On the distribution (mod 1) of the normalized zeros of the Riemann Zeta-function

ON THE DISTRIBUTION (MOD 1) OF THE NORMALIZED ZEROS OF THE RIEMANN ZETA-FUNCTION J. ARIAS DE REYNA Abstract. We consider the problem whether the ordinates of the non-trivial zeros of ζ(s) are uniformly distributed modulo the Gram points, or equivalently, if the normalized zeros (xn) are uniformly distributed modulo 1. Odlyzko conjectured this to be true. This is far from being proved, even assuming the Riemann hypothesis (RH, for short). Applying the Piatetski-Shapiro 11/12 Theorem we are able to show that, for 0 < κ < 6/5, the mean value 1 NPn≤Nexp(2πiκxn) tends to zero. The case κ= 1 is especially interesting. In this case the Prime Number Theorem is sufficient to prove that the mean value is 0, but the rate of convergence is slower than for other values of κ. Also the case κ= 1 seems to contradict the behavior of the first two million zeros of ζ(s). We make an effort not to use the RH. So our Theorems are absolute. We also put forward the interesting question: will the uniform distribution of the normalized zeros be compatible with the GUE hypothesis? Let ρ=1 2+iα run through the complex zeros of zeta. We do not assume the RH so that αmay be complex. For 0 <κ<6 5we prove that lim T→∞ 1 N(T)X 0<Re α≤T e2iκϑ(α)= 0 where ϑ(t) is the phase of ζ(1 2+it) = e−iϑ(t)Z(t). 1. Introduction. This paper deals with the distribution of the ordinates of the non-trivial zeros of ζ(s). This distribution has received much attention. On the assumption of the RH Rademacher [18] (also [17, p. 434–459]) showed that the sequence (γn) is uniformly distributed modulo 1. Here ρn=βn+iγnare the zeros of ζ(s) in the upper half plane, counted with multiplicity, ordered by increasing γn, and ties being broken by ordering βnfrom smallest to largest. Given a sequence of non-negative real numbers (γn) and another strictly increasing sequence of non-negative real numbers (gn) with limngn=∞, it is said that the (γn) are uniformly distributed modulo (gn) if the numbers yndefined by yn=γn−gm gm+1 −gm ,where γn∈[gm, gm+1) are uniformly distributed in [0,1], (compare [6, p. 4]). Our problem is whether the ordinates (γn) of the zeros of ζ(s) are uniformly distributed modulo the Gram points (gn). In a non published report, Odlyzko [14, p. 60] conjectured Date: March 28, 2014. Author supported by MINECO grant MTM2012-30748. 1 arXiv:1403.7126v1 [math.NT] 27 Mar 2014 that the ordinates of the zeros γnare not related to the Gram points for γnlarge. Hence he conjectured that the normalized zeros will be uniformly distributed modulo 1. This is far from being proved, even assuming the RH. Since the Gram points are defined by the equation πn =ϑ(gn), our problem is equivalent to the question as to whether the normalized zeros xn:= 1 πϑ(γn)≈γn 2πlog γn 2πe are uniformly distributed mod 1. By the asymptotic properties of ϑ(t) this is equivalent to the uniform distribution mod 1 of the numbers γn 2πlog γn 2πe (see Theorem 1.2 in [6, p.3]). The question is especially interesting because the normalized zeros have average spacing 