Relations Among the Riemann Zeta and Hurwitz Zeta Functions, As Well As Their Products

Home , Gamma function, Hurwitz zeta function

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Article Relations among the Riemann Zeta and Hurwitz Zeta Functions, as Well as Their Products

A. C. L. Ashton 1,* and A. S. Fokas 2 1 Department of Applied Mathematics and Theoretical Physics, University of Cambridge, Cambridge CB3 0WA, UK 2 Viterbi School of Engineering, University of Southern California, Los Angeles, CA 90089-2560, USA; [email protected] * Correspondence: [email protected]

Received: 2 May 2019; Accepted: 31 May 2019; Published: 4 June 2019 Abstract: In this paper, several relations are obtained among the Riemann zeta and Hurwitz zeta functions, as well as their products. A particular case of these relations give rise to a simple re-derivation of the important results of Katsurada and Matsumoto on the mean square of the Hurwitz zeta function. Also, a relation derived here provides the starting point of a novel approach which, in a series of companion papers, yields a formal proof of the Lindelöf hypothesis. Some of the above relations motivate the need for analysing the large α behaviour of the modiﬁed Hurwitz zeta function ζ1(s, α), s ∈ C, α ∈ (0, ∞), which is also presented here.

Keywords: Hurwitz zeta function; Riemann zeta function; asymptotics

MSC: 11M35; 11L07

1. Introduction Let s = σ + it, σ ∈ R, t ∈ R. (1)

The Riemann zeta function ζ(s) and the modified Hurwitz zeta function ζ1(s, α) are defined, respectively, by ∞ 1 ( ) = > ζ s ∑ s , σ 1, (2) n=1 n ∞ 1 ζ (s, α) = , σ > 1, α ≥ 1, (3) 1 ∑ ( + )s n=1 n α and by analytic continuation for s ∈ C, s 6= 1. The modified Hurwitz zeta function is simply related to the Hurwitz zeta function, ζ(s, α):

1 ζ(s, α) = + ζ (s, α), s ∈ C, α > 0. (4) αs 1

In this paper, we present certain relations between ζ(s) and ζ1(s, α), as well as between products of these functions. The following results are presented in Sections2–5 in more detail. In Section2, it is shown that the modiﬁed Hurwitz zeta function satisﬁes the identity

Z 2 1 Γ(s + z) Γ(s¯ + z) |ζ1(s, α)| = ζ1(2σ, α) + + Γ(−z)ζ(−z)ζ1(2σ + z, α) dz 2πi (c) Γ(s) Γ(s¯) σ > 1, t > 0, max(−σ, 1 − 2σ) < c < −1, (5)

Symmetry 2019, 11, 754; doi:10.3390/sym11060754 www.mdpi.com/journal/symmetry Symmetry 2019, 11, 754 2 of 17 where Γ(s), s ∈ C, denotes the Gamma function and (c) denotes the vertical line in the complex z-plane on which Re (z) = c. It is shown in [1] that (5) yields a singular integral equation for |ζ(s)|2, 0 < σ < 1, t > 0, and this equation provides the starting point for the proof of the analogue of Lindelöf’s hypothesis for a certain Riemann zeta-type double exponential sum describing the leading asymptotics of |ζ(s)|2; namely, it −σ−it −σ−it includes m1 m2 , but the limits of the summation are different from the double exponential sum relevant to |ζ(s)|2—see [1]. We recall that Lindelöf’s hypothesis concerns the growth of ζ(s) as t → ∞ 1 e along the critical line σ = 1/2, and states that ζ( 2 + it) = O(t ) for every positive e. The Riemann hypothesis implies Lindelöf’s hypothesis, and conversely, Lindelöf’s hypothesis implies that very few zeros could disobey Riemann’s hypothesis [2]. Signiﬁcant progress has been made by developing 2πi f (n) ingenuous ways of estimating exponential sums ∑n e using generalisations of the Vinogradov method [3]. Until recently, the best result in this direction was obtained by Huxley [4], where it is 1 32/205+e proved that ζ( 2 + it) = O(t ). Just recently, Bourgain announced a further improvement [5] where the exponent 32/205 was reduced to 53/342. In Section3 it is shown that there exists the following asymptotic relation between the Riemann zeta function and the modiﬁed Hurwtiz zeta function:

