On the exact location of the non-trivial zeros of Riemann’s zeta function

ON THE EXACT LOCATION OF THE NON-TRIVIAL ZEROS OF RIEMANN’S ZETA FUNCTION J. ARIAS DE REYNA AND J. VAN DE LUNE Abstract. In this paper we introduce the real valued real analytic function κ(t) implicitly defined by e2πiκ(t)=−e−2iϑ(t)ζ0(1 2−it) ζ0(1 2+it),(κ(0) = −1 2). By studying the equation κ(t) = n(without making any unproved hypotheses), we will show that (and how) this function is closely related to the (exact) position of the zeros of Riemann’s ζ(s) and ζ0(s). Assuming the Riemann hypothesis and the simplicity of the zeros of ζ(s), it will follow that the ordinate of the zero 1/2 + iγnof ζ(s) will be the unique solution to the equation κ(t) = n. Contents 1. Introduction. 1 2. Phase and argument of a function. 3 3. The function ϑ(t). 6 4. The function κ(t). 7 5. Hypothesis P and its consequences. 13 6. Connection of κ0(t) with the zeros of ζ0(s). 15 7. Counting the zeros of ζ(s). 18 8. Connections between the zeros of ζ(s) and ζ0(s). 22 9. The functions E(t) and S(t). 24 10. Extension to other L-functions. 25 References 27 1. Introduction. The functional equation of Riemann’s zeta function ζ(s) implies that ζ(1 2+it) = Z(t)e−iϑ(t) where Z(t) and ϑ(t) are real valued and real analytic functions and the phase −ϑ(t) is a rather simple function depending only on Euler’s gamma function Γ(s). An analogous decomposition is valid for any meromorphic function. We give a formal definition of the phase of a real analytic function in Section 2. We will define some functions related to the zeros of ζ(s) and the phase of related functions. Of course, these functions have appeared in the literature but only in an implicit way and Date: June 5, 2013. First author supported by MINECO grant MTM2012-30748. 1 arXiv:1305.3844v2 [math.NT] 4 Jun 2013 have not been studied for their own sake. For example, Levinson and Montgomery [11] define J(1 2+it) := ζ(1 2+it) + ζ0(1 2+it)hh0(1 2+it) h(1 2+it)+h0(1 2−it) h(1 2−it)i−1 where h(s) = π−s/2Γ(s/2), and assert that the determination of the number of zeros of ζ(s) in [0 <]σ < 1 2can be conveniently ascertained from the variation of arg J(1 2+it). They do not use the simplified form J(1 2+it) = −e−2iϑ(t)ζ0(1 2−it) 2ϑ0(t). With our notations we would have ph J(1 2+it) = π−2ϑ(t)−ph ζ0(1 2+it) = π 2+πκ(t)−ϑ(t)=2π−E(t). Here κ(t) is the main function introduced here. It is closely connected with the zeros of ζ(s), and is implicitly used in Levinson [10, equation (1.6)] to prove that more than 1/3 of the zeros of ζ(s) are on the critical line. In our paper we seldomly assume the RH, and use the standard notations of the subject. Therefore we shall denote the zeros of ζ(s) on the upper half-plane by βn+iγn(where βnand γnare real numbers) and 0 < γ1≤γ2≤ ···. If a zero is multiple with multiplicity m, then it appears precisely mtimes consecutively in the above sequence. [14, Chapter 9, p. 214]. We shall need to introduce another related sequence of real numbers (0 <)ξ1< ξ2<··· defined so that the set {ξn:n∈N}:= {t > 0 : ζ(1 2+it)=0}. Here only the ordinates of the zeros on the critical line appear. These ξndo not repeat by any circumstance. The two sequences (ξn) and (γn) coincide if and only if the RH is true and all the zeros of ζ(s) on the critical line are simple. Even in case the RH were not true, we will show that κ(t) is related to the zeros of ζ(s) on the critical line. We will show that κ(ξn) = nfor all natural numbers, independently of any hypothesis. The relations between the zeros of ζ(s) and ζ0(s) has been the object of much study. Starting with Speiser [13] who showed that the RH is equivalent to ζ0(s) having no zeros in 0< σ < 1 2, Levinson and Montgomery [11] give a quantified version of Speiser’s theorem, and Berndt [1] gives an estimation of the number of zeros of ζ0(s) to a given height. Great interest in the zeros of ζ0(s) is related to their horizontal distribution, in which many questions remain open (see Levinson Montgomery [11], Conrey and Ghosh [4], Soundararajan [12], Zhang [15], Garaev and Yıldırım [7]). Here we get a new way to study these relationships by means of our function κ(t). The number of zeros of ζ(s) on an interval of the critical line not counting multiplicities is related to the increment of κ(t) in this interval. Assuming the RH this function will be strictly increasing, so κ0(t)≥0. The connection is by means of equation (30) which represents this function in terms of the zeros of ζ0(s). Therefore, κ(t) is related to the zeros of ζ(s) (Prop. 17), and κ0(t) is fully determined by the zeros of ζ0(s) (Prop. 30). The relationship of κ0(t) with the zeros of ζ(s) is also direct and double (Prop. 41 and equation (45)). See figure 5 for a graphical description of these relations. In Section 2 we give the definition and (some simple) properties of the decomposition in phase and signed modulus of a real analytic function. In particular, in Proposition 6 we will write the phase as a convergent integral. After this we devote Section 3 to some properties 2 of the phase −ϑ(t) of ζ(1 2+it). Since we will use its convexity for all t > 0, we give a simple derivation of this fact. Section 4 is devoted to the introduction of κ(t). The definition in Proposition 16 e2πiκ(t)= 1 + 2ϑ0(t)ζ(1 2+it) ζ0(1 2+it), κ(0) = −1 2. is possible because the function in the right hand side makes a circular movement for t∈ R. We study the relationship of κ(t) with ph ζ0(1 2+it) and ϑ(t). The function κ(t) is a complicated function, its behavior being connected with the RH. We show here the equation κ(ξn) = n which determines the set of real numbers twith ζ(1 2+it) = 0. Proposition 23 may come as a surprise. It relates the points where κ(t) is half an integer with the zeros of Z0(t). Assuming the RH the function κ(t) will be strictly increasing and between γnand γn+1 there would be only one zero of Z0(t), situated just at the point where κ(t) = n+1 2. In the next section we show what of this remains true if we do not assume the RH, and see the first application of the function κ0(t). The main result of Section 6 is a formula for κ0(t) in terms of the zeros of ζ0(s) (see Proposition 30). Therein appears a constant Awhich we relate in equation (32) with the zeros of ζ0(s). In Section 7 we obtain the value A=1 2log 2. We give a proof that relates this constant to the difference in the counting of zeros of ζ(s) given by Riemann and the one for the zeros of ζ0(s) given by Berndt. Also we include a proof that the RH implies κ0(t)>0 for t > aκ. Section 8 establishes the connection of κ0(t) with the zeros of ζ(s). We know from Section 4 that for n < m we have Rξm ξnκ0(t)dt =m−n. We show that κ0(ξn) = ϑ0(ξn)/ω where ωis the multiplicity of the zero 1 2+iξnof ζ(s). In Proposition 42 we apply these relationships to give, assuming the RH, a new proof of a strengthening of a Theorem of Garaev and Yıldırım [7] (which they prove unconditionally). In Section 9 we introduce a related function E(t) and show its relationship with the classical function S(t) and with a function RH(t) which counts the failures up to height tof both the RH and the simplicity of the zeros of ζ(s). This is almost the function considered by Levinson and Montgomery. Most of the functions appearing in this paper were found some years ago (in 1997) by one of us (JvdL) while searching for a formula (or equation) for the exact location of the non-trivial zeros of the Riemann zeta function. 2. Phase and argument of a function. The results in this section are easy but we did not find any proper references. We include the simple proofs and introduce our notations about phase and argument of a real analytic function. Definition 1. A function f:R→Cis called real analytic if for every t0∈Rthere exists a convergent power series P(z) = P∞ k=0 ckzksuch that f(t) = P(t−t0)for all tin a neighborhood of t0. In other words: A function f:R→Ccalled real analytic if fhas an analytic extension to a neighborhood of R. Proposition 2. If f:R→Cr{0}is a real analytic function, then there exists a real analytic function gsuch that