Lower Bounds of Discrete Moments of the Derivatives of the Riemann Zeta-Function on the Critical Line Tome 25, No 2 (2013), P

Thomas CHRIST et Justas KALPOKAS Lower bounds of discrete moments of the derivatives of the Riemann zeta-function on the critical line Tome 25, no 2 (2013), p. 285-305.

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cedram Article mis en ligne dans le cadre du Centre de diffusion des revues académiques de mathématiques http://www.cedram.org/ Journal de Théorie des Nombres de Bordeaux 25 (2013), 285–305

Lower bounds of discrete moments of the derivatives of the Riemann zeta-function on the critical line

par Thomas CHRIST et Justas KALPOKAS

Résumé. Nous établissons des bornes inférieures incondition- nelles pour certains moments discrets de la fonction zêta de Rie- mann et de ses dérivées dans la bande critique. Nous utilisons ces moments discrets pour donner des bornes inférieures incondition- R T (l) 1 2k nelles pour les moments continus Ik,l(T ) = 0 |ζ ( 2 + it)| dt, où l est un entier positif et k ≥ 1 un nombre rationnel. En parti- culier, ces bornes inférieures sont de l’ordre de grandeur attendu pour Ik,l(T ).

Abstract. We establish unconditional lower bounds for certain discrete moments of the Riemann zeta-function and its derivatives on the critical line. We use these discrete moments to give unconditional lower bounds for the continuous moments Ik,l(T ) = R T (l) 1 2k 0 |ζ ( 2 + it)| dt, where l is a non-negative integer and k ≥ 1 a rational number. In particular, these lower bounds are of the expected order of magnitude for Ik,l(T ).

1. Introduction and statement of the main results In this paper, we investigate the value-distribution of the Riemann zeta- 1 function ζ(s) on the critical line s = 2 + iR. Recall the functional equation of the zeta-function, s s−1 πs (1.1) ζ(s) = ∆(s)ζ(1 − s), where ∆(s) := 2 π Γ(1 − s) sin( 2 ). 1 It follows immediately that ∆(s)∆(1 − s) = 1; hence ∆( 2 + it) lies on the unit circle for real t. For a given angle φ ∈ [0, π), we denote by tn(φ), n ∈ N, the positive roots of the equation 2iφ 1 e = ∆( 2 + it) in ascending order. These roots correspond to intersections of the curve 1 iφ t 7→ ζ( 2 + it) with straight lines e R through the origin (see Kalpokas and Steuding [10] for more details). In particular, the points tn(0) that

The first author would like to thank the Hanns-Seidel-Stiftung for their support. The second author was partially supported by grant No MIP-066 from the Research Council of Lithuania. 286 Thomas Christ, Justas Kalpokas are obtained by the special choice of φ = 0 correspond to intersections of 1 t 7→ ζ( 2 + it) with the real axis and are called Gram points (named after Gram [5] who observed that the first of these points seperate ordinates of consecutive zeros on the critical line). For fixed φ ∈ [0, π), the number Nφ(T ) of points tn(φ) which lie in the intverval (0,T ] is asymptotically given by T T N (T ) = log + O(log T ). φ 2πe 2πe For a proof we refer to Kalpokas and Steuding [10]. In the following, we investigate the growth behaviour of discrete moments 2k X (l) 1 Sk,l(T, φ) := ζ 2 + itn(φ) , 0

Theorem 1.1. For any rational k > 1 and any non-negative integer l, uniformly for φ ∈ [0, π), as T → ∞,

k2+2kl+1 Sk,l(T, φ) l,k T (log T ) . Theorem 1.1 generalizes a result of Kalpokas, Korolev and Steuding [11], who obtained the lower bound for Sk,l(T, φ) in the case l = 0. Under the assumption of the Riemann hypothesis, the authors proved in [1] that for non-negative integers k and l, resp. non-negative real k if l = 0, uniformly for φ ∈ [0, π), as T → ∞,

k2+2kl+1+ε Sk,l(T, φ) l,k,ε T (log T ) with any ﬁxed ε > 0. Thus, T (log T )k2+2kl+1 seems to be the true order of magnitude for the moments Sk,l(T, φ) as T → ∞. Essentially, the discrete moments Sk,l(T, φ) act, after some suitable nor- malization, like a Riemann sum approximating the continuous moments Z T 2k (l) 1 Ik,l(T ) := ζ 2 + it dt. 1 Thus, we can deduce from the estimate for the discrete moments in Theorem 1.1 the following estimate for the continuous ones.