R 1 R 1 −σ/2 ζ(s) = 0 BN(α)ζ1(s, α) dα + χ(s) 0 BN(−α)ζ1(1 − s, α) dα + O(t log t), 0 < σ < 1, t → ∞, (6) where BN(α) is deﬁned by 2πinα BN(α) = ∑ e , α > 0 (7) 1≤n≤N √ and N = t/2π. A direct consequence of (6) is the following theorem (c.f. [6–8] and references therein):

Theorem 1. Let Ik(t) denote the 2k-th power mean of the modiﬁed Hurwitz zeta function, namely

Z 1 1 2k Ik(t) = |ζ1( 2 + it, α)| dα, k ∈ N. (8) 0

Then, 1 1/4k 1/2k |ζ( 2 + it)| t Ik(t) , k ∈ N (9) in which the implicit constant is independent of k.

This result immediately implies that Lindelöf’s hypothesis is true, provided that for each e > 0 e Ik(t) k,e t for each k ∈ N. In connection with (9), we recall that an equivalent formulation of the Lindelöf hypothesis involves estimating the 2k-th power mean of the Riemann-zeta function

Z T 2k 1 1 Jk(T) = ζ( 2 + it) dt. (10) T 0

e It can be shown ([9] Th. 13.2) that the Lindelöf hypothesis holds true if, and only if Jk(T) = O (T ) for each e > 0 and for each k ∈ N. In Section4, the following identities are presented:

Z 1 1 Z ∞ Z ∞ −u1 −u2 ζ1(u1, α)ζ1(u2, α) dα = + α ζ1(u2, α) dα + α ζ1(u1, α) dα, (11) 0 u1 + u2 − 1 1 1 Symmetry 2019, 11, 754 3 of 17

Z 1 1 Z ∞ −u1 ζ1(u1, α)ζ1(u2, α)ζ1(u3, α) dα = + α ζ1(u2, α)ζ1(u3, α) dα 0 u1 + u2 + u3 − 1 1 Z ∞ Z ∞ −u2 −u3 + α ζ1(u1, α)ζ1(u3, α) dα + α ζ1(u1, α)ζ1(u2, α) dα 1 1 Z ∞ Z ∞ Z ∞ −u1−u2 −u2−u3 −u3−u1 + α ζ1(u3, α) dα + α ζ1(u1, α) dα + α ζ1(u2, α) dα, (12) 1 1 1 and

Z 1 4 1 Z ∞ ζ (u , α) dα = + α−(u1+u2+u3)ζ (u , α) dα ∏ 1 i + + + − ∑ 1 4 0 i=1 u1 u2 u3 u4 1 perms 1 Z ∞ Z ∞ −(u1+u2) −u1 + ∑ α ζ1(u3, α)ζ1(u4, α) dα + ∑ α ζ1(u2, α)ζ1(u3, α)ζ1(u4, α) dα, (13) perms 1 perms 1 where Re (u1 + u2) > 1, Re (u1 + u2 + u3) > 1, and Re (u1 + u2 + u3 + u4) > 1, respectively. The above formulae can be generalised in a straightforward way. As a direct application of (11), we present in Section4 a new derivation of the following exact identity in [10]:

Z 1 1 Γ(1 − v) Γ(1 − u) ζ1(u, α)ζ1(v, α) dα = + Γ(u + v − 1)ζ(u + v − 1) + 0 u + v − 1 Γ(u) Γ(v) Z 1 Z 1 ζ(u) − 1 ζ(v) − 1 u 1−v v 1−u + + + α ζ1(u + 1, α) dα + α ζ1(v + 1, α) dα (14) v − 1 u − 1 v − 1 0 u − 1 0 valid for Re u < 2 and Re v < 2. From this, one can obtain the estimate

t 1 I (t) = log + γ + O , t → ∞.. (15) 1 2π t2

It is shown in [11] that (12) plays a crucial role for the derivation of an interesting identity between certain double exponential sums. Indeed, it is well-known that if 0 ≤ α < 1, then the large t-asymptotics of ζ1(s, α) is dominated by the sum S1, deﬁned by