f(t) = eg(t)for every t∈R. 3 Proof. For every t0∈Rlet ∆(t0) be a disk with center at t0such that f(t) = P(t−t0) for t∈∆(t0)∩R, and such that P(z−t0)6= 0 for z∈∆(t0). The union G=St0∆(t0) is a simply connected domain and fcan be extended to Gas an analytic function. Since f(z)6= 0 for z∈G, there exists an analytic function gon Gsuch that f(z) = eg(z)for all z∈G. Corollary 3. If f:R→Cr{0}is a real analytic function, then there exists a real analytic function ϕ:R→Rsuch that f(t) = |f(t)|eiϕ(t). We write in such a case ϕ(t) = arg f(t). This is an analytic (and hence continuous) determination of the argument of f. Two such functions differ only by an integral multiple of 2π. Proposition 4. If f:R→Cis a real analytic function, then there are two real analytic functions U:R→Rand ϕ:R→Rsuch that f(t) = U(t)eiϕ(t). Given two such representations, f=U1eiϕ1and f=U2eiϕ2, we have either U1=U2and ϕ1−ϕ2= 2kπ or U1=−U2and ϕ1−ϕ2= (2k+ 1)πfor some integer k. Proof. If fdoes not vanish, then |f|is real analytic and by Corollary 3 there exists a real analytic function ϕ:R→Rsuch that f|f|−1=eiϕ, and we can take U=|f|in this case. Now assume that fhas real zeros. Let anbe the real zeros of f(t) listed with multiplicities. We may assume that a1=··· =am= 0 and all the others non-zero. By Weierstrass’ factorization theorem there exists an entire function g(z) = zmY n>m En−1(z/an), z ∈C whose zeros are the numbers an, and the En(z) = (1 −z)ez+z2/2+···+zn/n are the canonical factors. Observe also that this function is real for real z=t. By the previous argument there exist real analytic functions hand ϕsuch that f/g =heiϕ. Thus f= (gh)eiϕ, and U=gh. This proves that the claimed decomposition exists. Finally, if f=U1eiϕ1=U2eiϕ2, then U1/U2is a real analytic function without zeros. Also |U1/U2|=|ei(ϕ2−ϕ1)|= 1 and it follows that U1/U2is either equal to 1 or to −1. In the first case ei(ϕ2−ϕ1)= 1 and ϕ2= 2kπ +ϕ1for some integer k. The other case may be treated similarly.  Definition 5. Given a real analytic function f:R→Cwe call phase of fany real analytic function ph f:R→Rsuch that f(t) = U(t)eiph f(t)with U:R→Ra real analytic function. If g1and g2are two such functions there exists an integer ksuch that g1(t) = g2(t) + kπ for every t∈R. Observe that the above definition is not standard. We are making use of the word phase with a peculiar mathematical meaning. The main difference between the phase of a real analytic function and its argument is that for some t∈Rthe value ph f(t) may not be equal to one of the arguments of the complex number f(t). We will have only ph f(t) equal to this argument modulo π. 4 Example 1.It is easy to check that (1) cos π 2(1 2+it) = 1 √2√cosh πt e−iarctan(tanh πt 2). Example 2.One of the most interesting examples is that of the zeta function on the critical line. In this case we have (see Edwards [5, p. 119]) (2) ζ(1 2+it) = Z(t)e−iϑ(t) where Z:R→Rand ϑ:R→Rare real analytic. Z(t) is the Riemann-Siegel function (sometimes called Hardy function [8]). Example 3.The phase −ϑ(t) in Example 2 is related to the phase of Γ(1 4+it 2) by (3) Γ(1 4+i1 2t) = Γ(1 4+i1 2t)ei(ϑ(t)+ t 2log π). (For more details see [14, (4.17.2)]). Example 4.We have not found any reference for our next example: (4) Γ(1/2 + it) = rπ cosh πt expi(2ϑ(t) + tlog(2π) + arctan tanh πt 2). This may be shown using only properties of Γ(s) but we present a proof based on the functional equation of ζ(s). Let Φ(s) = 1 2ζ(s)ζ(1 −s). Then, by the functional equation (5) Φ(s) = cos πs 2(2π)−sΓ(s)ζ(s)2. Substituting (1) and (2) into this equation, we get with s=1 2+it (6) 1 2Z(t)2=rcosh(πt) 2e−iarctan(tanh πt 2)·1 √2πe−it log(2π)·Γ(s)·e−2iϑ(t)Z(t)2 from which (4) follows for Z(t)6= 0. But since the argument in (4) is real analytic it is true for all t. Proposition 6. If fis a non-constant real analytic function, then for every t∈Rwe have (7) ph f(t) = ph f(0) + Zt 0 Im f0(x) f(x)dx. Proof. The function ph f(t) is real analytic, so that ph f(t) = ph f(0) + Zt 0 (ph f)0(x)dx. There exists a real analytic function Usuch that f(t) = U(t)eiph f(t). Therefore, if f(x)6= 0 then f0(x) f(x)=U0(x) U(x)+i(ph f)0(x) so that (ph f)0(x) = Im f0(x) f(x). 