Micah B. Milinovich
ABSTRACT
Assuming the Riemann hypothesis, we obtain an upper bound for the 2kth moment of the derivative of the Riemann zeta-function averaged over the non-trivial zeros of ζ(s) for every positive integer k . Our bounds are nearly as sharp as the conjectured asymptotic formulae for these moments.
1.
Introduction and statement of the main results
Let ((s) denote the Riemann zeta-function. This article is concerned with estimating discrete moments of the following form:
where k ∈ N and the sum runs over the non-trivial (complex) zeros ρ = β + iγ of ζ(s). As usual, the function
denotes the number of zeros of ζ(s) up to a height T counted with multiplicity. It is an open problem to determine the behavior of J k ( T ) as k varies. Independently, Gonek [7] and Hejhal [10] have conjectured that
for fixed k ∈ R as T → ∞. Though widely believed for positive values of k , there is evidence to suggest that this conjecture is false for k ≤ −3/2. Until recently, estimates in agreement with (1.3) were only known in a few cases. Assuming the Riemann hypothesis (which asserts that β = 1 for each non-trivial zero of ζ(s)), Gonek [5] has shown that J 1 (T) ～ ↑ 2 (log T) 3 and Ng [17] has proved that J 2 (T) × (log T) 8 . Confirming a conjecture of Conrey and Snaith [1, Section 7.1], the author [14] has calculated the lower- order terms in the asymptotic expression for J 1 ( T ). Under the additional assumption that the zeros of ζ(s) are simple, Gonek [7] has shown that J−1(T) ≫ (logT)−1 and conjectured [9] that J_1(T) ~ (6/π 2 )(logT)~1 . In addition, there are a few related unconditional results where the sum in (1.1) is restricted to the simple zeros of ζ(s) with β = ξ . See, for instance, [3, 4, 13, 21]. By modeling the behavior of the Riemann zeta-function and its derivative on the critical line using the characteristic polynomials of random matrices, Hughes, Keating and O'Connell [12] have refined Gonek's and Hejhal's conjecture in (1.3). In particular, they conjectured a precise constant D k such that J k ( T ) ～ D k (log T ) k ( k +2) as T → ∞ for fixed k ∈ C with R k > − 3/2. Their conjecture is consistent with the results mentioned above. Very little is known about the moments J k (T) when k > 2. However, assuming the Riemann hypothesis, one may deduce from well-known results of Littlewood ([23, Theorems 14.14 A-B]) that, for σ ≥ 1/2 and t ≥ 10, the estimate
holds for some constant C > 0. It immediately follows that for any k ≥ 0. The goal of this paper is to improve this estimate by obtaining a conditional upper bound for J k ( T ) (when k ∈ N ) very near the conjectured order of magnitude. In particular, we prove the following result. THEOREM 1.1. Assume the Riemann hypothesis. Let k ∈ N and let ε > 0 be arbitrary. Then for sufficiently large T we have where the implied constant depends on k and ε . Under the assumption of the generalized Riemann hypothesis for Dirichlet L-functions, Ng and the author [15] have shown that J k (T) ≫ (log T ) k (k+2) for each fixed k ∈ N . Combining this result with Theorem 1.1 lends strong support for the conjecture in (1.3) when k is a positive integer. Our proof of Theorem 1.1 is based upon a recent method of Soundararajan [22] that provides upper bounds for the frequency of large values of |ς( ž + it )|. His method relies on obtaining an inequality for log | ζ( ξ + it ) | involving the real part of a 'short' Dirichlet polynomial which is a smoothed approximation to the Dirichlet series for log ((s). Using mean-value estimates for high powers of this Dirichlet polynomial, he deduces upper bounds for the measure of the following set:
and from this, for arbitrary positive values of k and ε , is able to conclude that
distribution of values of log ζ( ½ + it ) in the complex plane. Since log ζ ′ (s) does not have a Dirichlet series representation, it is not clear that the function log | ζ ′ ( ž + it ) | can be approximated by a Dirichlet polynomial. † For this reason, we do not study the distribution of the values of ζ ′ ( ρ ) directly, but instead examine the frequency of large values of |ζ(ρ + α)|, where α ∈ C is a small shift away from a zero ρ of ζ(s). This requires deriving an inequality for log | ζ ( σ + it ) | involving a short Dirichlet polynomial that holds uniformly for values of σ in a small interval to the right of, and including, σ = ½ . Using a result of Gonek (Lemma 4.1 below), we estimate high power moments of this Dirichlet polynomial averaged over the zeros of the zeta-function and are able to derive upper bounds for the frequency of large values of |ζ(ρ + α)|. Using this information we prove the following theorem. T HEOREM 1.2.  Assume the Riemann hypothesis. Let α ∈ C with |α | ≪ 1 and | Rα | ≪ (logT) − 1. Let k ∈ R with k > 0 and let ε > 0 be arbitrary. Then, for sufficiently large T, the inequality
holds uniformly in α . Our argument, modified in a straightforward manner, actually implies that the constant ε appearing in power of log T in the inequalities in Theorems 1.1 and 1.2 is O (1/log log log T). Comparing the result of Theorem 1.2 with the estimate in ( 1.4 ), we see that our theorem provides essentially the same upper bound (up to the implied constant) for discrete averages of the Riemann zeta-function near its zeros as can be obtained for continuous moments of |ζ( ž + it )| by using the methods in [22]. There has been some previous work on discrete mean- value estimates of the zeta-function that are of a form that is similar to the sum appearing in Theorem 1.2. For instance, see the results of Fujii [2], Gonek [5], and Hughes [11]. We deduce Theorem 1.1 from Theorem 1.2 since, by Cauchy's integral formula, we can use bounds for ζ(s) near its zeros to recover bounds on the values for ζ ′ (ρ). For a precise statement of this idea, see Lemma 7.1 below. Our proof only allows us to establish Theorem 1.1 when k is a positive integer despite the fact that Theorem 1.2 holds for all k > 0.
