c0 x for every x > e500 . Applying this inequality to
(7), we get
                                   π(x) > g0 (x)                               (52)
for every x ≥ e500 , where g0 (x) = c0 (li(x) − li(e500 ) + e500 /500). If we show that
g0 (x) > f (x) for every x satisfying e540 ≤ x ≤ e1680 , we can use (52) to see that the
required inequality (14) holds for every x with e540 ≤ x ≤ e1680 . Since g00 (x) > f 0 (x)
for every x so that 9 ≤ x ≤ e1680 , it remains to show that g0 (x0 ) > f (x0 ), where
                                  P6
x0 = e540 . First, we note that t k=1 (k−1)!/ logk t < li(t) < 1.003t/ log t, where the
left-hand side inequality holds for every t ≥ 565 and the right-hand side inequality
is valid for every t ≥ e500 . Therefore,
                                               6
                g0 (x0 ) − f (x0 )      X (k − 1)! 0.003c0   f (x0 )
                                   > c0           −        −         .
                        x0                  540k     e40       x0
                                              k=1

Since the right-hand side of the last inequality is positive and we conclude that the
required inequality holds for every x with x0 ≤ x ≤ e1680 . The final step of the
proof consists in the verification of the required inequality for every x ≥ x1 , where
x1 = e1680 . If we combine (7) with Lemma 4, we can see that
                                            Z x           Z x
                            x     xa(x)           dt           a(t)
                  π(x) ≥        −        +              −         8 dt.           (53)
                         log x log7 x        2  log 2
                                                      t    2  log  t

Integration by parts in (53) provides that
                        5                                            Z x
                        X           k!             (720 − a(x))x         5040 − a(t)
         π(x) ≥ C + x                         +                  +                   dt,
                        k=0
                              log   k+1
                                          x            log7 x         x1    log8 t

where
                            Z x1                        6
                                    5040 − a(t)        X       k!
                       C=                       dt − 2               .
                              2        log8 t          k=1
                                                           log k+1
                                                                   2
Since 0 < a(t) ≤ a(x1 ) for every t ≥ x1 (cf. Lemma 5), it turns out that
                    5                                                            Z x
                    X         k!  (720 − a(x1 ))x                                      dt
     π(x) ≥ C + x               +                 + (5040 − a(x1 ))                     8 .
                  k=0
                      log k+1
                              x       log7 x                                      x1 log t


Note that 5040 − a(x1 ) < 0. Hence
                       5
                       X       k!      (720 − a(x1 ))x
        π(x) > C + x                 +                 + (5040 − a(x1 ))E(x1 ),
                       k=0
                           log k+1
                                   x       log7 x
INTEGERS: 24 (2024)                                                                        21

where                                                                           !
                                                              7
                                   1                          X (k − 1)!x
                          E(x) =                  li(x) −                           .
                                 5040
                                                              k=1
                                                                       logk x
Since C + (5040 − a(x1 ))E(x1 ) < 0, we obtain that
            5
                                                                                  log7 x1
                                                                                         
            X        k!            x
 π(x) > x                    +             720 − a(x1 ) + (5040 − a(x1 ))E(x1 ) ×           .
            k=0
                  logk+1 x       log7 x                                             x1

Now we use a computer to get that
                                            5
                                            X           k!            6918930x
                             π(x) > x                             −
                                            k=0
                                                  log   k+1
                                                              x         log7 x

for every x ≥ x1 . Now we set
                                 5
                                 X   k!    6918930          1
                     H(y) =              −         −                  .
                                   y
                                 k=0
                                     k+1      y 7    y − 1 − y1 − y32

It is easy to see that H 0 (y) < 0 for every y ≥ 859. Together with limy→∞ H(y) = 0,
it turns out that H(log x) ≥ 0 for every x ≥ e859 . If we combine the last inequality
with (6), we get that π(x) > f (x) for every x ≥ e1680 and we arrive at the end of
the proof.

Remark 11. The method employed in the proof of Theorem 4 can also be used
to find further lower bounds for π(x) given by truncating the asymptotic expansion
(11) at later terms (with logn x in the denominator, where n ≥ 3). However, these
bounds will only hold when x is exceptionally large. For instance, (11) provides the
even sharper inequality
                                                         x
                          π(x) >
                                       log x − 1 − log1 x − log32 x − log133 x

for all sufficiently large values of x. Similar to the proof of Theorem 4, we get that
this inequality holds for every x satisfying 11, 471, 757, 461 ≤ x ≤ e57.820987 and
every x ≥ e3661.424 .