5 It follows that Im f0(x) f(x)is in fact a real analytic function, the possible singularities at the points where f(x) = 0 being removable.  Example 5.By Examples 2 and 3 we have (8) ϑ(t) = −Zt 0 Re ζ0(1 2+ix) ζ(1 2+ix)dx =−t 2log π+1 2Zt 0 Re Γ0(1 4+ix 2) Γ(1 4+ix 2)dx. 3. The function ϑ(t). In this section we recall some properties of the function ϑ(t) introduced in Example 2. We need to show that ϑ(t) = 0 has only one solution for t > 0. To this end we must show explicit formulae for ϑ(t) for small t, which are seldomly considered. Indeed, after having introduced ϑ(t), most authors immediately start discussing its asymptotic expansion (compare Edwards [5, p. 119] and Gabcke [6, p. 4]). Proposition 7. For ϑ(t)we have the following series expansion ( convergent for all t∈R) (γbeing Euler’s constant ) (9) ϑ(t) = −1 2γ+ log π+ 3 log 2 + π 2t+ ∞ X k=02t 4k+ 1 −arctan 2t 4k+ 1. Proof. From the Weierstrass product for Γ(s) we obtain −Im log Γ1 4+it 2) = = argn(1 4+it 2)eγ(1 4+it 2) ∞ Y k=11 + 1 4+it 2 ke−(1 4+it 2)/ko+ 2`π = arctan(2t) + γ 2t+ ∞ X k=1arctan 2t 4k+ 1 −t 2k−2kπ for some `,k∈Z. Since ϑ(t) = arg Γ(1 4+it 2)−t 2log πwe have ϑ(t) = −γ+ log π 2t−arctan(2t)− ∞ X k=1arctan 2t 4k+ 1 −t 2k+ 2kπ so that, taking t= 0, we find that k= 0. We rewrite the last series as follows ∞ X k=1arctan 2t 4k+ 1 −t 2k=−t 2 ∞ X k=11 k−1 k+1 4+ ∞ X k=1arctan 2t 4k+ 1 −2t 4k+ 1. The first series in the right hand side can be summed explicitly ∞ X k=11 k−1 k+1 4= ∞ X k=1Z1 0 uk−1du −Z1 0 uk−1+1 4du=Z1 0 1−u1/4 1−udu = = 4 Z1 0 1−v 1−v4v3dv =Z1 0 4v3dv (1 + v)(1 + v2)= 4 −3 log 2 −π 2. Combining these equations we obtain (9).  6 Proposition 8. For every t∈Rwe have (10) ϑ0(t) = −1 2(γ+ log π)−2 1+4t2− ∞ X k=12(4k+ 1) (4k+ 1)2+ 4t2−1 2k. Corollary 9. The function ϑ(t)is convex on (0,+∞), and there exists a unique positive real number aϑwhere ϑ0(aϑ)=0. By differentiation of (10) we get (11) ϑ00(t) = 16t ∞ X k=0 4k+ 1 ((4k+ 1)2+ 4t2)2 from which the Corollary follows. This Corollary is proved in [9, Lemma 11, Lemma 12]. We have aϑ= 6.28983 59888 36902 77966 50901 00821 85339 66583 12945 19278 95453 57765 . . . 4. The function κ(t). The next Proposition is included in Titchmarsh [14, p. 291], but we write the proof below, because we are also interested in the formulas used. Proposition 10. If ζ0(1 2+ia) = 0 for a real a, then ζ(1 2+ia)=0. Proof. We start from ζ(1 2+it) = e−iϑ(t)Z(t). Differentiation with respect to tyields iζ0(1 2+it) = −iϑ0(t)e−iϑ(t)Z(t) + e−iϑ(t)Z0(t). Multiplying this by −ieiϑ(t)we get (12) eiϑ(t)ζ0(1 2+it) = −ϑ0(t)Z(t)−iZ0(t) and taking real parts we obtain (13) −ϑ0(t)Z(t) = Reeiϑ(t)ζ01 2+it which may also be written as (14) −2ϑ0(t)Z(t) = eiϑ(t)ζ01 2+it+e−iϑ(t)ζ01 2−it. Let us assume that ζ01 2+it= 0 for some real t. Since ζ0(1 2−it) = 0 we may assume that t > 0 and we get ϑ0(t)Z(t) = 0. Since ϑ0(t) = 0 only for t=aϑ≈6.29 where ζ0(1 2+iaϑ)6= 0, we get Z(t) = 0. Therefore ζ01 2+it= 0 implies ζ1 2+it= 0  Recall that we denote, as usual, by βn+iγnthe non-trivial zeros of ζ(s), ordered in such a way that (0 <)γ1≤γ2≤ ···, repeating each term according to its multiplicity. We will need another related sequence. Let (0 <)ξ1< ξ2<··· be the sequence of real numbers t such that ζ(1 2+it) = 0, counted without multiplicities. Hence the ξnonly denote zeros on the critical line. If we assume the RH and the simplicity of the zeros, we would, of course, have ξn=γn. Proposition 11. For every real t6=±ξnwe have (15) 1 + 2ϑ0(t)ζ(1 2+it) ζ0(1 2+it)=−e−2iϑ(t)ζ0(1 2−it) ζ0(1 2+it). 7 Proof. Multiplying (14) by e−iϑ(t)we get −2ϑ0(t)ζ(1 2+it) = ζ01 2+it+e−2iϑ(t)ζ01 2−it. Since t6=±ξn, and using Proposition 10 we have ζ0(1 2+it)6= 0, so that we can divide by ζ0(1 2+it) and obtain our result.  