On certain sums over ordinates of zeta-zeros II

# On certain sums over ordinates of zeta-zeros II

Andriy Bondarenko & Aleksandar Ivić & Eero Saksman & Kristian Seip
###### Abstract

Let denote the imaginary parts of complex zeros of . The problem of analytic continuation of the function to the left of the line is investigated, and its Laurent expansion at the pole is obtained. Estimates for the second moment on the critical line are revisited. This paper is a continuation of work begun by the second author in [9].

Dedicated to the memory of S. Srinivasan (1943 – 2005)

\@footnotetext

Saksman’s research was supported in part by the Lars Onsager Professorship at NTNU and in part by the Finnish Academy CoE “Analysis and Dynamics”. The research of Bondarenko and Seip was supported in part by Grant 227768 of the Research Council of Norway.

• 2010 Mathematics Subject Classification. 11M06
Keywords. Riemann zeta-function, Riemann hypothesis, analytic continuation, Laurent expansion, second moment

## 1. Introduction: the function G(s)

We will be concerned with the Dirichlet series

 G(s):=∑γ>0γ−s, (1.0)

where denotes the ordinates of the complex zeros of the Riemann zeta-function , counted as usual with multiplicities. Thus converges absolutely and represents a holomorphic function in the half-plane , in view of the classical Riemann–von Mangoldt formula

 N(T)=T2πlogT2π−T2π+O(logT) (1.0)

for the zero counting function (see [8, p. 17] or [15, p. 214]). The condition in (1.) is natural, since the property ensures that is a zero of if is a zero.

The function is mentioned in the work of Delsarte [3] and, in a perfunctory way, in the works of Chakravarty [1, 2]. A related zeta function, namely

 ∑γ>0γ−ssin(αγ)(α>0),

was studied by Fujii [4], but its properties are different from those of , and we shall not consider it here. Both Chakravarty and Delsarte (as well as Fujii) assume the Riemann hypothesis (that all complex zeros of satisfy ) when dealing with . Delsarte [3, p. 430] obtains its analytic continuation to by employing Poisson summation and a sort of a modular relation.

We discuss analytic continuation of Delsarte’s function in Section 2. We give a different proof of Delsarte’s result on the analytic continuation under the Riemann hypothesis. Moreover, we note that meromorphic extension to a larger half-plane would have strong consequences related to the density hypothesis. In Section 3., we provide an explicit Laurent expansion of at . Section 4. is devoted to estimating the second moment . There the growth of the second moment is connected explicitly to the fluctuations of the function (see (2.0) below), and we provide lower and upper bounds for the growth. Finally, in Section 5. we observe how the connection of to the “super zeta functions” studied in the monograph [16]111The study of such zeta functions goes back to Mellin [10] in 1916 and seems not to be widely known. leads to yet another approach to Delsarte’s result.

## 2. Analytic continuation of G(s)

The second author introduced and studied in [9]. Starting from the classical formula

 N(T)=T2πlogT2π−T2π+78+S(T)+f(T), (2.0)

it was shown that, for and a suitable constant ,

 G(s)=12π(s−1)2−log(2π)2π(s−1)+C1+s∫∞1(S(x)+f(x))x−s−1dx. (2.0)

Here, if is not an ordinate of a zero,

 S(T)=1πargζ(12+iT)=1πIm{logζ(12+iT)}≪logT, (2.0)

where the argument of is obtained by continuous variation of the argument of along the straight lines joining the points 2, , , starting with the value 0 at . If is an ordinate of a zero, then . The function is smooth and admits an asymptotic expansion in terms of negative odd powers of , namely for any given integer one has

 f(T)=N∑j=1ajT2j−1+ON(1T2N+1), (2.0)

where the are explicit constants, and means that the implied -constant depends only on . From (2.0)–(2.0) one obtains the analytic continuation of for .

Further analytic continuation will follow by integrating the integral in (2.0) by parts. This will give, for and a suitable constant ,

 G(s) =12π(s−1)2−log2π2π(s−1)+C1+s∫∞1f(x)x−s−1dx (2.0) +s(s+1)∫∞1∫x1S(u)du⋅x−s−2dx,

since we have the (unconditional) bound (see [15, pp. 221–222])

 ∫T0S(t)dt=O(logT). (2.0)

To study the analytic continuation of for we need some notation. Following [5], we define

 ~S0(T):=S(T)(T≠γ),

and for ,

 ~Sm(T):=∫T0~Sm−1(t)dt+Cm, (2.0)

where and

 C2k−1:=(−1)k−1π∫∞12∫∞σ1⋯∫∞σ2k−2log|ζ(σ)|dσ2k−1⋯dσ1

for every positive integer . When , we put

 ~Sm(T):=12(~Sm(T+0)+~Sm(T−0)).