Corollary 6. For every x ≥ 10, 384, 261, we have
                                      x     x     2x    6x
                          π(x) >         +     +     +       .
                                    log x log x log x log4 x
                                             2     3


Proof. It suffices to combine Proposition 6, Theorem 4, and [6, Theorem 2].
INTEGERS: 24 (2024)                                                                 22

7. Proof of Theorem 5

In this section, we want to find unrestricted effective estimates for the sum of
the reciprocals of all prime numbers not exceeding x For this purpose, we use
the method investigated by Rosser and Schoenfeld [63, p. 74]. They derived a
remarkable identity which connects the sum of the reciprocals of all prime numbers
not exceeding x with Chebyshev’s ϑ-function by showing that
                                     Z ∞
                          ϑ(x) − x       (ϑ(y) − y)(1 + log y)
                 A1 (x) =          −                           dy,            (54)
                           x log x    x        y 2 log2 y
where                                 X1
                           A1 (x) =             − log log x − B.
                                            p
                                      p≤x

Here, the constant B is defined as in (16). Applying (2) to (54), Rosser and
Schoenfeld [63, p. 68] refined the error term in Mertens’ result (15) by giving
                     √
A1 (x) = O(exp(−a log x)) as x → ∞, where a is an absolute positive constant.
Then [63, Theorem 5] they used explicit estimates for Chebyshev’s ϑ-function to
show that
                                   1                   1
                             −          < A1 (x) <          ,
                               2 log2 x            2 log2 x
where the left-hand side inequality is valid for every x > 1 and the right-hand side
inequality holds for every x ≥ 286. Meanwhile there are several improvements of
(7) (see, for instance, [30, Theorem 5.6] and [5, Proposition 7]). In Theorem 5, we
give the current best unconditionally effective estimates for A1 (x). The proof is
now rather simple.

Proof of Theorem 5. It suffices to combine (54) with Proposition 1.

Remark 12. Note that the positive integer N0 = 1, 757, 126, 630, 797 might not be
the smallest positive integer N so that the inequality given in Theorem 5 holds for
every x ≥ N .
Remark 13. Rosser and Schoenfeld [63, Theorem 20] used the calculation in [1]
to see that A1 (x) > 0 for every 1 < x ≤ 108 and raised the question whether this
inequality hold for every x > 1. Robin [62, Théorème 2] proved that the function
A1 (x) changes the sign infinitely often, which leads to a negative answer to the obove
question. By adapting a method for bounding Skewes’ number, Büthe [13, Theorem
1.1] found that there exists an x0 ∈ [exp(495.702833109), exp(495.702833165)] such
that A1 (x) is negative for every x ∈ [x0 − exp(239.046541), x0 ].
Remark 14. Under the assumption that the Riemann hypothesis is true, Schoen-
feld [66, Corollary 2] found some better estimate for the sum of the reciprocals of
all prime numbers not exceeding x. This result was recently improved by Dusart
[31, Theorem 4.1].
INTEGERS: 24 (2024)                                                              23

  Using the definition (16) of B, we get
                               Y       1
                                          
                        γ
                       e log x      1−      = e−S(x)−A1 (x) ,                   (55)
                                        p
                                 p≤x

where
                         X                     ∞
                                     1     1      X   1X 1
                S(x) =       log 1 −     +     =−              .                (56)
                         p>x
                                     p     p      n=2
                                                      n p>x pn

By Rosser and Schoenfeld [63, p. 87], we have
                                     1.02
                             −                 < S(x) < 0                       (57)
                                 (x − 1) log x

for every x > 1. Hence, the asymptotic formula (15) gives A2 (x) = O(1/ log2 x) as
x → ∞, where
                                    e−γ    Y        
                                                   1
                          A2 (x) =       −     1−      .
                                   log x           p
                                             p≤x

In [63, Theorem 7], Rosser and Schoenfeld found that

            e−γ                                   e−γ
                               Y                                
                          1                1                    1
                   1−           <     1−       <         1+            ,
           log x      2 log2 x         p≤x
                                           p     log x      2 log2 x

where the left-hand side inequality is valid for every x ≥ 285 and the right-hand
side inequality holds for every x > 1. We use (55) combined with Theorem 5 to
obtain the following refinement of [5, Proposition 9].

