arXiv:1809.03134v2 [math.NT] 20 Jul 2020 1]esnilysoe htif that showed essentially [14] fteerrtr ntepienme theorem number prime the in term zeta-function error Riemann the the of of region zero-free The Introduction 1 eraigfnto,te n has one then function, decreasing where ∗ † upre yACDsoeyPoetD10092adESCGra EPSRC and DP160100932 Project Discovery ARC by Supported upre yACDsoeyPoetD10092adACFutur ARC and DP160100932 Project Discovery ARC by Supported hst ogsadn rbe ocriga nqaiyst inequality previo an was concerning which problem that long-standing of a square-root to t this roughly of version term a gives error which Pintz, with of theorem a explicit make We h ro emi h rm ubrtheorem number prime the in term error The h nvriyo e ot ae abra Australia Canberra, Wales South New of University The nvriyo rso,Bitl UK Bristol, Bristol, of University ψ ( ζ ω x ( colo Mathematics, of School ) x [email protected] ( s x [email protected] ioh .Trudgian S. Timothy − a ozre ihra part real with zeroes no has ) =min := ) x colo Science of School ai .Platt J. David ≪ uy2,2020 21, July t ≥ exp 1 Abstract { η  ( 1 t − log ) ψ ( 2 1 x (1 ) ζ x ( ∼ − s log + sitmtl once ihtesize the with connected intimately is ) x ∗ ǫ ) where ω ( † t x } ) .  de yRamanujan. by udied elwhpFT160100094. Fellowship e epienme theorem number prime he ψ tEP/K034383/1. nt , σ sykon eapply We known. usly ( x ≥ = ) 1 − P η p ( m t ,where ), ≤ x log p Ingham . η ( t sa is ) (2) (1) The classical zero-free region has η(t)=1/(R log t), for some positive constant R. This gives t = exp log x/R as the optimal choice for t in (2) and gives { } p ψ(x) x log x − exp (1 ǫ) . (3) x ≪ − − r R ! Explicit versions of (3) have been given in [11, 23, 24, 26]. These have the form

ψ(x) x log x − A(log x)B exp , (x x ), x ≤ − R ≥ 0 r !

B with specific values of A, B and x0. The (log x) factor is not very important, but does give an explicit version of the ǫ in (3). Pintz [18] showed that Ingham’s bound is a substantial overestimate: we can delete the factor of 1/2 in (1), which leads to replacing the 1 ǫ in (3) by 2 ǫ. Therefore, making − − Pintz’s theorem explicit obtains an error term in the theorem that is almost the square-root of the current bound. We prove this in the following theorem.

Theorem 1. Let R =5.573412. For each row X,A,B,C,ǫ from Table 1 we have { 0} ψ(x) x log x B log x − A exp C (4) x ≤ R − R   r !

and ψ(x) x ǫ x | − | ≤ 0 for all log x X. ≥

We also state the following obvious corollary for θ(x)= p x log p. ≤ Corollary 1. For each row X,A,B,C from Table 1 we haveP { } θ(x) x log x B log x − A exp C for all log x X, x ≤ 1 R − R ≥   r !

where A1 = A +0.1. Proof. This follows trivially (and wastefully) from the work of Dusart [11, Cor. 4.5] or the authors [20, Cor. 2].

7 1/3 ψ(x) θ(x) < (1+1.47 10− )√x +1.78x . (5) − ·

2 We remark that the values of ǫ0 in Table 1 improve on those given by Faber and Kadiri in [12] when log x 3000. See also Broadbent et al. [3] for some recent results, and an excellent survey of the≥ field. We also give an explicit estimate of another version of the . Corollary 2. For all log x 2000 we have ≥ π(x) li(x) 235x(log x)0.52 exp( 0.8 log x). (6) | − | ≤ − This improves on results by Dusart [10, Thm 1.12] and byp Trudgian [26, Thm 2] for all log x 2000. We note that, with a little work, the coefficient 0.8 in (6) could be increased but at≥ present not beyond 0.8471 ....

Table 1: Values of X,A,B,C,ǫ0 in Theorem 1. XσABC ǫ0 5 1000 0.98 461.9 1.52 1.89 1.20 10− · 10 2000 0.98 411.4 1.52 1.89 8.35 10− · 13 3000 0.98 379.6 1.52 1.89 4.51 10− · 16 4000 0.98 356.3 1.52 1.89 7.33 10− · 19 5000 0.99 713.0 1.51 1.94 9.77 10− · 21 6000 0.99 611.6 1.51 1.94 4.23 10− · 23 7000 0.99 590.1 1.51 1.94 3.09 10− · 25 8000 0.99 570.5 1.51 1.94 3.12 10− · 27 9000 0.99 552.3 1.51 1.94 4.11 10− · 29 10000 0.99 535.4 1.51 1.94 6.78 10− · We collect some lemmas in 2 that allow us to prove Theorem 1 in 3. We then prove § § Corollary 2 in 4. While there are many applications for Theorem 1 we focus on just one in 5, where we prove§ the following theorem. § Theorem 2. The inequality ex x π2(x) < π log x e   is true for all 38358837682 < x exp(58) and for x exp(3 915). ≤ ≥ Throughout this paper, we use the notation f(x)= ϑ(g(x)) to mean that f(x) g(x). | | ≤ 2 Preparatory lemmas