Proposition 12. There exists a unique real analytic function κ:R→Rsuch that (16) e2πiκ(t)= 1 + 2ϑ0(t)ζ(1 2+it) ζ0(1 2+it), κ(0) = −1 2. Proof. By Proposition 11 the function f:R→Cdefined by f(t) = 1 + 2ϑ0(t)ζ(1 2+it) ζ0(1 2+it) satisfies |f(t)|= 1 for t6=ξn. By definition, and Proposition 10, fis real analytic and satisfies |f(ξ)|= 1, so that there exists a real analytic κ:R→Rsuch that f(t) = e2πiκ(t). This function is uniquely defined by its value at any one point. Since ϑ(0) = 0 we have f(0) = −1 (see (15)) and we can take κ(0) = −1 2. Applying Proposition 4 to ζ0(1 2+it) we arrive at two real analytic functions ρ:R→R and ph ζ0(1 2+it). Observing that ζ0(1 2)<0 we may choose ζ0(1 2+it) = ρ(t)eiph ζ0(1 2+it), ρ(0) = |ζ0(1 2)|,ph ζ0(1 2) = π. If we assume that ζ(s) has no multiple zero on the critical line, then ζ0(1 2+it)6= 0 and we will have ρ(t) = |ζ0(1 2+it)|and ph ζ0(1 2+it) = arg ζ0(1 2+it) (where arg ζ0(1 2+it) is meant to be a continuous function of tin R). −10 −30 −50 10 20 30 40 50 phζ!(1 2+it) Figure 1. ph ζ0(1 2+it) Proposition 13. For all t∈R (17) κ(t) = 1 2−1 πϑ(t) + ph ζ0(1 2+it). 8 Proof. By the definition of κ(t) and (15) we have exp2πiκ(t)=−e−2iϑ(t)ζ0(1 2−it) ζ0(1 2+it)= expπi −2iϑ(t)−2iph ζ0(1 2+it). Hence there exists an integer nsuch that 2πiκ(t) = πi −2iϑ(t)−2iph ζ0(1 2+it)+2πin. For t= 0 we get n= 0 and (17) follows.  Corollary 14. For every real twe have (18) κ(t) = −1 2−1 πZt 0ϑ0(x) + Re ζ00(1 2+ix) ζ0(1 2+ix)dx. Proof. In formula (17), we replace ph ζ0(1 2+it) by the integral expression given by (7).  Observing that −ϑ(t) is the phase of ζ(1 2+it) we also obtain (see (8)) (19) κ(t) = −1 2+1 πZt 0 Reζ0(1 2+ix) ζ(1 2+ix)−ζ00(1 2+ix) ζ0(1 2+ix)dx. 1 2 3 4 5 6 10 20 30 40 aϑ κ(t) Figure 2. κ(t) From (19) we see that κ(t) + 1 2is an odd function. Proposition 15. Choosing the phase of the real analytic function t7→ Z0(t)−Z(t)ϑ0(t)to be =π/2at t= 0 we will have (20) κ(t) = −1 πph(Z0(t)−iZ(t)ϑ0(t)). 9 Proposition 30. We have (30) πκ0(t) = A+f(t) + X ρ0=β0+iγ0 β0−1/2 (1/2−β0)2+ (t−γ0)2 where Ais a constant and f(t)is a bounded continuous function such that f(t) = O(t−1)as t→+∞. Remark 31.The exact definition of f(t) is given in (33). Remark 32.In Corollary 39 we will prove that A=1 2log 2. Proof. From (18) we get (31) πκ0(t) = −ϑ0(t)−Re ζ00(1 2+it) ζ0(1 2+it). Now in (28) we put s= 1/2 + it and take real parts Re ζ00(1/2 + it) ζ0(1/2 + it)=a+4 1+4t2+ ∞ X n=11/2−an (1/2−an)2+t2+1 an +X ρ0=β0+iγ01/2−β0 (1/2−β0)2+ (t−γ0)2+β0 β02+γ02. Hence from (31) and (10) we get πκ0(t) = A−2 1+4t2+ ∞ X n=12(4n+ 1) (4n+ 1)2+ 4t2−1/2−an (1/2−an)2+t2 +X ρ0=β0+iγ0 β0−1/2 (1/2−β0)2+ (t−γ0)2 where (32) A=1 2(γ+ log π)−a− ∞ X n=1 2β0 n β0 n2+γ0 n2− ∞ X n=11 2n+1 an. We put (33) f(t) = −2 1+4t2+ ∞ X n=12(4n+ 1) (4n+ 1)2+ 4t2−1/2−an (1/2−an)2+t2. Now observe that the terms of the sum can be written as (2n+ 1/2) (2n+ 1/2)2+t2−1/2−an (1/2−an)2+t2=Z1/2−an 2n+1/2 x2−t2 (x2+t2)2dx. 16 The intervals (2n+ 1/2,1/2−an) do not intersect, so that for |t|< T the absolute values of the terms of the sum are bounded by ∞ X n=12(4n+ 1) (4n+ 1)2+ 4t2−1/2−an (1/2−an)2+t2≤Z+∞ 5/2 |x2−t2| (x2+t2)2dx ≤Z+∞ 5/2 T2+x2 x4dx < +∞. This proves that f(t) is a continuous function. Also for t > 1 we will have (34) |f(t)| ≤ 2 1+4t2+Z+∞ 0 |x2−t2| (x2+t2)2dx =2 1+4t2+1 t.  Remark 33.It can be shown that the zero an−1contained in (−2n, −2n+ 2) satisfies 2n+ an−1∼1/log(n/π). This information can be used to show that f(t) = O(1/t log t). In this way we may improve the error term in (36) from O(log t) to O(log log t). We introduce some notation: if t > 0 and n≥1 let ϕ(t, ρ0 n) = ϕn(t) be the angle at ρ0 n of the triangle with vertices at ρ0 n=β0 n+iγ0 n, 1/2−it and 1/2 + it. We consider this angle expressed in radians positive if β0 n>1/2 and negative if β0 n<1/2, and we put ϕn(t) = 0 when β0 n=1 2. In other words with s=1 2+it and ρ0=β0+iγ0we have (35) ϕ(t, β0+iγ0) = arctan t−γ0 β0−1 2 + arctan t+γ0 β0−1 2 = Arg s−ρ0 s−ρ0(β06=1 2). Proposition 34. For t > 0we have (36) πκ(t) = At + ∞ X n=1 ϕn(t) + O(log t) where the sum is extended over all zeros ρ0 n=β0 n+iγ0 nof ζ0(s)with γ0 n>0. Proof. By (18) and (30) we have πκ(t) = −π 2+At +Zt 0 f(x)dx +Zt 0X ρ0=β0+iγ0 β0−1 2 (1 2−β0)2+ (x−γ0)2dx. Observe that if β0=1 2then the corresponding term does not contribute to the sum. Thus πκ(t) = −π 2+At +Zt 0 f(x)dx +X ρ0=β0+iγ0narctan t−γ0 β0−1 2 + arctan γ0 β0−1 2o where the terms with β0 n=1 2should be omitted. It is easy to see that the sum of the terms corresponding to ρ0 n=β0 n+iγ0 nand ρ0 n=β0 n−iγ0 nadd up to exactly ϕn(t). (This is the reason we made the convention about the sign of ϕn(t)). Thus we arrive at πκ(t) = −π 2+At +Zt 0 f(x)dx + ∞ X n=1 ϕn(t). 