Proposition 7. For every x ≥ 1, 757, 126, 630, 797, we have

      e−γ                Y               e−γ
                                                                        
                                  1                             1.02
           exp(−f (x)) <     1−       <        exp f (x) +                  ,
     log x                        p      log x              (x − 1) log x
                           p≤x

where f (x) denotes the right-hand side of (17).

Proof. First, we apply the left-hand side inequality of Theorem 5 to (55) and see
that
                                      e−γ
                     Y          
                               1
                          1−       <       exp(−S(x) + f (x))                (58)
                               p     log x
                     p≤x

for every x > 1, 757, 126, 630, 797. Now it suffices to apply the right-hand side
inequality of (57) to (58) and we get the required right-hand side inequality. One
the other hand, we have S(x) < 0 by (57). Applying this and (17) to (55), we arrive
at the end of the proof.
INTEGERS: 24 (2024)                                                                24

Remark 15. Note that the positive integer N0 = 1, 757, 126, 630, 797 in Proposition
7 might not be the smallest positive integer N so that the inequality given holds
for every x ≥ N .

Remark 16. Under the assumption that the Riemann hypothesis is true, Schoen-
feld [66, Corollary 3] found that the inequality

                                            3 log x + 5
                              |A2 (x)| <        √
                                           8πeγ x log x

holds for every x ≥ 8. This was slightly improved by Dusart [31, Theorem 4.4] in
2018.

Remark 17. Rosser and Schoenfeld [63, Theorem 23] found that A2 (x) > 0 for
every 0 < x ≤ 108 and stated [63, p. 73] the question whether this inequality also
hold for every x > 108 . In [62, Proposition 1], Robin answered this by showing that
the function A2 (x) changes the sign infinitely often.

  Now we can use Proposition 7 to derive the following effective estimates for
                                Y        1
                                            
                                     1+       ,
                                          p
                                   p≤x

where p runs over primes not exceeding x.

Corollary 7. For every x ≥ 1, 757, 126, 630, 797, one has

     6eγ
                                              Y         
                            1.02                         1
         exp   −f (x) −                 log x <      1 +
      π2                (x − 1) log x                    p
                                                p≤x

                                                6eγ
                                                           
                                                          1
                                              < 2 1+          exp(f (x)) log x,
                                                 π       x

where f (x) denotes the right-hand side of (17).

Proof. Since 1 + 1/p = (1 − 1/p2 )/(1 − 1/p) and ζ(2) = π 2 /6, it suffices to combine
Proposition 7 and [31, Lemma 4.3].

Remark 18. Note that the positive integer N0 = 1, 757, 126, 630, 797 might not be
the smallest positive integer N so that the inequality given in Corollary 7 holds for
every x ≥ N .

   Let us briefly study S(x), defined as in (56), in more detail. In the proof of
the left-hand side inequality in (57), Rosser and Schoenfeld used the inequality
ϑ(x) < 1.02x which is valid for every x > 0 (see [63, Theorem 9]). If we use
approximations for ϑ(x) of the form (5), we get the following result.
INTEGERS: 24 (2024)                                                                    25

Proposition 8. Let k be a positive integer and let ηk and x0 = x0 (k) be positive
real numbers with x0 > 1 so that |ϑ(x) − x| < ηk x/ logk x for every x ≥ x0 . Then,
we have
                    ∞
                       li(x−n )
                                                                   
                   X                  ηk                     x
            S(x) −              <            (x + 1)  log          − 1
                   n=1
                        n+1        logk+1 x                 x−1

for every x ≥ x0 .

  In order to prove this proposition, we first establish the following lemma.

Lemma 6. Let n be a positive integer with n ≥ 2. Under the assumptions of
Proposition 8, we have
                                                                                 
                                   X 1                  ηk                   n
                     li(x1−n ) +              <                        1+
                                   p>x
                                         pn       x n−1 logk+1 x            n−1

for every x ≥ x0 .