1. Hardy and Littlewood [7, p. 162–177] proved that for any a > 0, uniformly for x∈J, where Jis any compact interval of positive numbers, we have X 0<γ<T eaρ log(−iρ)xρρ−a 2=O(T1 2+a 2) from which, assuming the RH, they derive that for any a,θ > 0, we have X 0<γ<T eaiγ log(γθ)=O(T1 2+a 2). Recall that gn∼2πn/ log n. Fujii [3] proved that for any a > 0 and b > 0, the sequence (γn) is uniformly distributed modulo ((log n)a·bn/ log n). In [4] Fujii proves, under the RH, that for any positive κand awe have X 0<γ≤T eiκγ log κγ 2πea =−eπi 4√a κX n≤(κT 2πa )κ Λ(n) n1 2−1 2κ e−2πian 1 κ+O(Tκ 2log Tlog log T) + O(T1 2−κ 2log T). Our main result is an absolute version of this equation. Since we do not assume the RH, we define the numbers αin such a way that the zeros of zeta are given by ρ=1 2+iα. Here the numbers αmay be (non-real) complex numbers, if the Riemann hypothesis fails. Then we show without any assumption the following. Theorem 3.1. For κ > 0we have (1) X 0<Re α≤T e2iκϑ(α)=−eπi 4(1−κ) √κX n≤(T/2π)κ Λ(n) n1 2−1 2κ e−2πiκn1/κ +Oκ(T1−κ 2log T)+Oκ(Tκ 2log2T). Applying, for κ > 1, the Piatetski-Shapiro 11/12 Theorem, we get lim T→∞ 1 N(T)X 0<Re α≤T e2iκϑ(α)= 0,0< κ < 6 5. The case κ= 1 is especially interesting, since it seems to contradict the behavior of the first two million zeros of ζ(s). Our results are also very similar to some of Schoißengeier [20], who extended the cited analysis of Hardy and Littlewood. In the case κ= 1 our formula gives the following 2 Corollary. The Von Mangoldt function can be approximated in the following way ψ(T/2π) = X n≤T/2π Λ(n) = −X 0<Re α≤T e2iϑ(α)+O(T1 2log2T). This is equivalent to one of the results of Schoißengeier, but obtained here by a simpler analysis. 1.1. Computational data. Our interest in the distribution of the γnwith respect to the Gram points originates from the observation that the angle of the curve Re ζ(s) = 0 at a zero 1 2+iγ (on the critical line) with the positive real axis is equal to ϑ(γ) mod π. Odlyzko’s list of first 2,001,052 zeros of zeta has been used to generate the following pictures of the distribution of 1 πϑ(γ) mod 1. 0.2 0.4 0.6 0.8 1.0 0.9 1.0 1.1 0.2 0.4 0.6 0.8 1.0 0.9 1.0 1.1 Figure 1. Distribution mod 1 of the normalized first 2 001 052 zeros. The curve we have drawn, approximately fitting the data, is the density function ρ(x) = 1−3 17 cos(2πx). If the RH is true so that αn=γnfor all nand this initial distribution is maintained then we have lim T 1 N(T)X 0<Re α≤T e2iϑ(α)= lim N 1 N N X n=1 e2πixn≈Z1 0 e2πix1−3 17 cos 2πxdx =−3 34. Hence, our Corollary 3.2 shows that the trend in the behavior of γnseen in the above figures will not be maintained for larger values of T. Odlyzko observed that the distribution of the normalized zeros is nearer to a uniform distribution for higher zeros. But this seems to be at odds with Titchmarsh [21, Theorem 10.6] who shows that the mean values of Z(g2n) and Z(g2n+1) are equal to 2 and −2, respectively. Hence we ask the question: It is true that the normalized zeros are uniformly distributed modulo 1? We also comment here about another conjecture regarding the distribution of the zeros of zeta. Assuming the RH, Montgomery [13] proved his result about the correlation of pairs of zeros and stated his pair correlation conjecture. 