Throughout the remainder of this article, we use s = σ + it to denote a complex variable and use p to denote a prime number. We let λ 0 = 0.5671... be the unique positive real number satisfying e-λ o = λ o . Also, we put σχ = σχ,x = 1/2 + λ/log x and let
As usual, we denote by Λ( - ) the arithmetic function defined by Λ( n ) = log p when n = p k and Λ( n ) = 0 when n̸= p k . The main result of this section is the following lemma. Lemma 2.1. Assume the Riemann hypothesis. Let τ = | t | + 3. Then, for λ 0 ≪ λ ≪ ½ log x and 3 ≤ x ≤ τ², the estimate
holds uniformly forｶ≪ σ ≪ σ ≪ｼIn [22], Soundararajan proved an inequality similar to Lemma 2.1 for log | ς( ž + it ) | . In his case, when ζ( ξ + it)̸= 0, an inequality slightly stronger than (2.1) holds with the constant λ 0 replaced by δ 0 = 0.4912…, where δ 0 is the unique positive real number satisfying e − δ 0 = δ 0 + ž δ 2 . Our proof of the above lemma is a modification of his argument. Proof of Lemma 2.1. We assume that | ζ (σ + it) | ≥ 1, as otherwise the lemma holds for a triv- ial reason. In particular, we are assuming that ζ ( σ + it )̸= 0. Assuming the Riemann hypothesis, we denote a non-trivial zero of ζ(s) as ρ = ξ + iγ and define a function F(s) as follows:
Note that F(s) ≥ 0 whenever σ ≥ 1/2 and s ≠ ρ . The partial fraction decomposition of ζ′(s)/ζ(s) (see [23, equation (2.12.7)]) says that, for s̸= 1 and s not coinciding with a zero of ζ(s), we have
where the constant B = —発= log 2π − 1 − ž γ o , in which γ o denotes Euler's constant.
for σ < ³ , say. Stirling's asymptotic formula for the gamma function implies that
uniformly for ½ ≪ σ ≪ ¾. Consequently, the inequality
holds uniformly for f ≪ σ ≪ σ λ ≪ } . Here, while using the inequality ( 2.5 ) in the second line of (2.6), we have implicitly assumed that λ < ξ log x to ensure that σ λ < ξ . To complete the proof of the lemma, we require an upper bound for log | ζ (σ λ + it) | which, in turn, requires an additional identity for ζ ′ ( s )/ζ ( s ). Specifically, for s̸= 1 and s not coinciding
using the assumption that 3 < x < τ 2 , we deduce from the above identity that
We now estimate the second and third terms on the right-hand side of this expression. Using the first inequality in ( 2.5 ), it follows that
for 0 ≤ λ ≤ ± log x . Also, by observing that
and then combining (2.9) and (2.10) with (2.8), we see that
If λ ≥ λ 0 , then the term on the right-hand side involving F (σ λ + it) is less than or equal to zero, and hence omitting it does not change the inequality. Thus, when λ ≥ λ 0 , we have
Since we have assumed that | ζ(σ + it)| ≥ 1, the lemma now follows by combining the inequalities in ( 2.6 ) and ( 2.11 ) and then taking absolute values. □
3.