−2πinα s−1 S1(σ, t, α) = ∑ e m , 0 < σ < 1, t > 0. (16) 1≤n

However, if 1 ≤ α < ∞, the large t-asymptotics of ζ1(s, α) is dominated by the sum S1 deﬁned 0 in (16), as well as by a different sum, S2(σ, t, α) [12] (c.f. [13]). Thus, the large t-asymptotics of 0 Equation (12) provides a relation between two double sums generated from S1 and S2, and the explicit formulae obtained from the large t-asymptotics of the linear and quadratic terms. Similarly, Equation (13) yields novel relations between cubic exponential sums. Before using Equations (12) and (13) in the cases of Re (u1 + u2 + u3) < 1 and Re (u1 + u2 + u3 + u4) < 1, respectively, it is necessary to regularize the terms involving α → ∞; this regularisation is discussed in Section4. Finally, in Section5, by considering the Fourier series of the product ζ1(u, α)ζ1(v, α) with complex numbers u, v satisfying Re u > 0, Re v > 0, by using certain elementary estimates for the resulting coefﬁcients, and by employing Theorem1 together with Parseval’s identity, we obtain the following asymptotic result.

Theorem 2. For each η > 0, we have

Z t/2π+η 2 − + − |ζ( 1 + it)|4 t1/2 α 1/2 itζ ( 1 + it, α)e 2πinα dα . 2 η ∑ 1 2 |n|≤t/π 1 Symmetry 2019, 11, 754 4 of 17

e 1 1/8+e In particular, if the sum is Oe(t ) for each e > 0, then |ζ( 2 + it)| = Oe(t ).

2. An Identity Involving the Hurwitz Zeta Functions In order to derive (5), we let

α ≥ 0, u ∈ C, v ∈ C, Re u > 1, Re v > 1. (17)

The deﬁnition of the modiﬁed Hurwitz zeta function, namely Equation (3), implies

ζ1(u, α)ζ1(v, α) = ζ1(u + v, α) + f (u, v, α) + f (v, u, α) (18) where ∞ ∞ f (u, v, α) = ∑ ∑ (n + α)−v(n + m + α)−u. (19) n=1 m=1 Assuming that | arg(−w)| < π, −Re a < b < 0, (20) we observe the Mellin-Barnes type integral identity

1 Z Γ(a)(1 − w)−a = Γ(z + a)Γ(−z)(−w)z dz. (21) 2πi (b)

Letting in Equation (21) m w = − , a = u, n + α we ﬁnd 1 Z Γ(u)(n + α)u(m + n + α)−u = Γ(u + z)Γ(−z)mz(n + α)−z dz. 2πi (c) Thus,

1 Z Γ(u + z) (n + α)−v(n + m + α)−u = Γ(−z)mz(n + α)−u−v−z dz. 2πi (c) Γ(u)

Summing over m and n, we obtain

1 Z Γ(u + z) f (u, v, α) = Γ(−z)ζ(−z)ζ1(u + v + z, α) dz. (22) 2πi (c) Γ(u)

Substituting (22) into (18) and then letting u = s, v = s¯ in the resulting expression, we ﬁnd (5).

3. An Asymptotic Relation between the Riemann and Hurwitz Functions The approximate functional equation for the Riemann zeta function provides the starting point ( ) for the estimation of the ζ s along the critical√ line. In this section,√ we derive a weak analogue of this equation. Throughout, we will set N = b t/2πc and assume t/2π ∈/ N so that N2 < t/2π. We refer to the sum eiπ(N+1)α sin(Nπα) B (α) = e2πinα ≡ . N ∑ ( ) 1≤n≤N sin πα This function is similar to the classical Dirichlet kernel that arises in the Fourier analysis. As such, we have the following well-known estimates.