3 The differences xn+1 −xnof the normalized zeros satisfy xn+1 −xn=1 πZγn+1 γn ϑ0(t)dt. Since ϑ0(t) = 1 2log t 2π+O(t−2) we have xn+1 −xn≈γn+1 −γn π 1 2log γn 2π=δn. So, Montgomery’s conjecture is also a conjecture about our normalized zeros. A natural question here is: Is the pair correlation conjecture compatible with a uniform distribution (mod 1) of the normalized zeros? 2. Notations and tools. When possible we follow the standard notations. As in Titchmarsh [21, section 9.4] the zeros β+iγ of ζ(s) with γ > 0 are arranged in a sequence ρn=βn+iγnso that γn+1 ≥γn. We will not assume the Riemann hypothesis (RH for short), and following Riemann [19] define αnby ρn=1 2+iαn. The numbers αnare the zeros of Ξ(t) with positive real part. The RH is equivalent to the equality αn=γnfor all natural numbers n. We denote by N(T), where T > 0, the number of zeros of ζ(s) in the rectangle 0 < σ < 1, 0 ≤t≤T. The functional equation of ζ(s) can be written as (2) ζ(s) = χ(s)ζ(1 −s), χ(s) = πs−1 2Γ(1 2−1 2s) Γ(1 2s)= 2s−1πssec 1 2πs/Γ(s). For treal we have |χ(1 2+it)|= 1, and there exist two real and real analytic functions ϑ(t) and Z(t) (see Titchmarsh [21, section 4.17]) such that (3) ζ(1 2+it) = e−iϑ(t)Z(t), χ(1 2+it) = e−2iϑ(t). The function χ(s) has poles for s= 2n+ 1 with n= 0, 1, . . . and zeros for s=−2n. Let Ω be the complex plane Cwith two cuts along the half-lines (−∞,0] and [1,∞). The function χ(s) is analytic on the simply connected Ω and does not vanish there. So we may define log χ(s) in Ω in such a way that (4) −log χ(1 2+it) = 2iϑ(t), ϑ(0) = 0. The function ϑ(t) is extended in this way to all −i(Ω −1 2) as an analytic function. Also we fix the meaning of (5) χ(s)−κ:= e−κlog χ(s)=e2iκϑ(τ)where s=1 2+iτ ∈Ω. Definition 2.1. For any non-trivial zero ρ=β+iγ =1 2+iα we define the normalized zero as (6) x=1 πϑ(α). Also let xnbe the normalized zero corresponding to ρn=βn+iγn. 4 The function ϑ(t) is strictly increasing for t > 6.28984 . . . For integral k≥ −1 the Gram point t=gk(>7) is defined as the unique solution of ϑ(t) = kπ (see [2, p. 126]). In the interval [0, T] there are approximately as many Gram points as zeros β+iγ of ζ(s) with 0 < γ ≤T. Gram’s “law” (to which there are many exceptions) states that in each Gram interval (gn, gn+1) there is a zero of ζ(s). If Gram’s law were generally true, then the RH would be true, the zeros would be simple and γnwould be an element of (gn−2, gn−1). Of course Gram’s law is not true, but it is still a good heuristic to locate the zeros of ζ(s) for relatively small twhich can be reached by our computers. Also, it is well known that in each interval (T, T + 1), with T≥2 we can select a number T0such that if γis the ordinate of any zero of ζ(s) then |T0−γ|  1 log T. From the book by Huxley we quote two lemmas which will be essential in our proof [9, p. 88]. Lemma 2.2 (First Derivative Test).Let f(x)be real and differentiable on the open