A variation of Lemma 2.1
In this section, we prove a version of Lemma 2.1 in which the sum over n on the right-hand side of the inequality is restricted just to the primes. A sketch of the proof of the lemma appearing below has been given previously by Soundararajan (see [22, Lemma 2]). Our proof is different and the details are provided for completeness. Lemma 3.1. Assume the Riemann hypothesis. Consider τ = | t | + 100. Then
uniformly for σ ≥ ½ and | t | ≥ 1. As a consequence, for any λ satisfying λ 0 ≪ λ ≪ ½ log x and 3 ≤ x ≤ τ², the estimate
holds uniformly forヵ≤ σ ≤ σ λ and | t | ≥ 1.
otherwise
changing the order of summation and integration (which is justified by absolute convergence), it follows that
Here we have made use of the following standard identity:
which is valid for c>0. By moving the line of integration in the integral left to Rs = − u,
follows from the Riemann hypothesis. To estimate the integral on the right-hand side of the above expression, we use [23, Theorem 14.5], namely, that if the Riemann hypothesis is true, then
Inserting this estimate into equation (3.2) and dividing by log z , it follows that
see [16, Section 13.2]. Using the estimates in (3.5) and recalling that we are assuming that
This establishes ( 3.1 ) and, thus, the lemma.
In this section we prove an estimate for the mean-square of a Dirichlet polynomial averaged over the zeros of ((s). Our estimate follows from the Landau–Gonek explicit formula.
log p if x is a positive integral power of a prime p, and Λ( x ) = 0 otherwise.
Assume the Riemann hypothesis and let ρ = ξ + iγ denote a non-trivial zero LEMMA 4.2. of ζ(s). For any sequence of complex numbers ♟ = { α n } ∞ 1 and for ξ ≥ 1, we define
Proof. Assuming the Riemann hypothesis, we note that 1 − ρ = ρ for any non-trivial zero ½ + iγ of ζ(s). This implies that ρ =
and, moreover, that
that
Appealing to Lemma 4.1, we find that
where
and
Again using the assumption that Rα ≥ 0, we find that
Here we have made use of the standard estimate ∑ m ξ ε Λ( m )/m ≪ log ξ . We can replace Rα by 0 in each of the sums Σ i (for i = 2,3 or 4), as doing so will only make the corresponding estimates larger. Thus, using the assumption that 3 < ξ < T/ logT, it follows that
Next, turning to Σ 3 , we find that
Writing n as qm + ℓ with −m/2 < ℓ ≤ m/2, we have
where, as usual, [ x ] denotes the greatest integer less than or equal to x . Now 〈q + ℓ / m 〉 = |ℓ|/m if q is a prime power and ℓ̸= 0, otherwise ⟨ q + ℓ/m ⟩ is at least ﾗ. Using the estimate Ence Λ(n) ≪ ξ, we now find that
It remains to consider the contribution from Σ 4 , which is much less than
Now, by combining the estimates, we obtain the lemma.
5. The frequency of large values of |ζ(ρ + α)| Our proof of Theorem 1.2 requires the following lemma concerning the distribution of values of |ζ(ρ + α)|, where ρ is a zero of ζ(s) and α ∈ C is a small shift. In what follows, log 3 ( · ) stands for  log log log( • ). Lemma 5.1. Assume the Riemann hypothesis. Let T be large, let V ≥ 3 be a real number and let α ∈ C with |α | ≪ 1 and 0 ≪ Rα ≪ (log T ) − 1 . Consider the set
where ρ = ξ + iγ denotes a non-trivial zero of ζ (s). Then, the following inequalities for #Sα(T;V), the cardinality of Sα(T;V), hold.
Here, as usual, the function N(T) ～ (T/2π) logT denotes the number of zeros ρ of ζ(s) with 0 < γ < T . Since λ 0 < 3/5, by taking x = (log τ ) 2 − ε in Lemma 3.1 (where ε > 0 arbitrary) and Proof. estimating the sum over primes trivially, we find that
(2/5)(log T /log log T ), for otherwise the set S α ( T ; V ) is empty. We define a parameter
< T1/2 set x = TA/V  and put z = x 1 / 10g 10g T . We now observe that since λ 0 ≥ ±, x < TA/V and 0 ≤ Rα ≤ (log T )− 1 , it follows that
Therefore, if ρ ∈ Sα(T;V), then either
Let N 1 (T;V) be the number of ρ with 0 < γ < T such that | S 1 (ρ)| ≥ V 1 and let N 2 (T;V) be the number of ρ with 0 < γ ≪ T such that | S 2 ( ρ ) | ≫ V 2 . We prove the lemma by obtaining upper bounds for the size of the sets N i ( T ; V ) for i = 1 and 2 using the inequality
which holds for any positive integer k . With some restrictions on the size of k , we can use Lemma 4.2 to estimate the sums appearing on the right-hand side of this inequality. We first turn our attention to estimating N 1 ( T ; V ). If we define the sequence α k ( n ) = α, (n, x, z) by