Corollary 1.1. For any rational k > 1 and any non-negative integer l, as T → ∞, k2+2kl Ik,l(T ) l,k T (log T ) . Moments of the zeta-function and its derivatives 287

For I1,l(T ) with a non-negative integer l and I2,0, there are classical asymptotic extensions by Hardy & Littlewood [6] and Ingham [8] which are in agreement with the estimates above. Furthermore, we must note that in the case l = 0, the bounds of Corollary 1.1 were proved by Heath-Brown [7] for any positive rational k and under the assumption of the Riemann hypothesis by Ramachandra [16] for any positive real k. Milinovich [13, Theorem 3.2] showed under the assumption of the Rie- k2+2kl+ε mann hypothesis that Ik,l(T ) l,k,ε T (log T ) for non-negative integers l and k and any ε > 0. This and Corollary 1.1 suggest that k2+2kl T (log T ) is the true order of magnitude for the moments Ik,l(T ). Especially for l = 0, there are many works which give evidence for this conjectured order of magnitude: e.g. Soundararajan [19], Heath-Brown [7] and Radziwill [15]. The paper is organized as follows: In the next section we provide some preliminary results. In Section 3 we prove key Proposition 3.1 which leads to Theorem 1.1. In Section 4 we prove Corollary 1.1. In Section 5 we give an alternative proof for Corollary 1.1 and in Section 6 we close with a remark.

2. Preliminaries Recall the function ∆(s) defined by (1.1). By Striling’s formula, we get 1 |t| 2 −σ−it (2.1) ∆(σ + it) = exp(i(t + π ))(1 + O(|t|−1)) for |t| 1 2π 4 > uniformly for any σ from a bounded interval. Hence, ! 1 −e−2iφ ∞ (2.2) = = −e−2iφ 1 + X e−2kiφ∆(s)k ∆(s) − e2iφ 1 − e−2iφ∆(s) k=1 1 1 holds for σ > 2 . Obviously, ∆( 2 + it) is a complex number on the unit circle for t ∈ R. Moreover, note that, for t large enough, ∆(σ + it) lies on 1 the unit circle only if σ = 2 (see Spira [18] and Dixon & Schoenfeld [2]). 0 1 Furthermore, ∆ ( 2 +it) is non-vanishing for sufficiently large t as follows from the asymptotic formula ∆0 |t| (2.3) (σ + it) = − log + O(|t|−1), ∆ 2π which holds for |t| ≥ 1 uniformly for any σ from a bounded interval. By (1.1), we can write 1 −2iθ(t) (2.4) ∆( 2 + it) = e , where 1 t t (2.5) θ(t) = Im log Γ + i − log π, 4 2 2 288 Thomas Christ, Justas Kalpokas is the Riemann-Siegel theta function (see Edwards [3, page 119]) which is asymptotically given by t t π 1 7 (2.6) θ(t) ∼ log − + + + ... for t ≥ 1. 2 2πe 8 48t 5760t3 The function θ(t) is differentiable and according to (2.6) t (2.7) θ0(t) = 1 log + 1 + O(t−2) 2 2πe 2 holds for t ≥ 1. Hence, t t 1 log < θ0(t) < 1 log + 1 2 2πe 2 2πe for t sufficiently large. This implies that θ(t) is monotonically increasing for t large enough. 1 iφ Due to (2.4), the solutions of ∆( 2 + it) − e = 0 correspond to the solutions of θ(t) ≡ φ mod π. Next, we introduce certain Dirichlet polynomials x y (2.8) X(s) = X n ,Y (s) = X m , ns ms n6X m6Y where X,Y 6 T . Moreover, we define the following quantities X |xn| X |ym| X0 = max |xn|, Y0 = max |ym|, X1 = , Y1 = n6X m6Y n m n6X m6Y and set x y X (s) = X n ,Y (s) = X m . 1 ns 1 ms n6X m6Y We shall use a variation of Lemma 5.1 from Ng [14]. For a proof we refer to Kalpokas, Korolev and Steuding [11, Lemma 5]. P∞ −s Lemma 2.1. Suppose the series f(s) = n=1 αnn converges absolutely P∞ −σ −γ for Res > 1 and n=1 |αn|n (σ − 1) for some γ ≥ 0 as σ → 1 + 0. Next, let X(s) and Y (s) be Dirichlet polynomials as defined in (2.8). Then, uniformly for a ∈ (1, 2] we have 1 Z a+iT ∆0(s) J = f(s)X(s)Y (1 − s) ds 2πi a+i ∆(s) ! T T α x y Y a(log T )2X Y = − log X n m mn + O 0 0 , 2π 2πe mn (a − 1)γ+1 m6X mn6Y where the implicit constant is absolute.