interval (α, β)with f0(x)monotone and f0(x)≥µ > 0on (α, β). Let g(x)be real, and let Vbe the total variation of g(x)on the closed interval [α, β]plus the maximum modulus of g(x)on [α, β]. Then Zβ α g(x) exp(2πif(x)) dx≤V πµ. Lemma 2.3 (Second Derivative Test).Let f(x)be real and twice differentiable on the open interval (α, β)with f00(x)≥λ > 0on (α, β). Let g(x)be real, and let Vbe the total variation of g(x)on the closed interval [α, β]plus the maximum modulus of g(x)on [α, β]. Then Zβ α g(x) exp(2πif(x)) dx≤4V √πλ. The next two lemmas can be inferred from Levinson [12] and Gonek [5]. Lemma 2.4. Let κ0>0and K⊂Ra compact set be given. Then there exist constants c > 0,C > 0such that for any r > 1,a∈Kand κ≥κ0, we have Zr(1+c) r(1−c) xaexpn2πiκx log x erodx =κ−1 2ra+1 2eπi 4e−2πiκr +R, with |R| ≤ Cra. Lemma 2.5. Let κ0>0and K⊂Ra compact set be given. Then there exist constants c > 0,C > 0such that for any r > 1,κ≥κ0,a∈Kand r 2≤A<B≤2rwe have ZB A xaexpn2πiκx log x erodx =I0+R1+R2+R3, where |R1| ≤ Cra,|R2| ≤ Cra+1 |A−r|+r1/2,|R3| ≤ Cra+1 |B−r|+r1/2, and where I0=κ−1 2ra+1 2eπi 4e−2πiκr for A≤r≤Band 0in all other cases. 5 Now we state the best zero-free region known. A proof can be found in the book of Ivi´c [11, Thm. 6.1]. Theorem 2.6. There is an absolute constant C > 0such that ζ(s)6= 0 for (7) σ≥1−C(log t)−2 3(log log t)−1 3(t≥t0). Lemma 2.7. Let ρ=β+iγ with β∈(0,1) and γ > 0and define αby ρ=1 2+iα. Then for any κ > 0we have (8) e2iκϑ(α)=γ 2πκ(β−1 2)expniκγ log γ 2π−κγ −κπ 4o(1 + Oκ(γ−1)). Proof. This follows easily from Titchmarsh [21, eq. (4.12.3)].  We will use the following Theorem of Piatetski-Shapiro [15]. Theorem 2.8. For ε > 0,2 3< γ < 1and all kwith 1≤k≤x1−γlog2x, we have (9) X p≤x e2πikpγx11 12 +ε. The exponent 11/12 in this Theorem has been improved, but with a smaller range of γ. For our needs the range is important. Therefore, we will use this theorem as stated. 3. Main theorem Theorem 3.1. For κ > 0 (10) X 0<Re α<T e2iκϑ(α)=−eπi 4(1−κ) √κX n<(T/2π)κ Λ(n) n1 2−1 2κ e−2πiκn1/κ +Oκ(T1−κ 2log T) + Oκ(Tκ 2log2T). Proof. There exists T0such that T < T0< T + 1 with |T0−γ|  1/log Tfor any ordinate γof a zero of ζ(s) and such that X 0<Re α≤T e2iκϑ(α)=X 0<Re α≤T0 e2iκϑ(α)+Oκ(Tκ 2). In fact, here the difference between the two sums is composed of at most of Clog Tterms. Lemma 2.7 yields X T<Re α≤T0 e2iκϑ(α)≤X T<γ≤T0γ 2πκ(β−1 2). Applying Theorem 2.6 we have X T<Re α≤T0 e2iκϑ(α)≤Clog TT0 2πκ(1 2−c(log T0)−2 3(log log T0)−1 3) κC(log T)Tκ 2e−cκ(log T0)1 3(log log T0)−1 3=Oκ(Tκ 2). 6 Also X (T/2π)κ<n≤(T0/2π)κ Λ(n) n1 2−1 2κ e−2πiκn 1 κκ(log T)T1 2−κ 2(T0κ−Tκ) κ(log T)T1 2+κ 21−T+ 1 Tκκ(log T)Tκ 2−1 2. Therefore, replacing Tby T0if needed, we may assume for the rest of the proof that T satisfies |T−γ|  1/log Tfor any ordinate γof a zero of ζ(s). Since (cf. Davenport [1, p. 80]) (11) ζ0(s) ζ(s)=B−1 s−1+ ∞ X n=11 s+ 2n−1 2n+X ρ1 s−ρ+1 ρ we have by Cauchy’s Theorem, (12) U(T) := X 0<Re α<T e2iκϑ(α)=1 2πi ZCT ζ0(s) ζ(s)χ(s)−κds. Here the path of integration CTis the boundary of the rectangle (σ0, σ1)×(2π, T) with 1< σ1<3/2 and Twith |T−γn|  1/log Tfor all n. We