We start with an explicit version of the explicit formula, given in [8, Thm. 1.3]. Lemma 1. Let 50 e60 and x is half an odd integer. Then

ρ 1 2 ψ(x) x x − 2 log x − = + ϑ . (7) x − ρ T γ

8(1 σ)/3 5 2σ 2 N(σ, T ) C (σ)T − log − T + C (σ) log T. (8) ≤ 1 2 1 Table 1 of [16] gives values for C1 and C2 for various σ, based in part on a result of the first author [19] showing that the Riemann hypothesis holds to height H = 3.06 1010. · A reworking of [16] using Lemma 4 would improve the constants C1 and C2 and hence our results. The log2 T term in (8), while a nuisance to carry through the calculations, is utterly negligible in the final bounds. Finally, we note that Kadiri, Lumley, and Ng produce slightly superior versions of (8), but we have opted for the simpler version presented above.

3 Pintz’s method

We follow the argument given by Pintz [18, pp. 214-215]. We aim to derive a bound valid for all x > x0. To control the ϑ term in (7) we set T = exp(2 log x/R). We then break the sum in (7) into two pieces: those zeroes ρ with ρ < σ for some σ [1/2, 1) to be fixed ℜ p ∈ later depending on x0, and the rest. For the former we have

1 We also note that the first column in Table 1 in [16] should have σ in place of σ0, and that the condition ‘σ σ0’ should be deleted from the caption. We thank Habiba Kadiri for pointing this out to us. ≥ 4 ρ 1 2 x − σ 1 1 σ 1 1 T < x − x − log +1.8642 , (9) ρ γ ≤ 2π 2π γ T γ T | |   ! |Xρ<σ|≤ |X|≤ ℜ

by Lemma 2. Note that we could improve (9) by taking advantage of Lemma 4. Such an alteration to (9) makes only a negligible improvement to the bound in Theorem 1. We turn now to zeroes ρ with β σ. We shall estimate their contribution by using the ≥ 1 zero-free region σ 1 η(t)=1 (R log t)− and the zero-density estimate, but by Lemma 4, we are free to start≥ − counting from− any height H 3 1012. We can therefore write ≤ ·

T 1 ρ 1 R t x − x− log 2 dN(σ, t) ρ ≤ t γ T HZ |Xρ|≤σ ℜ ≥ 1 T 1 x− R log T d x− R log t 2 N(σ, T ) N(σ, t)dt . ≤  T − dt t !  HZ   Now the integrand only becomes negative once t > t0 = exp( log x/R) (which may or may not exceed H) so we can write p

1 T 1 ρ 1 R T R t x − x− log d x− log 2 N(σ, T ) N(σ, t)dt ρ ≤  T − dt t  γ T ! tZ0 |Xρ|≤σ ℜ ≥   1 x− R log T log x 2N(σ, T ) + exp 2 ≤ " T − r R !# 2.0025N(σ, T ) , ≤ T where we have used log x 1000 and T = exp(2 log x/R). We invoke Lemma 5 to give ≥ p

ρ 1 −σ x − 1 8(1 ) 5 2σ 2 2.0025 C (σ)T 3 log − T + C (σ) log T ρ ≤ T 1 2 γ T |Xρ|≤σ h i ℜ ≥ 5 2σ 10 16σ log x log x − 2.0025 C (σ) exp − 2 ≤ 1 3 R R "   r ! r ! log x log x +4C2(σ) exp 2 . − r R ! R #

5 We can now write

1 5 2σ − 10 16σ log x log x − k(σ, x )= exp − 0 0 , 0  3 R R    r ! r !   followed by

log x0 2 C3(σ, x0) = 2 exp 2 log x0 k(σ, x0) − r R !