We proceed with a modiﬁed version of Lemma 6 of Gonek [4]. Moments of the zeta-function and its derivatives 289

Lemma 2.2. Let l be a non-negative integer and |t| ≥ 1. Then, uniformly for σ from a bounded interval, l ! l−k (l) l X l (k) t σ −1+ ζ (1 − s) = (−1) ζ (s)∆(1 − s) log + O(t 2 ). k 2π k=0 Proof. First, we note that the estimates  1 −σ+ε |t| 2 , if σ ≤ 0,  (l) 1 (1−σ)+ε (2.9) ζ (σ + it) |t| 2 , if 0 < σ ≤ 1,  |t|ε, if σ > 1, hold for any ε > 0 as t → ∞. The estimates for the case l = 0 can be found in Titchmarsh [21, Chapter V]. The estimates for l ∈ N can be deduced from the case l = 0 by applying Cauchy’s integral formula for derivatives of analytic functions to the zeta-function in a small disc centered at s = σ+it. Next, taking the l-th derivative of both sides of the functional equation (1.1), we get according to Leibniz’s rule ! l l (2.10) ζ(l)(1 − s) = X (−1)kζ(k)(s)∆(l−k)(1 − s). k k=0 Initially, we will show by induction that for every non-negative integer ν ν (ν) t σ− 3 ν−1 (2.11) ∆ (1 − s) = ∆(1 − s) − log + O(t 2 (log t) ) 2π holds uniformly for σ from a bounded interval: The case ν = 0 is obviously true. Now, suppose that the assertion (2.11) is proved for ν = 0, ..., µ − 1. Diﬀerentiating the identity ∆0 ∆0(1 − s) = ∆(1 − s) (1 − s) ∆ yields that µ−1 ! µ − 1 ∆0 (µ−ν−1) ∆(µ)(1 − s) = X ∆(ν)(1 − s) (1 − s). ν ∆ ν=0 By (2.3) and Cauchy’s estimate for the derivatives of analytic functions applied to a small square centered at 1 − s, we ﬁnd that ∆0 (ν) (1 − s) |t|−1, for ν ≥ 1. ∆ By the estimate above, (2.1) and (2.3), we can conclude that the assertion (2.11) holds for ν = µ; and thus, inductively, for all non-negative integers ν. The assertion of the Lemma follows now immediatelly by combining (2.9), (2.10) and (2.11). 290 Thomas Christ, Justas Kalpokas

In the next four Lemmas, we will gather several properties of the generalized κ-th divisor function (see Heath-Brown [7, Section 2]): Let κ be a positive real number. The generalized κ-th divisor function dκ : N → R is deﬁned by the coeﬃcients dκ(n) of ∞ κ X −s ζ(s) = dκ(n)n , σ > 1. n=1

The function dκ(n) is multiplicative and on prime powers given by Γ(κ + j) d (pj) = . κ Γ(κ)j! If κ is a positive integer the definition above coincides with the definition of the divisor function X dκ(n) = 1. n1,...,nκ∈N n1···nκ=n The generalized κ-th divisor function satisfies the following properties: Lemma 2.3. Let κ be a positive real number and n a positive integer.

(1) For κ ≥ 0 and n ≥ 1 ,we have dκ(n) ≥ 0. (2) For fixed n, dκ(n) increases with respect to κ. (3) For fixed κ ≥ 0 and > 0, we have dκ(n) n . (4) If j is an integer, then X dκj(n) = dκ(n1)dκ(n2) . . . dκ(nj). n=n1n2...nj For a proof, we refer to Heath-Brown [7, Lemma 1]. Lemma 2.4. Let λ, µ be fixed positive real numbers. Then, X λµ−1 dλ(n)dµ(n) λ,µ x(log x) n≤x and, thus, X −1 λµ dλ(n)dµ(n)n λ,µ (log x) . n≤x The first assertion of Lemma 2.4 can be established by the Selberg- Delange method (see Tenenbaum [20, Chapter II.5]) based on Perron’s formula and contour integration. The second assertion then follows by Abel’s summation. Let ϕ(m) denote Euler’s totient function that is defined by ϕ(m) = X 1. n≤m (n,m)=1 Then, we have the following. Moments of the zeta-function and its derivatives 291

Lemma 2.5. Let λ, µ be ﬁxed positive real numbers. Then, ϕ(m)µ X d (m)d (m) x(log x)λµ−1 λ µ m λ,µ m≤x and, thus, ϕ(m)µ X d (m)d (m) m−1 (log x)λµ. λ µ m λ,µ m≤x As in Lemma 2.4, the ﬁrst assertion of Lemma 2.5 can be established by the Selberg-Delange method (see Tenenbaum [20, Chapter II.5]) based on Perron’s formula and contour integration. The second assertion then follows by Abel’s summation.