On the Riemann hypothesis, it is known that (see [15, pp. 350–354])

 ~Sm(T)≪mlogT(loglogT)m+1(m⩾1). (2.0)

Here means that the constant implied by the -symbol depends only on . Note that

 ∫∞1x−2j+1−s−1dx=12j+s−1(σ>1−2j,j∈N). (2.0)

Thus using (2.0) and (2.0), we see that

 s∫∞1f(x)x−s−1dx

in (2.0) represents a meromorphic function in , which has simple poles at . Hence repeated integrations by parts of the last term in (2.0) show, on using (2.0), that under the Riemann hypothesis the function admits meromorphic continuation to . Its only poles are: (order two) and (order one).

We next look at what are the consequences if we do not assume the Riemann hypothesis to hold, but instead assume that admits a meromorphic continuation across the line . The integration by parts argument used in the preceding discussion implies in particular that then the function

 H1(s):=∫∞1~S2(x)x−s−3dx (2.0)

continues meromorphically across the same line. By [5, Theorem 3], we have

 ~S2(T)=∑β>1/2,0<γ⩽T(β−1/2)2+O(logT)=A(T)+O(logT), (2.0)

where we have set . We therefore obtain that also the function

 H(s):=∫∞1A(x)x−s−3dx

extends meromorphically across .

Now let as usual be the number of zeros with and . We may then write

 A(x)=2∫11/2(σ−12)N(σ,x)dσ.

Namely the right-hand side above equals

 2∫11/2(σ−12)∑γ⩽x,β⩾σ⩾121dσ =2∑γ⩽x∫β12(σ−12)dσ =∑γ⩽x,β⩾12(β−12)2=A(x).

By Selberg’s uniform density estimate [15, Theorem 9.19 (C)],

 A(x)≪xlogx∫1/20yx−y/4dy≪xlogx.

Now

 ∫∞ex−ε−1logxdx=∫∞1e−εuudu⩽∫1/ε1duu+ε∫∞0e−εudu⩽log1ε+1.

Hence cannot have a pole at , and by our assumption it thus has to be analytic at . Then since is non-negative and increasing, a classical theorem of Landau [14, Theorem 6, Chapter II.1] implies that there is such that

 A(x)≪x1−δ.

It follows from the definition of and the fact that is a decreasing function that for ,

 N(σ,T)≪(σ−1/2)−2T1−δ, (2.0)

with the implicit constant depending only on . For close to , this assertion is stronger than the density hypothesis that .

We summarize the preceding discussion as follows:

###### Theorem 2.1

Unconditionally, has meromorphic continuation at least to ; it has only one pole in this half-plane, located at and of order two. If the Riemann hypothesis is true, then extends meromorphically to with additional poles that are all simple and located at . If continues meromorphically across , then the strong implication (2.0) holds for the density of the zeros off the critical line.

###### Remark 2.2

As noted in Section 1., the meromorphic continuation of to on the Riemann hypothesis was already obtained by Delsarte [3], but the above proof is simpler (see also Section 5. below). The last assertion of Theorem 2.1 indicates that establishing meromorphic continuation across the line is likely to be very difficult.

Notice that we may have meromorphic extension to the whole complex plane even if the Riemann hypothesis fails. This will for instance be the case if there are finitely many nontrivial zeros with . To see this, we write (2.0) in the form . Plugging this into (2.0), we see that the term gives rise to an additional simple pole at , while the term , by repeated integration by parts and use of [5, Theorem 1], yields an entire function.

An elaboration of this observation, along the lines of our discussion of possible meromorphic continuation across , shows that the distinctive feature of the meromorphic continuation obtained on the Riemann hypothesis is that will be analytic on the segment . This leads to the following statement.

###### Theorem 2.3

The Riemann hypothesis holds if and only if has meromorphic continuation to the half-plane with only two poles that are located at respectively and .

