YANNICK SAOUTER AND PATRICK DEMICHEL

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1 MATHEMATICS OF COMPUTATION Volume 79, Number 272, October 200, Pages S (0) Article electronically published on April 4, 200 A SHARP REGION WHERE π() li() IS POSITIVE YANNICK SAOUTER AND PATRICK DEMICHEL Abstract. In this article, we study the problem of changes of sign of π() li(). We provide three improvements. First, we give better esimates of error term for Lehman s theorem. Second, we rigorously prove the positivity of this difference for a region formerly conjectured by Patrick Demichel. Third, we improve the estimates for regions of positivity by using number theoretic results.. Previous work The problem of estimating the number of prime numbers goes back to Gauss. The function counting prime numbers is classically denoted π, i.e. π() = p. In 79, he conjectured that π() log. This result was proven in 896 by Hadamard and de la Vallée-Poussin. { In 849, Gauss suggested that the log-integral ε dt function, defined by li() = lim ε 0 0 log t + } dt +ε log t, should give a better approimation for π. Gauss also noted that the inequality π() < li() holds for any in the interval [2, ]. Since then, this property has been checked numerically up to 0 4 []. On the other hand, Littlewood proved in 94 that the property π() > li() holds infinitely often, but he did not give any eplicit value for such an. In 933, Skewes proved, assuming the Riemann hypothesis, that the latter inequality occurs at least once for a value < A great improvement was given in 966 by Lehman [2]. He established a theorem which enables one unconditionally to obtain much lower values. His theorem enabled him to show that there eists a region near where the difference π() li() admits positive values. Then, te Riele [3] in 987 discovered another region near ,andBays and Hudson [4] ehibited a region near in 999. In 2006, Chao and Plymen [5] gave an improvement on the error terms of Lehman s theorem. This enabled them to sharpen Bays and Hudson s region and established a new lower bound equal to Independently, in 2005, Demichel [6] made intensive computations on this problem and conjectured that this value could be improved to , without rigorously establishing the result. Another point is that this latter region is a new one, i.e. it is not included in that of Bays and Hudson, contrary to the result of Chao and Plymen. Our contribution to this problem is three-fold. First, we will show that Chao and Plymen s error term can be lowered. Second, by numerical computation, we will prove the validity of Demichel s region. Third, we Received by the editor January 8, 2009 and, in revised form, May 4, 2009 and July 3, Mathematics Subject Classification. Primary -04, A4, M26, N05, Y, Y35. c 200 American Mathematical Society Reverts to public domain 28 years from publication 2395

2 2396 YANNICK SAOUTER AND PATRICK DEMICHEL show, using some theorems in number theory, that the final results obtained by the classic approach to this subject can be improved. 2. Lehman s theorem The main tool to deal with this problem is Lehman s theorem: Theorem 2. (Lehman s Theorem). Let A be a positive number such that β = 2 for all zeros ρ = β + iγ of Riemann zeta function ζ(s) for which 0 <γ A. Letα, η and ω be positive values such that ω η> and the following conditions hold: (2.) 4A/ω α A 2, (2.2) 2A/α η ω/2. Let α (2.3) K(y) = 2 /2, (2.4) I(ω, η) = 2π e αy ω η Then for 2πe < T A, we have (2.5) I(ω, η) = K(u ω)ue u/2 {π(e u ) li(e u )}du. 0< γ T e iγω where R S + S 2 + S 3 + S 4 + S 5 + S 6,with (2.6) S = 3 ω η +4(ω + η)e (ω η)/6, (2.7) /2 S 2 = 2e αη2, 2παη ρ 2 e γ /2α + R (2.8) S 3 =0.08 αe αη2 /2, { S 4 = e T 2 /2α α πt 2 log T 2π +8log T + 4α } (2.9) T T 3, (2.0) S 5 = 0.05 ω η, (2.) S 6 = A log Ae A2 /2α+(ω+η)/2 {4α /2 +5η}. If the Riemann hypothesis holds, then conditions (2.) and (2.2) may be omitted and the term S 6 may be omitted in the upper bound for R. The complete proof can be found in [2]. The application of the previous theorem makes two essential assumptions. First, the Riemann hypothesis has to be checked up to height A. Second, eplicit values for the zeros of ζ have to be known up to height T. If we suppose that both conditions are met, we can estimate the integral (2.4) using the equation (2.5). Lehman s method amounts then to finding suitable values for α and ω such that the first two terms on the right-hand side of equation (2.5) sum to a positive value larger than the associated error term R. The integral (2.4) is then established to be positive and thus, by virtue of the positivity of K, the term {π(e u ) li(e u )} must admit some positive values for u in the interval [ω η, ω + η].