1 p 2p Lemma 2.6. For any rational k = q ≥ 0, m ≤ x and x suﬃciently large, we have d (n) 1 φ(m) k X k ≥ log x . n p m n≤x (m,n)=1 p Proof. Let k = q be a non-negative rational number. We consider the sum

d 1 (n) W := X q . n n≤ξ (m,n)=1 Taking the q-th power, we get d (n, ξ) W q = X 1 , n n≤ξq (m,n)=1 where the coeﬃcients d1(n, ξ) are given by X d1(n, ξ) = d 1 (n1)d 1 (n2) ··· d 1 (nq). q q q n1n2···nq=n n1,n2,...,nq≤ξ As q is an integer, we have, according to property (4) of Lemma 2.3, X d 1 (n1)d 1 (n2) ··· d 1 (nq) = d 1 (n) = d1(n) ≡ 1 q q q j ·j n1n2···nq=n for all positve integers n. Hence, we can deduce that

d1(n, ξ) = d1(n) = 1 if n ≤ ξ and d1(n, ξ) ≤ d1(n) = 1 if n > ξ. 292 Thomas Christ, Justas Kalpokas

Thus, we get 1 1 φ(m) X ≤ W q ≤ X ≤ 2q log ξ. n n m n≤ξ n≤ξq (m,n)=1 (m,n)=1 Using the inequality φ(m) 1 φ(m) log ξ ≤ X ≤ 2 log ξ, m n m n≤ξ (m,n)=1

1 which is valid for m ≤ ξ 2 and ξ suﬃciently large and can be established by standard techniques, we get φ(m) φ(m) log ξ ≤ W q ≤ 2q log ξ m m 1 for m ≤ ξ 2 . Therefore,

1 1 φ(m) q φ(m) q (2.12) log ξ ≤ W ≤ 2q log ξ m m 1 for m ≤ ξ 2 . Taking the p-th power of W yields that

d p (n, ξ) W p = X q n n≤ξp (m,n)=1 with coeﬃcients X d p (n, ξ) = d 1 (n1)d 1 (n2) ··· d 1 (np). q q q q n1n2···np=n n1,n2,...,np≤ξ By the same reasoning as above, we obtain that

d p (n) d p (n) X q ≤ W p ≤ X q n n n≤ξ n≤ξp (m,n)=1 (m,n)=1 Using the upper bound for W p from the above inequality and the lower bound for W from (2.12), we get p d (n) d p (n) φ(m) q X k = X q ≥ W p ≥ log ξ . n n m n≤ξp n≤ξp (m,n)=1 (m,n)=1

1 p for m ≤ ξ 2 . Setting x = ξ yields the assertion of the Lemma for m ≤ 1 x 2p Moments of the zeta-function and its derivatives 293

Lemma 2.7. Let l be a non-negative integer, r and k non-negative rational numbers. Then (log m)ld (m)d (mn) X r k (log x)l+kr+k. mn l,k,r m6x mn6x p Proof. Let k = q ≥ 0 be a rational number. We consider the sum (log m)ld (m)d (mn) (log m)ld (m) d (mn) W := X r k = X r X k . mn m n m x m≤x x 6 n6 m mn6x Certainly, the following estimates hold (log m)ld (m)d (m) d (n) W ≥ X r k X k m n m≤x x n6 m (m,n)=1 (log m)ld (m)d (m) d (n) ≥ X r k X k . 1 1 m 2p n 3p+1 2p+1 x ≤m≤x n6x 2p+1 (m,n)=1 Now, Lemma 2.6 yields

−k k −l 1 l+k X dr(m)dk(m) φ(m) W ≥ (3p + 1) p + 2 (log x) . 1 1 m m x 3p+1 ≤m≤x 2+1 By Lemma 2.5, we get l+kr+k W k,l,r (log x) and the Lemma is proved.

3. Proof of Theorem 1.1 In order to prove Theorem 1.1 we consider the discrete moments (3.1) X (l) 1 1 1 S1(T, φ) = ζ 2 − itn(φ) X 2 + itn(φ) Y 2 − itn(φ) 0

Proposition 3.1. Let X(s) and Y (s) be Dirichlet polynomials as deﬁned in (2.8). Then uniformly for φ ∈ [0, π), as T → ∞, (3.3) ! l l (log n)kx y T T S (T, φ) =e−2iφ X(−1)l+k X m mn P log 1 k mn 2π l−k+1 2π k=0 m6X mn6Y T T (log m)l y x + (−1)l log X m mn + O(R ), 2π 2πe mn 1 m6Y mn6X where Pn(x) is a polynomial of degree n and 1 + 1 1 1 + R1 = (X + Y )(T 2 X1Y1 + T X0Y0) + X 2 Y 2 T 6 X0Y0 + T X0Y1; moreover, uniformly for φ ∈ [0, π) as T → ∞,