Proof. By [63, p. 87], we have
                     X 1                           Z ∞
                                 ϑ(x)                    (1 + n log y)ϑ(y)
                             =− n       +                                  dy.        (59)
                     p>x
                         p n   x  log x             x       y n+1 log2 y

Since we have assumed that |ϑ(x) − x| < ηk x/ logk x for every x ≥ x0 , we see that
                                                      Z ∞
           X 1
                          1−n          ηk                 1 + n log y
                   ≤ −li(x    )+           k+1
                                                 + ηk                  dy       (60)
           p>x
               p n
                                 x n−1 log     x       x  y n logk+2 y

for every x ≥ x0 . Analogous to [63, Lemma 9], we get that
                    Z ∞
                         1 + n log y                n
                               k+2
                                     dy ≤                       .
                     x
                          n
                         y log     y      (n − 1)x n−1 logk+1 x

Applying this inequality to (60), we see that the required upper bound holds for
every x ≥ x0 . The proof of the required lower bound is quite similar and we leave
the details to the reader.

   Now we can combine the definition (56) with Lemma 6 to get the following proof
of Proposition 8.

Proof of Proposition 8. If we apply Lemma 6 to (56), it turns out that
                         ∞                             ∞ 
                           li(x−n )
                                                                   
                         X                     ηk     X         n     1
              S(x) −                      <     k+1
                                                           1+
                         n=1
                               n+1          log     x n=2     n − 1 nxn−1
INTEGERS: 24 (2024)                                                               26

for every x ≥ x0 . Now, it suffices to apply the identity
                  ∞                                         
                X           n         1                     x
                      1+                   = (x + 1) log         −1
                n=2
                          n − 1 nxn−1                      x−1

to complete the proof.

   If we combine (35) and (59), we find the following new necessary condition for
the Riemann hypothesis including the sum given in Lemma 6.
Proposition 9. Let n be a positive integer with n ≥ 2. Under the assumption that
the Riemann hypothesis is true, we have
                     X 1                                            
                                  1             2n                2
         li(x1−n ) +        <             1 +           log x +
                     p>x
                         pn   8πxn−1/2        2n − 1            2n − 1

for every x ≥ 599.

Proof. Instead of the assumption (5), we now use (35) in the proof of Lemma 6.



8. Proof of Theorem 6

Here we give the following proof of Theorem 6.

Proof of Theorem 6. Let the constant E be defined as in (19) and let
                                     X log p
                          A3 (x) =             − log x − E.
                                           p
                                     p≤x

By Rosser and Schoenfeld [63, p. 74], we have
                                           Z ∞
                              ϑ(x) − x         ϑ(y) − y
                    A3 (x) =             −              dy.                      (61)
                                   x        x     y2
Similarly to the proof of Theorem 5, we may combine (61) and Proposition 1 to get
that the desired both inequalities hold for every x ≥ 1, 757, 126, 630, 797.

Remark 19. Under the assumption that the Riemann hypothesis is true, Schoen-
feld [66, Corollary 2] found a better upper bound for |A3 (x)|. This result was later
improved by Dusart [31, Theorem 4.2].
Remark 20. Rosser and Schoenfeld [63, Theorem 21] also found that A3 (x) > 0 for
every 0 < x ≤ 108 . Again, they asked whether this inequality also holds for every
x > 108 . Robin [62, Proposition 1] showed that the function A3 (x) changes the
sign infinitely often, which leads again to a negative answer to the above question.
Unfortunately, until today no x0 is known so that A3 (x0 ) < 0.
INTEGERS: 24 (2024)                                                                              27

Acknowledgements. The author would like to express his great appreciation
to Kim Walisch, Tomás Oliveira e Silva, and Thomas Lessmann for the support
in writing the C++ codes used in this paper. Furthermore the author thanks
Samuel Broadbent, Habiba Kadiri, Allysa Lumley, Nathan Ng, and Kirsten Wilk,
whose paper has motivated him to deal with the present topic again. Moreover,
the author would also like to thank the two beautiful souls R. and O. for the never
ending inspiration. Finally, the author thanks the anonymous reviewer for the useful
comments and suggestions to improve the quality of this paper.