1−1/p Lemma 1. kBNkp = O(log N) if p = 1 and kBNkp = O(N ) for p > 1. Symmetry 2019, 11, 754 5 of 17

Our ﬁrst result expresses the approximate functional equation for ζ(s) as an integral equation involving the Hurwitz zeta function ζ1(s, α). The proof of Theorem1 will follow directly from this result.

Lemma 2. Let s = σ + it and BN, as previously deﬁned. Then, we have

Z 1 Z 1 ζ(s) = BN(α)ζ1(s, α) dα + χ(s) BN(−α)ζ1(1 − s, α) dα 0 0 Z 1 Z 1 −s s−1 − BN(α) ∑ (n + α) dα − χ(s) BN(−α) ∑ (n + α) dα 0 1≤n≤N 0 1≤n≤N + O t−σ/2 log t , (23) when 0 < σ < 1. Furthermore,

Z 1 Z 1 −σ/2 ζ(s) = BN(α)ζ1(s, α) dα + χ(s) BN(−α)ζ1(1 − s, α) dα + O t log t , t → ∞, (24) 0 0

s−1 s where χ(s) = 2 π sin(πs/2)Γ(1 − s) with Γ(s) denoting the Gamma function, and ζ1(s, α) denoting the modiﬁed Hurwitz zeta function.

Let us ﬁrst establish (23). The identity in (24) will be a consequence of this.

Proof. First, we recall the approximate functional equations for ζ(s) and ζ1(s, α) (see [14] and references therein) ζ(σ + it) = ∑ n−s + χ(s) ∑ ns−1 + O t−σ/2 , (25) 1≤n≤N 1≤n≤N and −s −2πinα s−1 −σ/2 ζ1(s, α) = ∑ (n + α) + χ(s) ∑ e n + O t , (26) 1≤n≤N 1≤n≤N uniformly in 0 < α < 1. The following identity is valid ! Z 1 z −2πimα z ∑ n = BN(α) ∑ e m dα, z ∈ C. (27) 1≤n≤N 0 1≤m≤N

Indeed, the left-hand side of (27) can be rewritten in the form

Z 1 z z 2πi(n−m)α ∑ δmnm = ∑ m e dα 1≤m,n≤N 1≤m,n≤N 0 ! ! Z 1 = ∑ e2πinα ∑ e−2πimαmz dα, 0 1≤n≤N 1≤m≤N which is the right-hand side of (27). Using z = s − 1 and employing (26), we ﬁnd " # Z 1 Z 1 s−1 −s h −σ/2 i χ(s) ∑ n = BN(α) ζ1(s, α) − ∑ (n + α) dα + O t BN(α) dα. (28) 1≤n≤N 0 1≤n≤N 0

We note that

Z 1 Z 1 h −σ/2 i −σ/2 −σ/2 O t BN(α) dα t |BN(α)| dα = O(t log t). 0 0 Symmetry 2019, 11, 754 6 of 17

Equation (27) implies ! Z 1 z 2πimα z ∑ n = BN(−α) ∑ e m dα, u ∈ C. (29) 1≤n≤N 0 1≤m≤N

Replacing s by 1 − s¯ in (26) and taking the complex conjugate of the resulting equation, we ﬁnd

s−1 2πinα −s −(1−σ)/2 ζ1(1 − s, α) = ∑ (n + α) + χ(1 − s) ∑ e n + O t . (30) 1≤n≤N 1≤n≤N

Using (29) with z = −s and employing (30), we ﬁnd " # Z 1 −s s−1 −σ/2 ∑ n = BN(−α) χ(s)ζ1(1 − s, α) − χ(s) ∑ (n + α) dα + O t log t . (31) 1≤n≤N 0 1≤n≤N Here, we have used χ(s) = O t1/2−σ and a similar estimate as before:

Z 1 Z 1 −(1−σ)/2 −σ/2 −σ/2 χ(s) O t BN(−α) dα t |BN(−α)| dα = O t log t . 0 0

Using Equations (28) and (31) in (25), we arrive at the result in the lemma.