deduce from this estimate that
It is now convenient to consider separately the case when V ≪ (log log T ) 2 and the case V > (log log T )². When V ≪ (log log T )² we choose k = [ V 2 / log log T ] where, as before, [ x ] denotes the greatest integer less than or equal to x . To see that this choice of k satisfies z k ≪ T (log T ) − 1 we note from the definition of A that V A ≤ max (V,  ½ (log log T) log 3 T) Therefore, we find that
Thus, by (5.2), we see that for V < (log log T ) 2 and T large we have
When V > (log log T )² we choose k = [10V]. This choice of k satisfies zk ≪ T (log T )− 1 since 210V = T10/log10gT ≪ T (logT )−1 for large T . With this choice of k , we conclude from (5.2) that
follows from (5.4) that N 1 (T;V) ≪ N (T) exp ( − 4V log V)
By combining (5.3) and (5.5), for any choice of V , we have shown that,
β k (n, x, z) by
then it can be seen that | β k ( n ) | ≪ k !. Thus, Lemma 4.2 implies that
1] satisfies x k ≪ T / log T when T is large. To see why, recall that A ≫ 1, x = T A /V  and V (2/5)(log T /log log T ). Therefore, we have
Also, observing that A < ½ log 3 (T) and recalling that V ≥ √ log log T , with this choice of k and T large, it follows from (5.1) that
Using our estimates for N 1 ( T ; V ) and N 2 ( T ; V ) we can now complete the proof of the lemma by checking the various ranges of V . By combining (5.6) and (5.8), we see that
(5.9) that
Certainly,

Proof of Theorem 1.2
6.
Using Lemma 5.1, we first prove Theorem 1.2 in the case where | α | ≪ 1 and 0 ≪ Rα ≪ (log T ) − 1. Then, from this result, the case when − (log T ) − 1 ≤ Rα < 0 can be deduced from the functional equation for ζ(s) and Stirling's formula for the gamma function. In what follows, k ∈ R is fixed and we let ε > 0 be an arbitrarily small positive constant that may not be the same at each occurrence. First, we partition the real axis into the intervals I 1 = ( − ∞, 3], I 2 = (3, 4k log log T ]  and I 3 = (4k log log T , ∞) and set
Using the trivial bound # S α ( T,ν ) ≪ N ( T ), which holds for every ν ∈ Z , we find that M 1 ≪ e 6k N(T). To estimate M 2 , we use the bound
When ν ∈ I 3 ∩ Z , the second two cases of Lemma 5.1 imply that ν) ≪ N(T)(logT) = Thus, we have
In light of ( 6.1 ), by collecting estimates, we see that
for every k > 0 when | α | ≪ 1 and 0 ≪ Rα ≪ (log T ) − 1 The functional equation for the zeta-function states that ζ(s) = χ(s)ζ(1 − s), where χ(s) = 2 s π s − 1 Γ(1 − s) sin(πs/2). Stirling's asymptotic formula for the gamma function (cf. [16, Theorem C.1]) can be used to show that uniformly for −1 < σ < 2 and | t | ≥ 1. Using the Riemann hypothesis, we see that for some absolute constant C > 0 when | α | < 1, | Rα | < (log T ) − 1 and 0 < γ < T . Consequently, for −(logT)− 1 ≤ Rα < 0, we have
for every k > 0 when |α| ≪ 1 and −(log T )− 1 ≪ Rα ≪ 0. The theorem now follows from the estimates in (6.2) and (6.4).
7.

Theorem 1.1 can now be established as a simple consequence of Theorem 1.2 and the following lemma. Assume the Riemann hypothesis. Let k , l ∈ N and let R > 0 be arbitrary. LEMMA 7.1. Then we have
Since the function ζ ( ℓ ) ( s ) is real when s ∈ R , it follows that ζ ( ℓ ) ( s ) = ζ ( t ) ( s ). Hence, Proof. assuming the Riemann hypothesis, the identity
holds for any non-trivial zero ρ of ζ(s) and any α ∈ C. For each positive integer k , let α k = (α 1 , α 2 , . . . , α 2 k ) and define
function Z(s; α k ) and then apply the inequality in (7.3). By Cauchy's integral formula and another application of (7.2), we see that
where, for each i = 1, . . . , 2k, the contour & denotes the positively oriented circle in the complex plane centered at 0 with radius R. Now, combining (7.3) and (7.4) we find that as claimed.
Let k ∈ N and set R = (log T ) − 1 . Then, it follows from Theorem 1.2 Proof of Theorem 1.1. and Lemma 7.1 that
(7.5) for any ℓ ∈ N and for ε > 0 arbitrary. Theorem 1.1 now follows by setting ℓ = 1.
Acknowledgement. This work constitutes a portion of my Ph.D. Thesis. I would like to thank my advisor, Steven M. Gonek, for his support and encouragement during my graduate studies. I would also like to thank Professor K. Soundararajan for some helpful conversations regarding his paper [22].