T T |x |2 (3.4) S (T, φ) = log X n + O(R ), 2 2π 2πe n 2 n6X where √ |x |2 R = X T (log T )2 X n + X(log T )3X 2. 2 n 0 n6X Proof of Proposition 3.1. A proof of statement (3.4) can be found in [11, Proposition 9, equation (10)]. Thus, it only remains to prove (3.3). We begin with the estimates (l) 1 1/6+ε ζ 2 + it t , √ 1 X |xn| X √ |xn| (3.5) X + it ≤ √ = n ≤ XX1, 2 n n n X n X √6 6 1 Y 2 + it ≤ Y Y1; the ﬁrst one follows from a well-known bound for the zeta-function on the critical line (see Titchmarsh [21, Chapter V]) in combination with Cauchy’s integral formula for the derivatives of an analytic function applied to a small disc centered at s, whereas the second and third assertion are straightfor- ward. Hence, in order to prove (3.3), it is suﬃcient to consider the sum over c < tn(φ) 6 T , where c > 32π is a large absolute constant. 1 Next, without loss of generality, we can assume that T = 2 (tν(φ) + tν+1(φ)) for some ν ∈ N since, by (3.5), for any T0 > T with T0 − T 1 we have X 1 1 1 (l) 1 1 1 2 2 6 + ζ ( 2 − itn(φ))X( 2 + itn(φ))Y ( 2 − itn(φ)) X Y T X0Y0. T

1 2iφ Since the points s = 2 + itn(φ) are the roots of the function ∆(s) − e , the sum in question can be rewritten as a contour integral:

X (l) 1 1 1 ζ ( 2 − itn(φ))X( 2 + itn(φ))Y ( 2 − itn(φ)) cintegrals over right and left sides of contour, and I2 and I4 be the integrals over the top and bottom edges of the contour. We may assume the constant c so large that the relations

t 1/2−a t −1/2 1 |∆(a + it)| = 1 + O(t−1) ≤ 2 < 2π 2π 2 hold for any t > c. Moreover, we observe that for s = a + it we have |x | |X(a + it)| ≤ X n X , na 6 1 n6X ma|y | (3.6) |Y (1 − a − it)| ≤ X m Y Y , m 1 m6Y ∞ −1/2 ∞ X −2ikφ k t X 1 −1/2 e ∆(a + it) ≤ 2 t . 2π 2k k=1 k=0 In view of (2.2) and Lemma 2.2 we have ! l l Z T τ l−k I = e−2iφ(−1)l+1 X log 1 k 2π k=0 c ! 1 Z a+iτ ∆0 ∞ × d ζ(k)(s)X(s)Y (1 − s) (s) 1 + X e−2ikφ∆(s)k ds 2π ∆ a+ic k=1 1 + + O(YT 2 X1Y1), where the error term comes from the application of (2.2), (3.6), and the error term of Lemma 2.2, i.e.

Z a+iT 0 ∞ 1 − 1 + ∆ X −2ikφ k O(t 2 )X(s)Y (1 − s) (s) 1 + e ∆(s) ds 2π ∆ a+ic k=1 1 + YT 2 X1Y1. 296 Thomas Christ, Justas Kalpokas

In order to evaluate I1, we ﬁrst consider j1 + j2 where Z a+iτ 0 1 (k) ∆ j1 = ζ (s)X(s)Y (1 − s) (s)ds 2πi a+ic ∆ and 1 Z a+iτ ∆0 ∞ j = ζ(k)(s)X(s)Y (1 − s) (s) X e−2ikφ∆(s)kds. 2 2πi ∆ a+ic k=1 By (3.6) we have Z τ (k) log tdt 1 + |j2| ζ (a)Y X1Y1 √ Y τ 2 X1Y1. c t

Applying Lemma 2.1 to j1, we get τ τ (log n)kx y j = (−1)k+1 log X m mn + O (Y τ X Y ) . 1 2π 2πe mn 0 0 m6X mn6Y Hence, ! l l (log n)kx y T T I =e−2iφ X(−1)l+k X m mn P log 1 k mn 2π l−k+1 2π k=0 m6X mn6Y 1 + + O(YT 2 X1Y1 + YT X0Y0 + T X1Y0), where T T Z T τ l−k τ τ Pl−k+1 log + O(1) = log d log 2π 2π c 2π 2π 2πe and Pn(x) is a polynomial of degree n. The additional error term for I1 comes from the bound

l ! k −2iφ X l+k l X (log n) xmymn e (−1) T X1Y0. k mn k=0 m6X mn6Y

In a similar way we may compute I3. We observe that Z T 0 1 (l) ∆ (1 − a + it) I3 = − ζ (a − it)X(1 − a + it)Y (a − it) 2iφ dt. 2π c ∆(1 − a + it) − e

This yields in combination with X(s) = X1(s), Y (s) = Y1(s) (see (2.8)) Z a+iT 0 1 (l) ∆ (1 − s) I3 = − ζ (s)X1(1 − s)Y1(s) −2iφ ds. 2πi a+ic ∆(1 − s) − e Moments of the zeta-function and its derivatives 297

In view of (2.2) we ﬁnd that 1 Z a+iT ∆0 ∞ I = − ζ(l)(s)X (1 − s)Y (s) (s) 1 + X e−2ikφ∆(s)k ds 3 2πi 1 1 ∆ a+ic k=1 = − (j3 + j4), where Z a+iT 0 1 (l) ∆ j3 = ζ (s)X1(1 − s)Y1(s) (s)ds 2πi a+ic ∆ and 1 Z a+iT ∆0 ∞ j = ζ(l)(s)X (1 − s)Y (s) (s) X e−2ikφ∆(s)kds. 4 2πi 1 1 ∆ a+ic k=1 By (3.6) we get 1 + |j4| XT 2 X1Y1. Using Lemma 2.1, we ﬁnd that T T (log m)l y x I = (−1)l log X m mn 3 2π 2πe mn m6Y mn6X 1 + + O XT 2 X1Y1 + XT X0Y0 .