3 A SHARP REGION WHERE π() li() IS POSITIVE Improvements Improvements on the error term R are possible. In fact, the dominating term in R is generally S. In his seminal work, Lehman [2] derived S from an upper bound for π() obtained by Rosser and Schoenfeld [7]. In their paper, Chao and Plymen derived a tighter bound by using recent results of Panaitopol [8]. Doing so, they could lower the constant 3 in the first term of S to In this part, we use one result obtained by Dusart [9, th..0] to show that this constant can be replaced by 2 (with some other terms in S ). Moreover, this value cannot be improved. Theorem 3. (Dusart s Theorem). If 32299, we have ( (3.) + log log +.8 ) log 2 π(). If 35599, we have (3.2) π() ( + log log ) log 2. Given this result, we prove the following theorem: Theorem 3.2. Under the hypothesis of Lehman s theorem and if ω η>25.57, equation (2.5) still holds if S is replaced by (3.3) S = 2 ω η (ω η) 2 +log2.(ω + η)e (ω η)/2 + 2 log 2 (ω + η)e (ω η)/6. Proof. We proceed as in Lehman s work. Let (3.4) Π() =π()+ 2 π(/2 )+ 3 π(/3 )+..., and let { } (3.5) Π 0 () = lim Π( + ε)+π( ε). ε 0 2 The Riemann-von Mangoldt formula states that for >, (3.6) Π 0 () = li() + li( ρ du )+ ρ (u 2 log 2, )u log u where ρ runs over the zeros of function ζ in the critical strip. We have (3.7) 2 π(/2 )+ 3 π(/3 ) π(/2 )+ log 3 π(/3 ). log 2 Then we use Dusart s theorem together with the classic bound π() Thus, if ,wehave (3.8) 2 π(/2 )+ ( 3 π(/3 )+... /2 + 2 log log ) log log 2 +2 log 2 and thus (3.9) 2 π(/2 )+ ( 3 π(/3 )+... /2 + 2 log log ) log log 2 /3. 2 log. /3 log,