## 3. The Laurent expansion of G(s) at s=1

From (2.0) one can obtain the (unconditional) Laurent expansion of at its pole . Namely, denoting for brevity, we have for ,

 ∫∞1F(x)x−s−1dx=∫∞1F(x)e−(s−1)logxx−2dx=∞∑j=0cj(s−1)j, (3.0)

say, with

 cj:=(−1)jj!∫∞1F(x)logjx⋅x−2dx(j=0,1,…). (3.0)

Here we could interchange integration and summation in view of absolute convergence; this follows from the bound which holds because . Writing the last expression in (2.0) as

 (s−1)∫∞1F(x)x−s−1dx+∫∞1F(x)x−s−1dx

and using (3.0), we obtain

###### Theorem 3.1

The Laurent expansion of at has the form

 G(s)=12π(s−1)2−log2π2π(s−1)+∞∑j=0bj(s−1)j,

where

 b0=C1+c0,bj=cj+cj−1(j=1,2,…),

and the are given by (3.0).

## 4. Estimates for the second moment

We will now provide upper and lower bounds for the second moment

 ∫T1|G(12+it)|2dt,

thus continuing the investigations started in [9, Section 2]. To this end, we follow [9] and write

 G(s)=∑γ⩽Xγ−s+R(s),

where

 R(s)=∫∞Xx−s2πlogx2πdx+∫∞Xx−sd(S(x)+f(x)).

Integrating by parts, we get

 R(s)=X1−s2π(s−1)logX2π+X1−s2π(s−1)2−X−s(S(X)+f(X))+s∫∞Xx−s−1(S(x)+f(x))dx, (4.0)

which to begin with is valid for , but which continues meromorphically for all . For , the integral on the right-hand side of (4.0) may be written as , where . Hence elementary estimates and Parseval’s identity together with Selberg’s bound yield

 ∫T1∣∣R(12+it)∣∣2dt≪Xlog2X+(T2/X)loglogX.

Choosing we get

 ∫T1∣∣R(12+it)∣∣2dt≪TlogT√loglogT. (4.0)

As we will see in (4.0) below, when for some positive number . Hence the second moment of the partial sum yields the dominant term which needs to be studied in more detail.

Let be a symmetric function in satisfying222We may for example choose .

 0⩽ψ(x)⩽χ[−1,1](x)and0

for all in . Let be an arbitrary subset of . We find that

 ∫T−T∣∣∑γ∈Aγ−1/2−it∣∣2dt ⩾∫R∣∣∑γ∈Aγ−1/2−it∣∣2ψ(t/T)dt=T∑γ,γ′∈A(γγ′)−1/2ˆψ(Tlog(γ/γ′)/2π) ≍T∑γ,γ′∈A(γγ′)−1/211+T2(log(γ′/γ))2.

Fix an arbitrary constant and replace by in the above computation. We obviously obtain an upper estimate up to a constant, but as , we infer that

 ∫T−T∣∣∑γ∈Aγ−1/2−it∣∣2dt ≍T∑γ,γ′∈A(γγ′)−1/211+T2(log(γ′/γ))2 (4.0) ≍T∑γ,γ′∈A(γγ′)−1/2ψ(Talog(γ′/γ)) (4.0) ≍T∑γ∈Aγ−1#{γ′∈(0,T)∩A:|γ′−γ|⩽γ/(aT)}.

As the parameter is arbitrary, we see that there is some flexibility in choosing the size of the “window” in (4.0) yielding the pairs of ordinates that contribute significantly to the second moment.

By considering only the diagonal in (4.0) and using the known average density of the ordinates, we obtain

 ∫T−T∣∣∑0<γ⩽Xγ−1/2−it∣∣2dt≫T∑0<γ≤Tbγ−1≫Tlog2T (4.0)

whenever for some exponent in .

Our next goal is to express (4.0) in terms of either of the functions and in the special case when . We start with the basic formula involving .

###### Lemma 4.1

Suppose that . Then

 (4.0)
• We start from (4.0) in the form

 ∫T−T∣∣∑0<γ⩽Xγ−1/2−it∣∣2dt≍T∑1<γ≤Xγ−1#{γ′:|γ′−γ|⩽γ/(aT)}. (4.0)

Here we discarded the condition that in the sum on the right-hand side. The right-hand side has thus been increased by a term of size which is admissible because the left-hand side is . We have also taken into account that implies .

Set and . We divide into intervals for ; it will be convenient to agree that For any real interval let denote the number of zeta zeros with ordinates in . Then for an arbitrary in we obtain

It follows that

 ∑γ∈(bk−1,bk]γ−1#{γ′∈(0,T):|γ′−γ|⩽γ/T}⩽ b−1k−1n(Ik)(n(Ik−1)+n(Ik)+n(Ik+1)) ≪ b−1k−1n(Ik−1)2+b−1kn(Ik)2+b−1k+1n(Ik+1)2.