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Appendix

Here we use Corollary 1 and Walisch’s primecount program [74] to note more weaker
upper bounds for π(x) of the form (4), where m is an integer with 0 ≤ m ≤ 2 and
a0 , . . . , am are suitable positive real numbers. We start with the case where m = 0.
INTEGERS: 24 (2024)                                                                    31

Proposition 10. One has
                                             x
                                    π(x) <
                                        log x − a0
for every x ≥ x0 , where a0 and x0 are given as in Table 7 and Table 8.

     a0           1.0344                     1.0345                  1.0346
     x0    98, 011, 218, 006, 714    90, 093, 726, 828, 053   82, 972, 765, 680, 514
     a0           1.0347                     1.0348                  1.0349
     x0    76, 292, 362, 570, 940    70, 363, 470, 737, 452   64, 716, 191, 738, 353
     a0            1.035                     1.036                    1.037
     x0    59, 667, 044, 596, 151    27, 086, 141, 056, 455   12, 806, 615, 320, 917
     a0            1.038                     1.039                    1.04
     x0    6, 317, 261, 904, 937     3, 231, 501, 496, 562    1, 697, 021, 254, 855
     a0            1.041                     1.042                    1.043
     x0     924, 640, 658, 874        519, 205, 451, 664       296, 735, 291, 225
     a0            1.044                     1.045                    1.046
     x0     175, 758, 684, 156        105, 640, 136, 371        65, 431, 161, 562
     a0            1.047                     1.048                    1.049
     x0      41, 022, 022, 044         25, 724, 702, 310        17, 231, 171, 472
     a0            1.05                      1.051                    1.052
     x0      11, 207, 440, 881          7, 538, 561, 672         5, 047, 295, 951
     a0            1.053                     1.054                    1.055
     x0       3, 745, 835, 388          2, 605, 443, 747         1, 810, 796, 757
     a0            1.056                     1.057                    1.058
     x0       1, 220, 594, 340           876, 542, 559            673, 828, 570
     a0            1.059                      1.06                    1.061
     x0        501, 155, 566             383, 446, 375            269, 585, 283
     a0            1.062                     1.063                    1.064
     x0        196, 894, 353             180, 220, 137            116, 749, 925
     a0            1.065                     1.066                    1.067
     x0        110, 166, 540             76, 223, 058             53, 431, 171
     a0            1.068                     1.069                    1.07
     x0        46, 097, 944              39, 706, 453             31, 027, 247

                      Table 7: Explicit values for a0 and x0 .
INTEGERS: 24 (2024)                                                                         32

     a0       1.071             1.072           1.073            1.074         1.075
     x0    22, 078, 017      18, 339, 738    13, 026, 859     12, 895, 928   8, 832, 927
     a0       1.076             1.077           1.078            1.079          1.08
     x0     7, 299, 254      7, 117, 256      5, 465, 656     4, 994, 010    3, 462, 478
     a0       1.081             1.082           1.083          1.08366         1.084
     x0     3, 455, 648      2, 279, 177      1, 529, 630     1, 526, 671    1, 525, 432
     a0       1.085             1.086           1.087            1.088         1.089
     x0     1, 515, 074      1, 200, 014      1, 195, 296      624, 878       618, 726
     a0        1.09             1.091           1.092            1.093         1.094
     x0      618, 058         445, 112         359, 804         356203        355, 990
     a0       1.095             1.096           1.097            1.098         1.099
     x0      355, 177         155, 935         155, 907         60, 297       60, 224

                          Table 8: Explicit values for a0 and x0 .

Proof. Let f (x) = x/(log x − 1.0344). Corollary 1 implies that
                                                   x
                                  π(x) <
                                            log x − 1.0344

for every x ≥ 106, 640, 139, 304, 611. If we compare the right-hand side of (8) with
the integral logarithm li(x), we can use Lemma 2 to see that the inequality (8)
also holds for every x with 98, 269, 667, 551, 459 ≤ x ≤ 106, 640, 139, 304, 611. We
conclude by direct computation.

Remark 21. The real number a0 = 1.08366 in Proposition 10 is mostly only of
historical value. On the basis of his study of a limited table of primes, Legendre
stated 1808 (see [46, p. 394]) that π(x) = x/(log x − A(x)), where limx→∞ A(x) =
1.08366. Clearly Legendre’s conjecture is equivalent to (8). However, from (11), it
follows that the best value of limx→∞ A(x) is 1. At this point it should be mentioned
that Panaitopol [52] claimed to have proved the inequality
                                                   x
                                  π(x) <                                                   (62)
                                            log x − 1.08366

for every x > 106 . In Proposition 10, it could be shown that N = 1, 526, 671 is the
smallest possible positive integer so that the inequality (62) holds for every x ≥ N .