will take σ1= 1 + 1 log T, but we maintain the simpler notation σ1. The restriction σ1<3/2 allows one to obtain explicit bounds (independent of σ1) on all pertinent inequalities. The value of σ0depends on κ. We will take σ0=1 2−2 κfor 0 <κ<4 3and σ0=−1 when κ≥4 3. In this way σ0≤ −1 in all cases. Then we have U(T) = 1 2πi Zσ1+2πi σ0+2πi 1 χ(s)κ ζ0(s) ζ(s)ds +1 2πi Zσ1+T i σ1+2πi ···− 1 2πi Zσ1+T i σ0+Ti ···− 1 2πi Zσ0+T i σ0+2πi ··· := U1(T) + U2(T)−U3(T)−U4(T). Lemmas 4.1, 4.2 and 4.3 yield U(T) = Oκ(1) + U2(T) + Oκ(Tκ 2log T) + Oκ(1). Now we apply Lemma 4.4 and we see that U2(T) is equal to the sum on the right in (10) plus the remainders Oκ(Tκ 2log2T) and Oκ(T1−κ 2log T). Therefore, we have our result.  Corollary 3.2. For 0< κ < 6 5we have (13) lim T→∞ 1 N(T)X 0<Re α≤T e2iκϑ(α)= 0. For κ= 0 the above limit is easily seen to be 1. Proof. We have N(T) = card{α: 0 <Re α≤T}=O(Tlog T). For 0 < κ < 1, the trivial bound yields X n≤(T/2π)κ Λ(n) n1 2−1 2κ e−2πiκn1/κ =Oκ(Tκ 2+1 2log T) 7 and the limit is easily shown to be 0. In the case κ= 1 we apply Theorem 3.1, and observe that in this case  eπi/4 √κX n≤(T/2π)κ Λ(n) n1 2−1 2κ e−2πiκn1/κ ≤X n≤T/2π Λ(n) = O(T). Therefore, 1 N(T)X 0<Re α≤T e2iκϑ(α)=O(1/log T). For 1 < κ < 3 2, the Piatetski-Shapiro Theorem 2.8 with k=κ, and γ= 1/κ yields, for any ε > 0, X n≤x Λ(n)e2πiκnγ=O(x11 12 +ε). Partial summation then yields X n≤(T/2π)κ Λ(n) n1 2−1 2κ e−2πiκn1/κ =O(T5κ 12 +1 2+εκ). It follows that 1 N(T)X 0<Re α≤T e2iκϑ(α)=O(T5κ 12 −1 2+εκ/log T) + O(Tκ 2−1log T). So the limit is 0 for 1 < κ < 6 5<3 2. 4. Bounds. 4.1. Bound of the bottom integral. Lemma 4.1. Uniformly for all σ1∈(1,3/2) (14) U1(T) = 1 2πi Zσ1+2πi σ0+2πi 1 χ(s)κ ζ0(s) ζ(s)ds =Oκ(1). Proof. U1(T) is a well defined and continuous function of σ1∈[1,3/2] (it does not depend on T).  4.2. Bound of the top integral. Lemma 4.2. Let Tbe such that |T−γn|  1/log T, and let σ1= 1 + 1 log T. Then (15) U3(T) = 1 2πi Zσ1+iT σ0+iT 1 χ(s)κ ζ0(s) ζ(s)ds =Oκ(Tκ/2log T). 8 Proof. We apply Titchmarsh [21, eq. (4.12.3)] so that χ(s) = 2π tσ+it−1 2ei(t+1 4π)1 + O(t−1) on any strip α≤σ≤βand for t→+∞. Therefore, we will have (16) χ(s)−κ=t 2πκ(σ+it−1 2)e−iκ(t+1 4π)1 + Oκ(t−1). It follows that for s=σ+iT with σ0< σ < σ1we will have |χ(σ+iT)|−κ≤CTκ(σ−1 2). We choose Tsatisfying |T−γ|  1/log T, so that by applying Theorem 9.6(A) of Titchmarsh we get on the segment s=σ+iT with −1< σ < σ1that ζ0(s) ζ(s)=O(log2T). By Ingham [10, Theorem 27 p. 73] this extends to the entire segment σ0< σ < σ1. Then, with the constant Cdepending on κ. |U3(T)| ≤ CZσ1 σ0 Tκ(σ−1/2) log2(T)dσ ≤CTκ(σ1−1/2) κlog Tlog2T. Taking σ1= 1 + 1 log Twe get |U3(T)|=Oκ(Tκ/2log T).  4.3. Bound of the left integral. Lemma 4.3. For 0≤κwe have (17) U4(T) = 1 2πi Zσ0+iT σ0+2πi 1 χ(s)κ ζ0(s) ζ(s)ds =Oκ(1). Proof. We integrate along the line s=σ0+it with 2π<t<T. So we may apply Ingham [10, Theorem 27 p. 73] so that ζ0(s) ζ(s)=O(log t). Also, applying (16) we get (18) U4(T)κZT 2π tκ(σ0−1 2)log t dt. The choice of σ0(σ0=1 2−2 κfor 0 <κ<4 3, and σ0=−1 when κ≥4 3) guarantees that κ(σ0−1 2)≤ −2. Therefore, the integral is bounded.  9