In order to estimate I2 we ﬁrst note that the following inequalities hold along the line segment of the integration:

(l) 1 + X |xn| 1−σ 1−σ |ζ (1 − s)| T 2 , |X(s)| ≤ n X X , n 1 n6X |y | |Y (1 − s)| ≤ X n nσ Y σY , n 1 n6Y and, ﬁnally, σ (l) 1 + Y |ζ (1 − s)X(s)Y (1 − s)| T 2 X Y X 1 1 X 1−a a 1 + Y Y XT 2 X Y + 1 1 X X 1 + (X + Y )T 2 X1Y1. Next, by (2.3) we get ∆0(s) ∆0(s) e2iφ 1 = 1 + (log T ) 1 + . ∆(s) − e2iφ ∆(s) ∆(s) − e2iφ |∆(s) − e2iφ| 298 Thomas Christ, Justas Kalpokas

The second term in the brackets is bounded by an absolute constant. Indeed, 1 1 T −1 in the case σ > 2 + 3 log 2π we have by (2.1) for sufficiently large T T 1/2−σ 1 |∆(σ + iT )| = 1 + O(T −1) ≤ e−1/31 + O(T −1) < , 2π 2 2iφ 1 and hence |∆(s) − e | > 1 − |∆(s)| > 2 . Similarly, in the case σ 6 1 1 T −1 2 − 3 log 2π we get for sufficiently large T 4 4 1 |∆(σ + iT )| ≥ e1/31 + O(T −1) > , |∆(s) − e2iφ| > − 1 = . 3 3 3 Finally, let 1 1 T −1 1 1 T −1 − log < σ < + log . 2 3 2π 2 3 2π Then, using the relations 1 −2iϑ(T ) −2iϑ(T ) −1 ∆ 2 + iT = e , ∆(σ + iT ) = τe 1 + O(T ) , where τ = T/(2π)1/2−σ and ϑ = ϑ(T ) denotes the increment of any fixed continuous branch of the argument of π−s/2Γs/2 along the line segment 1 1 −1/3 1/3 with end-points s = 2 and s = 2 + iT , we have e 6 τ 6 e and 2iφ 1 1 2iφ ∆(σ + iT ) − e = ∆(σ + iT ) − ∆ 2 + iT + ∆ 2 + iT − e = (τ − 1)e−2iϑ − 2iei(φ−ϑ) sin (φ + ϑ) + O(T −1) = e−iϑ(τ − 1) cos ϑ + 2 sin (ϑ + φ) sin φ− − i((τ − 1) sin ϑ + 2 sin (ϑ + φ) cos φ) + O(T −1). Thus, we obtain that 2iφ 2 2 2 −1 ∆(σ + iT ) − e = (τ − 1) + 4τ sin (ϑ + φ) + O(T ) 2 −1 > 4τ sin (ϑ + φ) + O(T ). 1 Using the fact that T = 2 (tν(φ) + tν+1(φ)) for some ν, we finally get π π 3 sin2(ϑ + φ) = sin2 πν + + O(T −1) ≥ sin2 = 2 3 4 for sufficiently large T , and hence, 3 |∆(σ + iT ) − e2iφ|2 ≥ 4 · e−1/3 + O(T −1) > 2. 4 2iφ 1 Thus, |∆(s) − e | > 3 for any s under consideration. Hence 1 + I2 (X + Y )T 2 X1Y1.

The integral I4 can be estimated in a similar way and, thus, relation (3.3) is proved. Moments of the zeta-function and its derivatives 299

Now we can proceed to proof Theorem 1.1.

p Proof of Theorem 1.1. Suppose that k = q is a rational number with p > q > 1 and (p, q) = 1. Let l be a non-negative integer. We set r := p − q and choose ξ := T 1/(4p). First, we define fixed coefficients for the Dirichlet polynomials X(s) and Y (s) in (2.8) via

d 1 (n)p d p (n; ξ) X(s) = X q = X q , ns ns n6ξ n6ξp

d 1 (n)r d r (n; ξ) Y (s) = X q = X q , ns ns n6ξ n6ξr where d m (n; ξ) is given by q X d m (n; ξ) = d 1 (n1) . . . d 1 (nm) q q q n=n1···nm n1,...,nm6ξ for m = p, r. From property (5) of Lemma 2.3 we can easily deduce that d m (n; ξ) = d m (n) for m ξ and 0 d m (n, ξ) d m (n) for m > ξ. q q 6 6 q 6 q Now, let S1(T, ϕ) and S2(T, ϕ) be the moments given by (3.1), resp. (3.2), with respect to the above chosen Dirichlet polynomials X(s) and Y (s). Hölder’s inequality assures that