17 Now, since f(t) = O(t−1) we can write this as πκ(t) = At + ∞ X n=1 ϕn(t) + O(log t).  7. Counting the zeros of ζ(s). The exact value of the constant Ain (36) can be obtained in two ways. One by computing the constants in the Mittag-Leffler expansion of related functions and the second, more interesting for us, by comparing two different counts of the number of zeros of ζ(s). We will present this second proof. We need some definitions: Let N−(T)=#{ρ=β+iγ |ζ(ρ) = 0, β < 1 2,0< γ ≤T} N0(T)=#{ρ=1 2+iγ |ζ(ρ)=0,0< γ ≤T} N0(T)=#{ρ0=β0+iγ0|ζ0(ρ0)=0,0< γ0≤T} N0 −(T)=#{ρ0=β0+iγ0|ζ0(ρ0)=0, β0<1 2,0< γ0≤T} N0 0(T)=#{ρ0=1 2+iγ0|ζ0(ρ0) = 0,0< γ0≤T} N0 +(T)=#{ρ0=β0+iγ0|ζ0(ρ0)=0, β0>1 2,0< γ0≤T}. In all these cases, as usual, we count the zeros with their multiplicities. But we also need to consider another count N00(T) which is the number of real numbers 0 < ξ ≤Tsuch that ζ(1 2+iξ) = 0, but in these cases we do not count multiplicities. Taking account of Proposition 10 it is clear that (37) N0(T)−N00(T) = N0 0(T) which equals the number of zeros of ζ0(s) on the critical line with 0 < γ0≤T. We know some relations between these counts: (i) Backlund refining previous work of von Mangoldt (see Edwards, [5, Section 6.7]) gave a complete proof of Riemann’s assertion (38) N(T) = N0(T)+2N−(T) = T 2πlog T 2π−T 2π+O(log T). (ii) Berndt [1] proved the corresponding result for ζ0(s) (39) N0(T) = N0 −(T) + N0 0(T) + N0 +(T) = T 2πlog T 4π−T 2π+O(log T). (iii) Levinson and Montgomery [11] showed that (40) N−(T) = N0 −(T) + O(log T). (iv) From our Proposition 21 we get that (41) κ(t) = N00(t) + O(1). Finally, in Proposition 37 we will prove a new relation (42). First we prove two lemmas about the zeros of ζ0(s). 18 Table 1. First non-trivial zeros of ζ(s) and ζ0(s) βn+iγnβ0 n+iγ0 n 0.5 + i14.13472 51417 2.46316 18694 + i23.29832 04927 0.5 + i21.02203 96387 1.28649 68222 + i31.70825 00831 0.5 + i25.01085 75801 2.30757 00637 + i38.48998 31730 0.5 + i30.42487 61258 1.38276 36057 + i42.29096 45545 0.5 + i32.93506 15877 0.96468 56227 + i48.84715 99050 0.5 + i37.58617 81588 2.10169 99009 + i52.43216 12451 0.5 + i40.91871 90121 1.89595 97624 + i57.13475 31990 Lemma 35. For t > 0we have X 0<γ0 n<t−1 1 t−γ0 n =O(log2t). Proof. Put N0(t) = t 2πlog t 4π−t 2π+R(t) where R(t) = O(log t) by Berndt’s Theorem. The first zero of ζ0(s) is ρ0≈2.46316+23.29832i so that N0(4π) = 0 and consequently R(4π) = 2. We have X 0<γ0 n<t−1 1 t−γ0 n =Zt−1 4π dN0(x) t−x=1 2πZt−1 4π log(x/4π) t−xdx +Zt−1 4π dR(x) t−x ≤log(t/4π)log(t−4π) 2π+R(t−1) + Zt−1 4π |R(x)| (t−x)2dx. Since R(x) = O(log x) all the above terms are O(log2t).  Lemma 36. For t→+∞we have X γ0 n>t+1 1 γ0 n2−t2=Olog2t t. Proof. Retaining the notations of the previous lemma we have X γ0 n>t+1 1 γ0 n2−t2=Z+∞ t+1 dN0(x) x2−t2=1 2πZ∞ t+1 log(x/4π) x2−t2dx +Z+∞ t+1 dR(x) x2−t2. For t > 4πthe first integral is less than or equal to Clog2t/t: 1 2πZ∞ t+1 log(x/4π) x2−t2dx ≤2 log t 2πZt2 t+1 dx x2−t2+Z+∞ t2 √x x2−t2dx =log t 2πt log t−1 t+ 1 + log(2t+ 1)+t−1/2Z+∞ t √y y2−1dy ≤(log t)2 πt +4 t. Now we bound the second integral Z+∞ t+1 dR(x) x2−t2=−R(t+ 1) 2t+ 1 +Z+∞ t+1 R(x) (x2−t2)22x dx. 19 For x>t+ 1 we have x/(x2−t2)<1, and |R(x)| ≤ Clog x. Thus we have Z+∞ t+1 dR(x) x2−t2≤c1 log t t+c2Z+∞ t+1 log x x2−t2dx. Finally, this integral is bounded exactly as the first integral.  