  Next, we obtain the following effective estimates for π(x) for the case where
m = 1. The proof is similar to the proof of Proposition 10 and is left to the reader.
INTEGERS: 24 (2024)                                                                    33

Proposition 11. We have
                                                 x
                                 π(x) <                a1
                                          log x − 1 − log x

for every x ≥ x1 , where a1 and x1 are given as in Table 9.

     a1             1.11                   1.1105                     1.111
     x1    62, 998, 850, 942, 976   55, 193, 608, 062, 217    49, 246, 036, 992, 716
     a1            1.112                    1.113                     1.114
     x1    38, 472, 138, 880, 411   30, 658, 643, 813, 468    23, 767, 640, 743, 883
     a1            1.115                    1.116                     1.117
     x1    19, 278, 513, 358, 342   15, 142, 627, 022, 527    12, 279, 648, 138, 508
     a1            1.118                     1.119                     1.12
     x1    9, 684, 114, 630, 824     7, 981, 446, 192, 206    6, 323, 967, 140, 812
     a1            1.121                     1.122                    1.123
     x1    5, 273, 225, 700, 761     4, 170, 462, 893, 841    3, 458, 549, 136, 539
     a1            1.124                     1.125                    1.126
     x1    2, 825, 539, 807, 244     2, 292, 448, 124, 593    1, 903, 596, 231, 542
     a1            1.127                     1.128                    1.129
     x1    1, 573, 767, 234, 188     1, 290, 096, 268, 844    1, 073, 403, 839, 693
     a1            1.13                     1.131                    1.132
     x1     889, 377, 392, 161        782, 989, 678, 664       608, 408, 258, 090
     a1           1.133                     1.134                    1.135
     x1     540, 050, 850, 157        452, 875, 824, 702       373, 479, 021, 700
     a1           1.136                     1.137                    1.138
     x1     335, 562, 521, 091        263, 728, 502, 964       242, 118, 904, 367
     a1           1.139                      1.14                    1.141
     x1     201, 924, 836, 111        161, 054, 192, 492       149, 061, 190, 565
     a1           1.142                     1.143                     1.144
     x1     125, 233, 112, 846        105, 053, 836, 224        86, 061, 321, 374
     a1            1.145                    1.146                     1.147
     x1      77, 278, 924, 451        61, 344, 524, 412         57, 720, 831, 343
     a1            1.148                    1.149                     1.15
     x1      46, 039, 922, 948        42, 575, 222, 481         38, 284, 442, 297

                      Table 9: Explicit values for a1 and x1 .

  Finally, we consider the case where m = 2 and find the following explicit estimates
INTEGERS: 24 (2024)                                                                          34

for π(x). Again, the proof is quite similar to the proof of Proposition 10 and we
leave the details to the reader.

Proposition 12. We have
                                                 x
                          π(x) <
                                    log x − 1 − log1 x − loga22 x

for every x ≥ x2 , where a2 and x2 are given as in Table 10.

     a2            3.49                       3.495                          3.5
     x2    83, 027, 761, 686, 134   63, 024, 307, 127, 421          50, 794, 512, 296, 846
     a2            3.51                       3.52                          3.53
     x2    30, 594, 003, 254, 258   17, 348, 455, 129, 950          11, 655, 963, 556, 138
     a2            3.54                       3.55                          3.56
     x2    5, 539, 984, 798, 515     4, 489, 052, 430, 063          2, 180, 930, 569, 481
     a2            3.57                       3.58                          3.59
     x2    1, 464, 200, 206, 021       882, 055, 689, 961            584, 256, 118, 105
     a2             3.6                       3.61                          3.62
     x2     437, 882, 804, 654         332, 203, 763, 508            201, 890, 631, 296
     a2            3.63                       3.64                          3.65
     x2     148, 632, 348, 138         102, 965, 110, 268             55, 102, 251, 180
     a2            3.66                       3.67                          3.68
     x2      38, 278, 086, 931         24, 178, 954, 639              21, 729, 109, 565

                      Table 10: Explicit values for a2 and x2 .