Lemma 3. With BN(α) deﬁned as before, we have

Z 1 i 1 1 1 B (α) (n + α)−s dα = + + O t−σ/2 log t . N ∑ Ns ∑ t ∑ t 0 1≤n≤N 2π 1≤n≤N 2πN − n 2πi 1≤n≤N 2π − n

Proof. Using the periodicity of BN(α) on R/Z, we ﬁnd

Z 1 Z n+1 −s −s BN(α) ∑ (n + α) dα = ∑ BN(α)α dα 0 1≤n≤N 1≤n≤N n Z N Z N+1 −s −s = BN(α)α dα + BN(α)α dα. (32) 1 N

We next estimate the ﬁrst integral on the right-hand side of (32), which we denote by IN(s):

Z N 2πinα−it log α −σ IN(s) = ∑ e α dα, 0 < σ < 1. (33) 1≤n≤N 1

The integral in the above sum does not have any stationary points. Indeed, candidates for stationary points are the points α∗, where t N2 α∗ = > 2πn n Thus, since 1 ≤ n ≤ N, α∗ is outside the range of integration. Hence, the above integral can be estimated using integration by parts:

Z N Z N d α1−σ e2πinα−it log αα−σ dα = i e2πinα−it log α dα 1 1 dα t − 2πnα ! 1 N−s 1 = − t − t + En(s), (34) 2πi 2πN − n 2π − n Symmetry 2019, 11, 754 7 of 17 where Z N 1−σ 2−σ d 2πinα−it log α (1 − σ)α 2πnα En(s) = e + dα. 1 dα (t − 2πnα)2 (t − 2πnα)3

The error term En(s) can be evaluated using the second mean value theorem for integrals. For instance, for some ξ ∈ (1, N), we have

1−σ 2−σ Z N (1 − σ)N 2πnN d 2πinα−it log α |Re En(s)| = + Re e dα (t − 2πnN)2 (t − 2πnN)3 ξ dα N1−σ nN2−σ = O + O , (t − 2πnN)2 (t − 2πnN)3 and similarly for Im En(s). It is now straightforward to show

−(1+σ)/2 ∑ En(s) = O t . (35) 1≤n≤N

We also have the elementary estimate

Z N+1 Z 1 −s −σ/2 −σ/2 BN(α)α dα t |BN(α)| dα = O t log t . N 0

This combined with (35) and (34) gives the desired result. The leading order terms in the above expansion are O t−σ/2 log t , thus they can be absorbed into the error term. Indeed, using Lemma3, it is now straightforward to see that

Z 1 −s −σ/2 BN(α) ∑ (n + α) dα = O t log t . 0 1≤n≤N

Using similar arguments, we also ﬁnd

Z 1 s−1 −σ/2 χ(s) BN(−α) ∑ (n + α) dα = O t log t . 0 1≤n≤N

Combining this observation with the result of Lemma2, we conclude that

Z 1 Z 1 −σ/2 ζ(s) = BN(α)ζ1(s, α) dα + χ(s) BN(−α)ζ1(1 − s, α) dα + O t log t , t → ∞ 0 0 for 0 < σ < 1. The proof to Theorem1 now follows from Lemmas2 and3 with s = 1/2 and the application of Hölder’s inequality with exponents

2k p = 2k, q = . 2k − 1

In particular, using the estimates in Lemma1, we have

Z 1 Z 1 1−1/2k Z 1 1/2k 2k/(2k−1) 2k BN(α)ζ1(s, α) dα ≤ |BN(α)| dα |ζ1(s, α)| dα 0 0 0 1/2k 1/2k N Ik(t) 1/4k 1/2k t Ik(t) .