4 2398 YANNICK SAOUTER AND PATRICK DEMICHEL Substituting in (3.6), we have (3.0) π() li() ( li( ρ ) /2 + 2 log log ) log 2 2 log 2 /3 log 2. ρ We put = e u,andthenifu>25.57, we have (3.) ue u/2 (π(e u ) li(e u )) ρ ue u/2 li(e uρ ) 2 u 0.04 u 2 2u log 2 e u/6 log 2.ue u/2. So, following Lehman s proof, we derive equation (2.5) with the same bounding terms S 2, S 3, S 4, S 5 and S 6.TermS 2 comes from bounding the two tail integrals ω η K(u ω)du and + ω+η K(u ω)du. Terms S 3, S 4, S 5 and S 6 come from the estimate of ρ ω η K(u ω)ue u/2 li(e uρ )du. Finally, the term corresponding to term S in our theorem comes from bounding the epression ( 2 (3.2) J = K(u ω) ω η u u 2 + 2u ) log 2 e u/6 +log2.ue u/2 du. Both terms in the previous integral are positive and + K(y)dy =,thuswehave (3.3) J 2 ω η (ω + η) + (ω η) 2 log 2 e (ω η)/6 +log 2.(ω + η)e (ω η)/2. 4. Numerical results As mentioned previously, the use of the previous theorems presupposes numerical verifications of the Riemann hypothesis up to height A. In his seminal paper, Lehman used a verification made on his own on the first zeros, giving A = Since then, a lot of work has been done to check numerically the Riemann hypothesis up to larger and larger heights. In 200, van de Lune [0] established that the conjecture is verified for the first zeros up to height A = This value, in fact, sets an upper bound for the value of A that can be used in Lehman s theorem. However, two more recent verifications are noteworthy. The first was performed by Gourdon and Demichel [] in 2004 using a fast multiple evaluation algorithm for ζ invented by Odlyzko. With their implementation, the conjecture has been verified up to the 0 3 -th zero. The second is the distributed ZetaGrid project [2], managed by Wedeniwski, which was active between 2002 and The official status of these verifications is not clear: Gourdon and Demichel s work has never been independently verified and, in the case of the ZetaGrid project, it was not established that all zeros were checked. For 0 <T A, we know that the real part of zeros ρ = β + iγ of ζ such that γ <T is equal to /2. Moreover, zeros of ζ in the critical strip occur as conjugate pairs, so the sum to evaluate is: (4.) 0< γ T e iγω ρ 2 e γ /2α = 0<γ T cos γω +2γsin γω 4 + e γ 2 /2α. γ2 For our numerical computations, we computed the first 22 million zeros of ζ. This was done in two phases. First, an approimation was computed by the classic

5 A SHARP REGION WHERE π() li() IS POSITIVE Figure. The Bays and Hudson region. Riemann-Siegel formula and then precision was improved up to 9 decimal digits using correction terms in this formula. The three additional correction terms we used were computed by formulae given in [3]. The last zero of our database gave us T = The relative precision that can be epected when computing the right-hand side of the previous equation is then bounded by ( ) ΔI = 0 9 cos γω +2γsin γω. 2 /2α (4.2). γ 0<γ T 4 + e γ γ2 In our computations, we set α =6 0 2, and in the range of our application we have ω We compute the associated precision given by equation (4.2), and it gives approimately We chose to make the computation eplicitly, instead of using bounds as in previous work, in order to obtain the best possible precision. Figure plots the value of I(ω, η) forω in [ , ]. Values are corrected by the error term R. This figure shows Bays and Hudson s region whose center is a bit larger than ω = In this figure, Chao and Plymen s region as well as our region are contained in the peak between and At the level of magnification of this figure, it is not very clear that the curve effectively cuts the line I(ω, η) = 0. Figure 2 depicts Chao and Plymen s region with a better level of magnification. In this figure, quite large areas of the curve are in the positive domain. Figure 3 depicts the new region with the same scale level as for Figure 2. The positive region is much sharper than the one of Figure 2. By looking at the ω ais, we notice that this region is left of Chao and Plymen s one. Finally, Figure 4 gives a closeup of the peak of the new region. Numerically, the least value for ω giving a positive value for I(ω, η) isω = By studying the remainder terms and especially S 6, we found that A = is the value minimizing the interval length. We then have η =2A/α = and conditions (2.) and (2.2) are met. Moreover, the condition ω η>25.57 of Theorem 3.2 is also verified. By computation, we then obtain 0< γ T eiγω /2α ρ e γ2 = , thus giving I (ω, η) = as an estimate of I(ω, η). We then have I(ω, η) I (ω, η) ΔI S S 2 S 3 S 4 S 5 S 6.

6 2400 YANNICK SAOUTER AND PATRICK DEMICHEL Figure 2. The Chao and Plymen region Figure 3. The new region on the same scale e e 06 5e e 06 6e e 06 7e e 06 8e 06 Figure 4. The maimum in the new region, scaled up.