σ 1 2 log x0 C (σ, x )= x − +1.8642 k(σ, x ) 4 0 0 π R 0   log x0 log x0 C5(σ, x0)=8.01 C2(σ) exp 2 k(σ, x0) · − r R ! R 5 2σ A(σ, x )=2.0025 2 − C (σ)+ C (σ, x )+ C (σ, x )+ C (σ, x ), 0 · · 1 3 0 4 0 5 0 so that we finally reach

5−2σ ψ(x) x log x 2 10 16σ log x − A(σ, x ) exp − , x ≤ 0 R 3 R   r !

for all σ [0.75 , 1) and log x 1000 large enough so that A(σ, x) is decreasing for x > x0. It is possible∈ to optimise the≥ above approach further, by keeping small negative terms in play. We have not done this since, as noted after Lemma 5, such an improvement ought to be coupled with a reworking of the constants C1(σ) and C2(σ) taking into account Lemma 4. These constants may be further improved using results in [6], and then spliced with other zero-density results in [15] and [25]. Finally, the constant ‘2’ in (7), while not significant for Dudek’s application in [8] plays a non-negligible role here and could well be improved. These options are all avenues for future research: we merely indicate here the power of using an explicit zero-density estimate such as (8). Writing 5−2σ log x 2 10 16σ log x ǫ (σ, x )= A(σ, x ) 0 exp − 0 , (10) 0 0 0 R 3 R   r ! we look for the σ that results in the minimum for a given x0, subject to the restriction that we need A(σ, x) to be decreasing for all x > x0. The bounds in (10), having used (7) are only valid when x is half an odd integer. However, as an example, the difference between 441 ǫ (0.98, exp(1000)) and ǫ (0.98, exp(1000) 0.5) is less than 10− . This approach leads 0 0 ⌊ ⌋ − to the values of σ in Table 1 and thus proves Theorem 1. We note that it seems likely that, for larger values of x0, taking σ larger than 0.99 would give an improvement, whence any extension to Table 1 of [16] would be useful in this regard.

6 4 Proof of Corollary 2

One can also consider the prime number theorem in the form π(x) li(x), where li(x)= x 1 ∼ 2 (log t)− dt. With a little more effort we could provide a version of Corollary 2 with tables of parameters, in the style of Theorem 1. R We begin by using partial summation and integration by parts to yield

θ(x) x 2 x θ(t) t π(x) li(x)= − + + − dt. (11) − log x log 2 t log2 t Z2 Let us write

1000 2000 x θ(t) t 563 e e x −2 dt = + + + = I1 + I2 + I3 + I4. t log t 1000 2000 Z2 Z2 Z563 Ze Ze

We estimate the integral I1 numerically. The combination of this and the 2/ log 2 term contributes at most 7.6 to (11). For I2 we use a result of Rosser and Schoenfeld [22], namely that for x 563 we have θ(x) x x/(2 log x). Although stronger bounds are known, we can afford≥ to be very | − | ≤ cavalier: the values of I1 and I2 are insignificant for large x. 5 For I3 we use Theorem 1 and (5). These show that that θ(x) x 1.2 10− x for x e1000. | − | ≤ · ≥ For I4 we follow the approach from [10]. For some α to be determined later, define

t exp C (log t)/R h(t)= {− α }. logp t We want to show that

exp C (log t)/R 2000 {− 2 B } h′(t), (t e ). (12) logp− t ≤ ≥ Since then, if (12) is true, we have

x x B exp C (log t)/R B B I4 A1R− {− 2 B } dt A1R− h′(t) < A1R− h(x). | | ≤ 2000 log − t ≤ 2000 Ze p Ze Now, to show that (12) is true it is sufficient to show that

B 1+α C 2000 log t (log t) − (log t)/R α> 0, (t e ). (13) − − 2 − ≥ p We therefore end up with

xA (log x)B 1 π(x) li(x) 1 − exp( C (log x)/R) 1 + ∆ , (x x ), | − | ≤ RB − { } ≥ 0 p 7 where

1 B α ∆ =(log x0) − −

B log x0 1 B R exp C R (log x0) − e1000 e2000 dt 5 dt +  q  7.6+ 3 +1.2 10− 2 . A1x0 2 log t · 1000 log t ( Z563 Ze ) (14) We require that 1 B<α< 2 B so that the first addend in (14) is decreasing, and so − − 2000 that (13) may be satisfied. We now take x0 = e and α = 0.47. We verify that (13) is true, and we find by (14) that ∆ 6.76. This proves the corollary. We briefly illustrate how one may≤ improve upon the bound (6), and to complete such a proof for all values of x 10, say. We can partition the interval [2, x] using the points 2, 563 ≥ and the entries in the first column of Table 1. Between consecutive values in Table 1 we use the ǫ0 coming from the smaller value. Indeed, we can take the smallest available ǫ0’s from our Table 1, Faber and Kadiri [12], and the results of B¨uthe [4, 5]. We therefore have