1/(2k) X (l) 1 2k |S1(T, φ)| ≤ ζ 2 + itn(φ) × 0

2k X 1 2k S1(T, φ) ζ 2 + itn(φ) ≥ . S (T, φ)2k−1 0

We proceed with bounding S1(T, ϕ) from below: by statement (3.3) in Proposition 3.1, we have

! l l T T S (T, φ) = X(−1)l+j P log Σ 1 j 2π l−j+1 2π 1 j=0 T T + log Σ + O(R ). 2π 2πe 2 1

By Lemma 2.4,

j (log n) d p (m; ξ)d r (mn; ξ) Σ = X q q 1 mn m6ξp,mn6ξr d r (n) r j X q X ≤ (log ξ ) d p (l) n q n≤ξr l|n

d r (n)d p +1(n) = (log ξr)j X q q n n6ξr ( p )2−1+j (log T ) q , and by Lemma 2.7,

l (log m) d r (m; ξ)d p (mn; ξ) Σ = X q q 2 mn m6ξr,mn6ξp l (log m) d r (m)d p (mn) ≥ X q q mn m6ξ ( p )2+l (log ξ) q .

The error term of S1(T, φ) is bounded by

d p (n; ξ) d r (m; ξ) p r 1 + X q X q p r 1 + R (ξ + ξ )T 2 + ξ ξ T 6 1 n m n6ξp m6ξr T 3/4+ T 4/5.

Thus, we obtain that

k2+l+1 |S1(T, φ)| T (log T ) . Moments of the zeta-function and its derivatives 301

Moreover, for S2(T, φ) we have by statement (3.4) of Proposition 3.1 and Corollary 2.4 that 2 d p (n; ξ) T T √ 2 |S (T, φ)| = log X q + Oξp T (log T )k +1 2 2π 2πe n n6ξp 2 T (log T )k +1. Altogether, we get 2k X 1 2k S1(T, φ) k2+2kl+1 ζ 2 + itn(φ) ≥ T (log T ) S (T, φ)2k−1 0 0 such that the Riemann-Siegel theta function θ(t) is monotonically increasing for t > c. Let T > c and Ti := tM+i(0) with i = 0, ..., N denote the Gram points that lie in the interval [T, 2T ]. We deﬁne a smooth function [c/π, ∞) 3 x 7→ tx via θ(tx) = π · x.

Then, tn+φ/π = tn(φ) for every positive integer n and every φ ∈ [0, π). Hence, we get

Z tn+1 Z tn+1 Z 1 0 (4.1) g(t)θ (t)dt = g(t)dθ(t) = g(tn+u)dθ(tn+u) tn tn 0 Z 1 Z 1 = g(tn+u)d(π(n + u)) = g(tn+u)πdu 0 0 Z π Z π = g(tn+φ/π)πd(φ/π) = g(tn(φ))dφ. 0 0 Therefore,

Z TN Z TN Z tn+1 g(t)θ0(t)dt = g(t)dθ(t) = X g(t)dθ(t) = T1 T1 M≤n≤M+N tn 302 Thomas Christ, Justas Kalpokas     Z π X Z π X =  g(tn(φ)) dφ =  g(tn(φ)) dφ. 0 0 M≤n≤M+N T1≤tn(φ)≤TN

Noting that the segments [T,T1] and [TN , 2T ] can be treated in a way analogue to (4.1), the assertion of the Proposition follows. We are now ready to prove Corollary 1.1. Proof of Corollary 1.1. Using the asymptotic extension (2.7) for θ0(t), Proposition 4.1 yields for any rational k ≥ 1 and any non-negative integer l Z 2T 2k 1 Z π 2k (l) 1 X (l) 1 ζ 2 + it dt ζ 2 + itn(φ) dφ. T log T 0 T ≤tn(φ)≤2T Combining this with Theorem 1.1, we get for any rational k ≥ 1 and any non-negative integer l Z T ∞ Z T/2j |ζ(l)( 1 + it)|2kdt ≥ X |ζ(l)( 1 + it)|2kdt 2 j+1 2 1 j=0 T/2 ∞ Z π X 1 X (l) 1 2k |ζ ( 2 + itn(φ))| dφ log T 0 T T j=0 ≤tn(φ)≤ 2j+1 2j 2 T (log T )k +2kl.

Thus, Corollary 1.1 follows.