7 A SHARP REGION WHERE π() li() IS POSITIVE 240 Numerically, we have ΔI = , S = , S 2 = , S 3 = , S 4 = , S 5 = , S 6 = Thus we obtain (4.3) I(ω, η) Thus, we proved that there eists a value in [ep( ), ep( )] for which the inequality π() > li() holds. More precisely, there are some values u in the interval [ , ] such that (4.4) π(e u ) li(e u ) > e u/2 /u > So, we can claim: Theorem 4.. There eists at least one value in the interval [ep( ), ep( )] for which π() > li() holds. Moreover, there are more than successive integers in the vicinity of ep( ) where the inequality holds. 5. Sharpening the interval The previous theorem gives us an upper bound of ep( ) for the first crossover. This value is better than the one obtained by Chao and Plymen. Nevertheless, it is possible to reduce again the length of the interval. Indeed, the integrand function decays very fast to 0 around its center and thus the meaningful part of the integral is in fact around ω. In order to reduce the interval, we need some information about the growth of π() li(). At this point, we split the study into two cases. First, we will consider the general case and second, we will suppose that the Riemann hypothesis holds. In the general case, we will first prove: Theorem 5.. If ep(8), we have (5.) 0 li() ( + log log + 2 ) log 2 C 2 log 4 + C 2, with C =li(2) 2 log 2 ( + log log 2 2 ) and C 2 = ep(8) 48dt 2 log 5 t 24 log 4 2. Proof. From the definition of li(), we have for 2, dt (5.2) li() =li(2)+ 2 log t. Then, after three successive integrations by parts, we have for 2, [ ( t (5.3) li() = li(2)+ + log t log t + 2 )] 6dt log 2 + t log 4 t, 2 2

8 2402 YANNICK SAOUTER AND PATRICK DEMICHEL and thus (5.4) li() ( + log log + 2 ) 6dt log 2 C = 2 log 4 t. At this point we obtain the first inequality of Theorem 5.. Another integration by parts on the right-hand side gives, for 2, [ ] 6dt 6t (5.5) 2 log 4 t = 24dt log 4 + t 2 2 log 5 t. Now, for ep(8), we have [ ] 6dt 6t ep(8) (5.6) 2 log 4 t 24dt log 4 t 2 2 log 5 t = 24dt ep(8) log 5 t. 24 But, for t ep(8), we have log 5 t So we obtain, for ep(8), log 4 t [ ] 6dt 6t ep(8) (5.7) 2 log 4 t 24dt log 4 t 2 2 log 5 t 6dt 2 ep(8) log 4 t 6dt 2 2 log 4 t. We obtain finally for ep(8), 6dt (5.8) 2 log 4 t 2 log 4 24 log which establishes the theorem. ep(8) 2 48dt log 5 t, The latter theorem could be further optimized but it will suffice for our purpose. Combined with Theorem 3., it gives: Theorem 5.2. If 35599, we have (5.9) 0.2 log 3 () π() li() 0.5 log 3 () log 4 () Moreover, if ep(40), then (5.0) π() li() 0.5 log 3 () Although this theorem was obtained by elementary methods, in the range of values for where we intend to use it, it gives a finer result than earlier theorems; for instance, that of Dusart [9, th..2]. With this theorem, we will now study the tail parts of the integral (2.4). Now, let η 0 be a real positive number such that η 0 <η. We then have, since ω>40, K(u ω)ue u/2 {π(e u ) li(e u )}du ω+η 0 (5.) K(u ω)ue u/2 { π(e u ) li(e u ) }du ω+η 0 { } 0.5e K(u ω)ue u/2 u ω+η 0 u du { } 0.5e u/2 K(u ω) ω+η 0 u ue u/2 du (η η 0 )K(η 0 ) {0.5 e(ω+η)/2 (ω + η 0 ) (ω + η)e (ω+η 0)/2 }.