1000 xǫ (x ) 2 563 θ(t) t 1 e dt π(x) li(x) := E(x)= 0 0 + + | − | dt + − log x log 2 t log2 t 2 log3 t Z2 Z563 e1500 e9500 x 1000 dt 9000 dt 9500 dt + ǫ0(e ) 2 + + ǫ0(e ) 2 + ǫ0(e ) 2 . 1000 log t ··· 9000 log t 9500 log t Ze Ze Ze Note that E(x) and the right-side of (6) are increasing in x. Therefore, if we wish to verify (6) in the range x x x we need only show that 0 ≤ ≤ 1 x A (log x )B 1 E(x ) 1.001 0 1 0 − exp( C (log x )/R). 1 ≤ RB − 0

p k k+0.1 Choosing pairs of points (x0, x1) sufficiently close together — e.g., x0 = e and x1 = e — and populating more entries in the relevant tables of ǫ0’s enables us to extend Corollary 2 to ‘moderate’ values of x. To extend it to ‘small’ values of x, we may invoke a result by B¨uthe [4], namely, that

√x 3.9 19.5 π(x) li(x) 1.95+ + , (x 1019). | − | ≤ log x log x log2 x ≤   We expect that following such a procedure, after taking advantage of the suggestions made at the end of 3 would give a ‘state of the art’ version of Corollary 2. § 5 Application to Ramanujan’s inequality

x 1 Using π(x) li(x)= (log t)− dt, and integrating by parts, Ramanujan noted that ∼ 2 R ex x π2(x) < π , (15) log x e   8 for sufficiently large x — see [2, Ch. 24]. It is an interesting, and difficult, problem to determine the last x for which (15) fails. Dudek and Platt [9] showed2 that (15) is true for all x exp(9658). This has recently ≥ been improved by Axler [1] to x exp(9032). Dudek and Platt gave good evidence that x = 38358837682 is the largest x≥for which (15) fails, and indeed, they showed this to be so under the Riemann hypothesis. The main obstacle in moving to an unconditional version is the size of the error term in the prime number theorem. The problem is therefore well within our wheelhouse given our result in Theorem 1. As an introduction to the method given by Dudek and Platt in [9] consider the expansion

4 j! x π(x)= x j+1 + O 6 . (16) j=0 log x log x X   Expanding, and keeping track of sufficiently many terms then gives us

ex x x2 x2 π2(x) π = + O , − log x e −log6 x log7     which is clearly negative for all x sufficiently large. Dudek and Platt then give an explicit version of (16) in their Lemma 2.1. This replaces the O(x/ log6 x) term in (16) by upper and lower bounds with constants ma and Ma (which we give in (19) and (20)), and a criterion for x being sufficiently large. Rather than duplicate the proof, we have included the critical pieces below. We need an a(x) such that

θ(x) x log5 x xa(x). (17) | − | ≤ By partial summation, and integrating li(x) by parts once, we have, from (11) and (17), that

x x x x dt a(t) π(x) + a(x) + + dt ≤ log x log6 x log2 t log7 t Z2 Z2 and x x x x dt a(t) π(x) a(x) + dt. ≥ log x − log6 x log2 t − log7 t Z2 Z2 We start with a lemma, which immediately gives us the lower bound in Theorem 2.

Lemma 6. For x (599, exp(58)] we have ∈ √x θ(x) x log2 x. | − | ≤ 8π 2We are grateful to Christian Axler who identified an error in the proof given in [9]. Fortunately it was easy to fix.

9 Proof. This follows directly from Theorem 2 of [4] coupled with Lemma 4. We are now free to use our improved bounds. We use

2 log 2 −2 2 < x 599 2 ≤ log x 599 < x exp(58)  8π√x ≤  1  8 log x 4 log x  17π 6.455 exp 6.455 exp(58) 0 so that a(x) is non-increasing for all x x and write ≥ a

xa xa 6 6 6 5 log xa 720 + a(t) log xa 720 a(t) log xa k! C1 = 7 dt, C2 = −7 dt, C3 =2 k+1 , xa log t xa log t xa log 2 Z2 Z2 Xk=1 8 6 1 7 2 7 log xa Ma(x) =120+ a(x)+ C1 + (720+ a(xa)) + ·2 + 8 , (19) log xa log x x log 2  a √ a  8 6 1 7 2 7 log xa ma(x) = 120 a(x) (C2 + C3) a(xa) + ·2 + 8 , (20) − − − log xa log x x log 2  a √ a  2 2Ma(x)+132 4Ma(x)+288 12Ma(x)+576 48Ma(x) Ma(x) ǫMa (x) = 72+2Ma(x)+ + + + + log x log2 x log3 x log4 x log5 x and 364 381 238 97 30 8 ǫma (x) =206+ ma(x)+ + + + + + . log x log2 x log3 x log4 x log5 x log6 x

We now need only find an x > xa such that

ǫ a (x) ǫ a (x) < log x. M − m 10 Choosing xa = exp(3914) and x = exp(3 915) will suffice, yielding the upper bound in Theorem 2.

Acknowledgements We wish to thank Habiba Kadiri, Nathan Ng, and Allysa Lumley, and the referees for very useful feedback.

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