Proposition 37. For t→+∞ (42) πκ(t) = At +πN0 +(t)−πN0 −(t) + O(log2t). Proof. By (36) we have to show that X γ0 n>0 ϕn(t) = πN0 +(t)−πN0 −(t) + O(log2t). To this end we will show that X γ0 n>0 β0 n>1 2 ϕn(t) = πN0 +(t) + O(log2t); X γ0 n>0 β0 n<1 2 ϕn(t) = −πN0 −(t) + O(log2t). To simplify the notation we will write X+to denote a sum restricted to β0 n>1 2and X− for a sum restricted to β0 n<1 2. We split the sums into three terms X+ γ0 n>0 ϕn(t) = X+ 0<γ0 n<t−1 ϕn(t) + X+ |γ0 n−t|≤1 ϕn(t) + X+ γ0 n>t+1 ϕn(t). The middle sum is O(log t) because each term is (in absolute value) less than πand the number of terms is O(log t). In the first sum the summands are approximately π(or −π). Thus we arrive at X+ γ0 n>0 ϕn(t) = πN0 +(t) + X+ 0<γ0 n<t−1ϕn(t)−π+X+ γ0 n>t+1 ϕn(t) + O(log t) and X− γ0 n>0 ϕn(t) = −πN0 −(t) + X− 0<γ0 n<t−1ϕn(t) + π+X− γ0 n>t+1 ϕn(t) + O(log t). It follows that X γ0 n>0 ϕn(t) = πN0 +(t)−πN0 −(t) + X 0<γ0 n<t−1ϕn(t)±π+X γ0 n>t+1 ϕn(t) + O(log t) where we must use the + sign when β0 n<1 2and the −sign when β0 n>1 2. Now for 0 < γ0 n< t −1 and β0 n>1 2we have 0< π −ϕn(t)<arctan β0 n−1 2 t−γ0 n + arctan β0 n−1 2 t+γ0 n <2 arctan β0 n−1 2 t−γ0 n <6 t−γ0 n and in the case β0 n<1 2analogously 0< π +ϕn(t)<arctan 1 2−β0 n t−γ0 n + arctan 1 2−β0 n t+γ0 n <2 arctan 1 2−β0 n t−γ0 n <1 t−γ0 n . 20 Also, for γ0 n> t + 1 and β0 n>1 2 0< ϕn(t) = arctan 2(β0 n−1 2)t (β0 n−1 2)2+γ0 n2−t2<6t γ0 n2−t2 and for β0 n<1 2the absolute value |ϕn(t)|is bounded by the same quantity. Thus applying the two above lemmas we find that X γ0 n>0 ϕn(t) = πN0 +(t)−πN0 −(t) + O(log2t).  Corollary 38. For t→+∞ (43) N00(t) = A πt+N0 +(t)−N0 −(t) + O(log2t). Proof. Combine (42) with (41).  Corollary 39. The constant Ais equal to 1 2log 2. Proof. Write f(t)$g(t) to denote that f(t)−g(t) = O(log2t). (In the same way as congruences we can operate with $as if it were an equality sign between equivalence classes). With this notation we have N(t)$t 2πlog t 2π−t 2πby (38) $N0(t)+2N−(t) trivially $N0(t)+2N0 −(t) by (40) $N0(t)−N00(t) + A πt+N0 +(t) + N0 −(t) by (43) $N0(t)−N00(t) + A πt−N0 0(t) + t 2πlog t 4π−t 2πby (39) $A πt+t 2πlog t 4π−t 2πby (37) We thus have t 2πlog t 2π−t 2π$A πt+t 2πlog t 4π−t 2π from which we deduce A πt$t 2πlog 2. Hence A=1 2log 2.  Proposition 40. The Riemann hypothesis implies the Hypothesis P. Proof. The Riemann hypothesis is equivalent to β0>1 2for every zero ρ0=β0+iγ0, and it follows by (30) that if the Riemann hypothesis is true, then πκ0(t)> A +f(t). Since A=1 2log 2, applying (34) we easily see that κ0(t)>0 for t > 3.4 if we assume the RH. It is clear that there is an aκ≥0 such that κ0(t)>0 for t>aκand aκ<3.4.  21 8. Connections between the zeros of ζ(s)and ζ0(s). Proposition 41. Let 1 2+iξ be a zero of ζ(s)of multiplicity ωon the critical line, then (44) κ0(ξ) = 1 πωϑ0(ξ). Proof. Since ζ(1 2+it) = e−iϑ(t)Z(t) the function Z(t) has a zero of multiplicity ωat t=ξ. Hence limt→ξZ(t) Z0(t)= 0, and for ω≥2, limt→ξZ(t) Z0(t) Z00 (t) Z0(t)=ω−1 ω. When ω= 1 this second limit is equal to 0 = ω−1 ω. Hence for 0 <|t−ξ|< δ we have Z(t), Z0(t) and Z00(t)6= 0 and by (26) we have lim t→ξκ0(t) = lim t→ξ 1 π Z Z0ϑ00 +ϑ0−Z Z0 Z00 Z0ϑ0 1 + Z Z0ϑ02=1 πϑ0(ξ)−ω−1 ωϑ0(ξ)=1 πωϑ0(ξ).  Assuming the RH and the simplicity of zeros we have Zγn aϑ κ0(t)dt =n. Hence the mean value of κ0(t) in [0, t] is N(t) twhich is approximately equal to ϑ0(t)/π. The above Proposition says that, assuming only the simplicity of zeros, at the points ξnthe value κ0(ξn) is just equal to this density. 1 0.5 1.5 20 40 60 80 100 ξ1ξ5ξ10 ξ15 ξ20 ξ25 κ!(t) 1 πϑ!(t) Figure 5. κ(t) Figure 5 illustrates two ways in which the zeros of ζ0(s) determine the ξn(assuming only simplicity of the zeros of zeta). First ξnis determined from κ0(t) by the equation (45) Zξn ξ0 κ0(t)dt =nor Zξn ξn−1 κ0(t)dt = 1. Second, the points ξnare intersections of the two curves κ0(t) and ϑ0(t)/π. Although, as we see in figure 5 not all these intersections correspond to points ξn. 22 We can see how two close ξncorrespond to a peak in the graph of κ0(t) that, according to equation (30) will be produced by one or more zeros β0+iγ0of ζ0(s) with a relatively small β0−1 2. Observe that equation (30) shows that κ0(t) is fully determined by the zeros of ζ0(s). Following these ideas we may improve (but assuming the RH) a theorem due to M. Z. Garaev, C. Y. Yıldırım [7]. For any given zero ρ0=β0+iγ0of ζ0(s) let γcbe of all ordinates of zeros of ζ(s), the one for which |γc−γ0|is smallest (if there are more than one such zero of ζ(s), take γcto be the imaginary part of any one of them). Garaev and Yıldırım prove unconditionally that |γc−γ0||β0−1 2|1 2. Proposition 42 (RH).Assuming the RH, we have for any zero β0+iγ0of ζ0(s) |γc−γ0| ≤ 1.9|β0−1 2|1 2. Proof. Assuming the RH, β0>1/2 so that by equation (30) we will have κ0(t)≥1 2πlog 2 + f(t) π+1 π β0−1 2 (β0−1 2)2+ (t−γ0)2. We will find an a > 0 so that κ(γ0+a)−κ(γ0−a)≥alog 2 π+2 πarctan a β0−1/2+1 πZγ0+a γ0−a f(t)dt > 1. Then there will be a point ξ∈[γ0−a, γ0+a] such that κ(ξ)∈Z. Then by Corollary 18 ζ(1 2+iξ) = 0, so that the ordinate γcof the nearest zero of ζ(s) will satisfy |γc−γ0| ≤ a. By (34), for t > 20 we have |f(t)|<41/40tand therefore for a/γ0<1/2 1 πZγ0+a γ0−a f(t)dt≤41 40πlog γ0+a γ0−a≤41 40π 8 3 a γ0≤a γ0, because log 1+x 1−x≤8x/3 for |x| ≤ 1/2. Therefore we want to choose asuch that κ(γ0+a)−κ(γ0−a)≥alog 2 π+2 πarctan a β0−1/2−a γ0>1 or alog 2 2−πa 2γ0≥π 2−arctan a β0−1/2= arctan β0−1/2 a. It suffices to take alog 2 2−π 2γ0≥β0−1/2 a Since γ0≥23 it suffices to take a= 1.9qβ0−1 2≥log 2 2−π 2γ0−1/2qβ0−1 2. Since always β0<3 and γ0>23 this asatisfies a/γ0<1/2, as used above.  23 9. The functions E(t)and S(t). In the theory of the zeta function we consider the function S(t) = π−1arg ζ(1 2+it) where the argument is obtained by its continuous variation along the straight lines joining 2, 2 + it,1 2+it starting with the value 0. If tis the ordinate of a zero, S(t) is taken equal to S(t+ 0). This function satisfies (see Edwards [5, p. 173]) (46) S(t) = N(t)−1−1 πϑ(t). If we assume the RH and the simplicity of the zeros, we will have N(t) = N00(t) = bκ(t)c (see Proposition 21). We introduce a real analytic version of S(t) that we will call E(t) (47) E(t) := π+ 2ϑ(t) + ph ζ0(1 2+it). By (17) this is equivalent to (48) E(t) = 3π 2+ϑ(t)−πκ(t) with E(0) = 2π. If 1 2+iξnis a simple zero of ζ(s) we will have E0(ξn) = 0 by Proposition 41. The converse is not true. For example at t0= 39.587 127 340 . . . the function E(t) has a local minimum with E(t0)=0.151 790 437 . . . It is also easy to show that E(t)−2πis a real analytic odd function. In fact E0(t) = ϑ0(t)−πκ0(t) so that the zeros of E0(t) are just the points where the graphs of 1 πϑ0(t) and κ0(t) intersect (see Figure 5). By equation (26) for Z0(t)2+ (Z(t)ϑ0(t))26= 0 we have (49) E0=ϑ0−πκ0=Z·Zϑ03−Z0ϑ00 +Z00ϑ0 (Z0)2+ (Zϑ0)2. 2π 100 200 E(t) Figure 6. E(t) 24 For the next Proposition we need a measure of the possible failure of the RH. Definition 43. For any t > 0we define RH(t)by (50) RH(t) := N(t)−N00(t). That is RH(t) is equal to the number of zeros β+iγ of ζ(s) with 0 < γ ≤tand β6=1 2, plus the number of zeros β0+iγ0of ζ0(s) with β0=1 2and 0 < γ0≤tall of them counted with their multiplicities. By Proposition 10 these zeros of ζ0(s) will be multiple zeros of ζ(s) on the critical line. We have RH(t) = 0 if and only if the zeros β+iγ of ζ(s) with 0 < γ ≤t are all on the critical line and are simple. Proposition 44. We have (51) −1 2+ RH(t)< S(t) + 1 πE(t)≤1 2+ RH(t), t > aϑ. Proof. By (46) and (48) we have S(t) + 1 πE(t) = N(t)−1−1 πϑ(t) + 1 π3π 2+ϑ(t)−πκ(t) =N(t) + 1 2−κ(t) = RH(t) + 1 2−κ(t) + N00(t) so that by (24) for t > aϑwe have (52) S(t) + 1 πE(t) = RH(t)−κ(t)−bκ(t)c− 1 2, t > aϑ from which the result follows.  Corollary 45. Assuming the RH and the simplicity of the zeros we will have (53) −1 2< S(t) + 1 πE(t)≤1 2, t > aϑ. Indeed, the hypotheses are equivalent to RH(t) = 0. By the well known Fourier series of e B1(x) = x−bxc− 1 2we get from (52), under the assumptions of the Corollary (54) S(t) + 1 πE(t) = 2 ∞ X n=1 sin(2πnκ(t)) 2πn , t > aϑ. 10. Extension to other L-functions. Most of the formulas and functions defined in this paper for ζ(s) can be generalized to other functions, including the Selberg class. The main thing we need is a functional equation. So let’s assume that we have a Dirichlet series f(s) = ∞ X n=1 an ns 25