9 A SHARP REGION WHERE π() li() IS POSITIVE 2403 Likewise, we obtain, since ω η > 40, ω η0 K(u ω)ue u/2 {π(e u ) li(e u )}du ω η (5.2) } (η η 0 )K( η 0 ) {0.5 e(ω η 0)/2 (ω η) (ω η 0)e (ω η)/2. We denote, respectively, T and T 2, the right-hand sides of the two previous final inequalities. The sum of the two tail integrals is then bounded above by T +T 2.Now, numerically, with the previous values we used and obtained in our computations, if we set η 0 = η/2.074, we obtain T = , T 2 = Those numeric values, together with the estimate (4.3), gives the following result: I(ω, η 0 ) This result then allows us to obtain a result finer than the one obtained in Theorem 4.. However, as we will see in the net part, more work can still be done to improve the final result. Thus, for the moment, we will state our result in a different way: Theorem 5.3. There eists one value in the interval [ep( ), ep( )] such that π() li() > In the following, we assume that the Riemann hypothesis holds. In this case, Theorem 5.2 is far from being optimal, even in the range of values we consider. A much finer result is given by Schoenfeld [4, p. 339]: Theorem 5.4. If the Riemann hypothesis holds, then for 2657, we have (5.3) π() li() < log. 8π If we denote T and T 2 upper bounds of the corresponding tail integrals, we then obtain (5.4) T = 8π K(η 0)(ω + η) 2, (5.5) T 2 = 8π K( η 0)(ω η 0 ) 2. Numerically, if we set η 0 = η/6.72, we obtain T = , T 2 = , and thus, I(ω, η 0 ) We can then state: Theorem 5.5. If the Riemann hypothesis holds, then there eists one value in the interval [ep( ), ep( )] such that π() li() >

10 2404 YANNICK SAOUTER AND PATRICK DEMICHEL 6. Interval of positivity Theorem 4. ehibits an interval of consecutive integers where π() li() is positive. Indeed, equation (4.4) states that there eists a point, such that π() li() > Let b be a positive integer. Then li( b) li(). Moreover, for any, we have π( ) π(), thus by recurrence and with the previous inequality, we can deduce that π( b) li( b) > b. Thus we can affirm that the successive integers preceding belong to the interval of positivity. This result is obtained by considering integers inferior to. However, as we will see, this result can be much improved by considering integers greater than. Infact,wehave: Theorem 6.. Let > and y>0, then we have +y dt li( + y) li() = log t y < log. With the previous theorem, we can state: Theorem 6.2. Let be a real positive number such that π() li() =A>0. Then if y is a real number such that 0 <y<a.log, we have π( + y) li( + y) > 0. Proof. Let y>0, since the function π() is increasing, we then have π( + y) li( + y) =(π( + y) π()) + (π() li()) + (li() li( + y)) >A y log. Then Theorem 5.3 enables us to state: Theorem 6.3. There are at least consecutive integers in the interval [ep( ), ep( )] such that π() li() > 0. The value obtained is a direct application of Theorem 6.2. However, we do not know where the first lies in the interval of Theorem 5.3. We only know that its maimal value is ep( ). But we have ep( ) ep( ) , which is larger than Thus we can affirm that the integers following belong to the interval [ep( ), ep( )]. In a similar way, Theorem 5.5 gives us the following result: Theorem 6.4. If the Riemann hypothesis holds, then there are at least consecutive integers in the interval [ep( ), ep( )] such that π() li() > 0. A final remark is in order about Theorem 6.4. The number of consecutive integers satisfying π() li() > 0 is approimately 50 times smaller than in Theorem 6.3. This theorem might then appear weaker, which would then make assuming the Riemann hypothesis pointless. In fact, this theorem is stronger than Theorem 6.3, since the length of the interval is three times shorter. The difference in terms of consecutive integers comes from the fact that the estimate for I(ω, η 0 ) is much sharper when the Riemann hypothesis holds. In turn, this fact is a consequence of having better upper bounds for tail integrals (see equations (5.) (5.2) and (5.4) (5.5)).

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