5. Alternative proof of Corollary 1.1 Using a method of Rudnick and Soundararajan [17], the assertion of Corollary 1.1 can be proved in a direct way without relying on the discrete moments Sk,l(T, φ) and Theorem 1.1. We will demonstrate this proof for the case l = 0: By Cauchy’s residue theorem we have

a+iT 1 +iT 1 +i a+i! 1 Z Z 2 Z 2 Z + + + ζ(s)X(s)Y (1 − s)ds = 0, 2πi 1 1 a+i a+iT 2 +iT 2 +i where X(s) and Y (s) are deﬁned by (2.8) and a = 1 + (log T )−1. We can conclude that 1 +iT 1 Z 2 ζ(s)X(s)Y (1 − s)ds 2πi 1 2 +i a+iT 1 +iT a+i! 1 Z Z 2 Z = + + ζ(s)X(s)Y (1 − s)ds. 2πi 1 a+i a+iT 2 +i Moments of the zeta-function and its derivatives 303

The second and the third integral on the right hand side are bounded by 1 + (X + Y )T 2 X1Y1, (see the proof of (3.3)). Thus, we get Z T 1 1 1 ζ( 2 + it)X( 2 + it)Y ( 2 − it)dt 1 Z a+iT 1 1 + = ζ(s)X(s)Y (1 − s)ds + O((X + Y )T 2 X1Y1). i a+i The integral on the right hand side can be evaluated as ∞ 1 x y 1 Z T k it X X m X k dt na ma k1−a 2π mn n=1 m≤X k≤Y 1 x y = T X m mn + O(Y (log T )4X Y ), mn 0 0 m≤X mn≤Y where the error term comes from the oﬀ-diagonal terms (see the proof of Lemma 5 in Kalpokas, Korolev and Steuding [11]). In a similar way we can show that Z T 2 1 2 X |xn| |X( 2 + it)| dt = T + O(R2), 1 n n6X where R2 is the same as in (3.4). Now, we follow the proof of Theorem 1.1, where k, p, q, r, X, Y, ξ are the same. We set

d 1 (n) A(s) := X q , ns n6ξ d 1 (n)r d r (n; ξ) X(s) := X q = X q , ns ns n6ξ n6ξr d 1 (n)p d p (n; ξ) Y (s) := X q = X q . ns ns n6ξ n6ξp Hence, A(s)k−1 = X(s) and A(s)k = Y (s). By Hölder’s inequality we get

Z T 1 1 k−1 1 k ζ( 2 + it)A( 2 + it) A( 2 − it) dt 1 1 2k−1 ! 2k ! 2k Z T Z T 2k 1 2k 1 k−1 1 k 2k−1 ≤ |ζ( 2 + it)| dt |A( 2 + it) A( 2 − it) | dt . 1 1 304 Thomas Christ, Justas Kalpokas

Thus, we have

2k T 1 1 1 T R ζ( + it)A( + it)k−1A( − it)kdt 2k Z 1 2 2 2 |S1| |ζ( 1 + it)|2kdt ≥ =: . 2 T 2k−1 2k−1 1 R 1 2k S2 1 |A( 2 + it)| dt

First, we bound |S1| from below. We have

r p d (m; ξ)d (mn; ξ) p 2 X q q ( ) |S | T T (log T ) q , 1 mn m≤X mn≤Y since the sum in |S1| is the same as Σ2 in the proof of Theorem 1.1. Next, we bound S2 from above. In the same manner as for |S2(T, φ)| in the proof of Theorem 1.1 we obtain that

( p )2 S2 T (log T ) q . Altogether it follows that Z T 1 2k k2 |ζ( 2 + it)| dt T (log T ) 1 holds for any rational number k ≥ 1.

6. Remark As a consequence of Proposition 4.1 we have for any non-negative integer l and any non-negative real k

2k Z 2T 2k X (l) 1 (l) 1 max ζ 2 + itn(φ) log T ζ 2 + it dt. φ∈[0,π) T T ≤tn(φ)≤2T

Now, using the unconditional lower bound for Ik,0(T ) by Heath-Brown [7], resp. the conditional one by Ramachandra [16], we can deduce that

2k 2 max X ζ 1 + it T (log T )k +1. φ∈[0,π) 2 T ≤tn(φ)≤2T holds for any rational k ≥ 0, resp. under the assumption of the Riemann hypothesis for any real k ≥ 0. Acknowledgements: Both authors gratefully thank professors R. Garunk- štis and J. Steuding for their support and leadership. Moreover both authors would like to thank the unknown referee for many valuable comments and suggestions. Moments of the zeta-function and its derivatives 305

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Thomas Christ Department of Mathematics Würzburg University Emil-Fischer-Str. 40, 97074 Würzburg E-mail: [email protected]

Justas Kalpokas Faculty of Mathematics and Informatics Vilnius University Naugarduko 24, 03225 Vilnius, Lithuania E-mail: [email protected]