On the other hand, we clearly have

 ∑γ∈(bk−1,bk]γ−1#{γ′∈(0,T):|γ′−γ|⩽2γ/T}⩾b−1kn(Ik)2.

Summing both the upper and lower estimates over and using (4.0), we get

 ∫T−T∣∣∑0<γ⩽Xγ−1/2−it∣∣2dt≍T∑kb−1kn(Ik)2. (4.0)

We may neglect the first interval and consider with . Divide each into three equally long subintervals. One of these three intervals contains at least ordinates, and there is then a subinterval of with such that for every in . Since for in we thus obtain

 Tb−1kn(Ik)2 ≪Tb−1k|Ik|−1∫Ik−1∪Ik(N(t+t/T)−N(t))2)dt ≪∫Ik−1∪Ik(N(t+t/T)−N(t))2)dtT2t2≪Tb−1k(n(Ik−1)+n(Ik)+n(Ik+1))2,

where the last inequality follows on noting that for interior points in . Summing over and using (4.0), we finally arrive at (4.0).

We may alternatively replace by in (4.0):

 ∫T1∣∣∣∑0<γ⩽Xγ−1/2−it∣∣∣2dt≍∫X1(S(t+t/T)−S(t)t/T)2dt. (4.0)

Indeed, by (2.0), (4.0) implies that

 ∫T1∣∣∣∑0<γ⩽Xγ−1/2−it∣∣∣2dt≪∫X1(S(t+t/T)−S(t)t/T)2dt+Xlog2X.

Since the left-hand side is , we get that the second term on the right-hand side can be removed. Since (2.0) implies that

 N(t+t/T)−N(t)⩾S(t+t/T)−S(t),

it is also clear that (4.0) trivially implies

 ∫T1∣∣∣∑0<γ⩽Xγ−1/2−it∣∣∣2dt≫∫X1(S(t+t/T)−S(t)t/T)2dt.

The above lemma as expressed by (4.0) shows that the asymptotics that we are interested in, depends crucially on the oscillations of in rather small intervals. To gain more insight into this asymptotics, we will next take a closer look at the size of the dyadic parts

 D(x):=∫T1∣∣∑x<γ⩽2xγ−1/2−it∣∣2dt.
###### Proposition 4.2

(i) Assume that . Then

 xlog2T≪D(x)≪Tlog2T.

(ii) Assume that Then

 Tlogx≪D(x)≪Tlogx√loglogx.
• We may assume that is an integer. Divide the interval into subintervals for . By (4.0), arguing as in the proof of (4.0), we then get

 D(x)≍Tx∑x⩽γ<2x∣∣∣{γ′:|γ−γ′|⩽xT}∣∣∣≍TxT∑j=1n(Ij)2. (4.0)

Since we know that , an application of the Cauchy–Schwarz inequality to (4.0) yields immmediately the lower bound To obtain the upper bound, we divide the ordinates in into 1-separated sets, and apply the Montgomery–Vaughan inequality [11, Corollary 2] to each of the corresponding sums.

In the proof of the upper bound of part (ii), we will use the following lemma.

###### Lemma 4.3

There is a universal constant such that if for a subinterval of , then

 |{t∈2I:|S(t)|⩾n(I)/8}|⩾c|I|.
• By the Riemann–von Mangoldt formula, the assumption implies that

 S(b)−S(a)⩾12|I|logx.

This implies that if , then in which case may choose a positive constant independent of such that the desired bound holds on the interval . On the other hand, if , then holds on the interval .

• We divide the interval into the same subintervals as in the preceding case. The lower bound of part is again proved by the Cauchy–Schwarz inequality. Indeed, observing that the number of non-zero summands in the sum over is at most of order , we find that

 D(x)≍Tx2logxxlogxT∑j=1n(Ij)2⩾Tx2logx(xlogx)2=Tlogx.

We turn to the proof of the upper bound. To begin with, note that since , we have . This means that satisfies the condition of Lemma 4.3 whenever . This fact allows us to make the following construction. Let be the smallest such that , and let be the largest such that

 n(k−1⋃m=0Ij1+m)⩾kxTlogx.

Now choose and inductively such that is the smallest such that and while is the largest such that

 n(k−1⋃m=0Ijℓ+m)⩾kxTlogx.

We terminate this iteration when . We set

 Jℓ:=kℓ−1⋃m=0Ijℓ+m

and notice that since

 n(I)=|I|2πlogx+O(logx).

It is also clear, by the maximality of , that

 n(Jℓ)=(1+o(1))|Jℓ|logx. (4.0)

Plainly by the construction, since , all the ordinates in will be covered by the intervals , and we thus have

 ∑jn(Ij)2⩽∑ℓn(Jℓ)2. (4.0)

Now, by a standard covering argument (see e.g. [6, Proof of Lemma 4.4]), we can find a subcollection of the intervals 333If is an interval, stands for the interval with same center and twice the length. such that no point in belongs to more than two intervals and such that the union of covers all the intervals . In particular, we have

 ∑ℓ|Jℓ|2≪∑J′|J′|2.

Combining this with Lemma 4.3 and (4.0), we get

 ∑ℓn(Jℓ)2 ≪logx∑ℓ(logx)|Jℓ|⋅|Jℓ|≪logx∑J′(logx)|J′|⋅|J′| ≪logx∫2xx|S(t)|dt≪xlogx√loglogx,

where in the last step we used the Cauchy–Schwarz inequality and Selberg’s mean value theorem444Alternatively, we could appeal to Ghosh’s asymptotic formula for the first moment of [7]. for . Recalling (4.0) and (4.0), we obtain the upper bound of part (ii).

As an immediate consequence of the preceding analysis, we now obtain a mean square estimate 555The upper bound was stated in [9] but with an incomplete proof. for the function on the critical line:

###### Theorem 4.4

• We choose the length of the partial sum to be so that the contribution from the remainder term is given by (4.0). By (4.0) (or directly by (4.0)) there is quasi-orthogonality between the dyadic parts of the sum (i.e., dyadic parts that are not adjacent are orthogonal with respect to a weighted norm, where the size of the weight is approximately ). We apply part (i) of Proposition 4.2 to suitable dyadic subintervals of and part (ii) to subintervals of . Summing over all these dyadic intervals, we obtain the desired upper bound. The lower bound is a consequence of (4.0) and (4.0).

We expect, partly in view of the very regular behavior of the ordinates predicted by the random matrix analogy, that the true order for dyadic summands is akin to something like

 D(x)≍{xlog2TforT/logT⩽x⩽T,Tlogxforx⩽T/logT. (4.0)

We note that this prediction corresponds to the lower bounds in Proposition 4.2. This leads us to the following

###### Remark 4.6

On the Riemann hypothesis, we may split the sequence of ordinates in the range in sequences that are all separated with separation constant . Using again the Montgomery–Vaughan inequality, we then obtain the conjectured bound for the sum over in this restricted range, namely

 ∫T1∣∣∑T/logT⩽γ⩽Tγ−1/2−it∣∣2dt≪Tlog2T.

## 5. On super zeta functions

In 1917 Mellin [10] considered a class of zeta functions that are defined by using the Riemann zeta zeros as the building block of new Dirichlet series. For example, he established meromorphic continuation to the whole complex plane for a class of such functions including the function

 Z(s):=∑ρρ−s,

where the sum is taken over all nontrivial zeros of the Riemann zeta function. Voros has devoted a monograph [16] on generalisations of such functions, for which he coins the name super zeta functions666We urge the reader to take a look at Frankenhuijsen’s MathSciNet review of the book, where Serge Lang’s point of view of “climbing the ladder” is explained..

For the convenience of the reader, we now sketch a proof of the existence of a meromorphic extension of the function

 Mα(s):=∑ρ(α−ρ)−s. (5.0)

to the whole complex plane. This function is initially well-defined in the half-plane , as one sets for in , using the convention that is real-valued for real . We will will consider only with , although we could consider the more general case. However, our main interest lies in the case , which under the Riemann hypothesis yields another approach to Delsarte’s result. We refer to [16, Section 7] which contains a detailed discussion of the analytic continuation, and also note that [16, Appendix D] provides an English translation of Mellin’s work [10] which was originally written in German.

Clearly (5.0) defines a function analytic in the half-plane . Set and let be the “Hankel contour” going from below the real line to , where it follows the semicircle of radius centred at to the left of to , and then going back to . Then the integral

 Eα(s):=12πi∫ηαζ′(w)ζ(w)(α−w)−sdw

defines an entire function , by the rapid decay of . For , the contour can be moved into the left half-plane and towards in the usual way and therefore

 ζα(s)=Eα(s)−∞∑n=0(α+2+2n)−s=Eα(s)−2−sζ(s,α/2+1), (5.0)

where as usual is the Hurwitz zeta function. The meromorphic continuation follows since the Hurwitz zeta function is known to be a meromorphic function in which is analytic in with a simple pole of residue at .

###### Remark 5.1

We notice that in the case , we may write

 Eα(s)=(α−1)−s+12πi∫η′α