This gives rise to the result in Theorem1. Symmetry 2019, 11, 754 8 of 17

4. Relations among Products of the Hurwitz Zeta Functions

4.1. Quadratic Formula

Lemma 4. Let ζ1(u, α), u ∈ C, α > 0, denote the modiﬁed Hurwitz function, i.e.,

∞ 1 ζ (u, α) = , α > 0 Re u > 1. (36) 1 ∑ ( + )u m=1 m α Then, for Re (u + v) > 2,

Z 1 Z ∞ Z ∞ 1 −v −u ζ1(v, α)ζ1(u, α) dα = + α ζ1(u, α) dα + α ζ1(v, α) dα. (37) 0 u + v − 1 1 1

Proof. Let q0(v, u) denote the LHS of Equation (37). Using the integral representation of the modiﬁed Hurwitz function, namely

1 Z ∞ e−αρρs−1 Re ζ1(s, α) = ρ dρ, α > 0, s > 1, (38) Γ(s) 0 e − 1 we ﬁnd

Z ∞ Z ∞ v−1 u−1 −(ρ1+ρ2) 1 ρ1 ρ2 1 − e q0(v, u) = dρ1 dρ2 ρ ρ . Γ(u)Γ(v) 0 0 ρ1 + ρ2 (e 1 − 1)(e 2 − 1) Inserting in this equation the identity

1 − e−(ρ1+ρ2) 1 1 = e−(ρ1+ρ2) 1 + + , (eρ1 − 1)(eρ2 − 1) eρ2 − 1 eρ1 − 1 we ﬁnd q0(v, u) = J0(v, u) + I0(v, u) + I0(u, v), (39) where Z ∞ Z ∞ v−1 u−1 −(ρ1+ρ2) 1 ρ1 ρ2 e J0(v, u) = dρ1 dρ2 , (40) Γ(u)Γ(v) 0 0 ρ1 + ρ2 and Z ∞ Z ∞ v−1 u−1 −(ρ1+ρ2) 1 ρ1 ρ2 e I0(v, u) = dρ1 dρ2 ρ . (41) Γ(u)Γ(v) 0 0 (ρ1 + ρ2)(e 2 − 1) Next, we will show that 1 J (v, u) = . (42) 0 u + v − 1 Indeed, using the integral representations of Γ(u) and Γ(v), we ﬁnd

Z ∞ Z ∞ u−1 v−1 −(ρ1+ρ2) Γ(u)Γ(v) = dρ1 dρ2 ρ1 ρ2 e . (43) 0 0

Replacing in the RHS of (43) ρ1 and ρ2 by αx1 and αx2, multiplying the resulting expression by α−(u+v), and integrating with respect to α from α = 1 to α = ∞, we ﬁnd

Z ∞ Z ∞ u−1 v−1 −(x1+x2) Γ(u)Γ(v) x1 x2 e = dx1 dx2 , u + v − 1 0 0 x1 + x2 which gives (42). Finally, we will show that Z ∞ −v I0(v, u) = α ζ1(u, α) dα. (44) 1 Symmetry 2019, 11, 754 9 of 17

Indeed, using the integral representations of Γ(v) and of ζ1(u, α), we ﬁnd

Z ∞ Z ∞ −αρ2 −ρ v−1 1 u−1 e Γ(v)ζ1(u, α) = e ρ dρ × ρ2 ρ dρ2. (45) 0 Γ(u) 0 e 2 − 1

Replacing in the RHS of the above equation ρ by αρ1, multiplying the resulting equation by α−v/Γ(v), and then integrating with respect to α from α = 1 to α = ∞, we ﬁnd (44). Inserting in Equation (39) the expressions for J0(v, u), for I0(v, u) and for I0(u, v) from Equations (42) and (44), we ﬁnd (37).

Equation (12) can be derived following the approach used in Lemmas4 and5, and thus it is omitted.

4.2. Quadruple Formula

Lemma 5. Let ζ1(s, α) be deﬁned as in (38). Then for Re ui > 1, i = 1, 2, 3, 4, the following identity is valid:

Z 1 4 1 Z ∞ ζ (u , α) dα = + α−(u1+u2+u3)ζ (u , α) dα ∏ 1 i + + + − ∑ 1 4 0 i=1 u1 u2 u3 u4 1 perms 1 Z ∞ Z ∞ −(u1+u2) −u1 + ∑ α ζ1(u3, α)ζ1(u4, α) dα + ∑ α ζ1(u2, α)ζ1(u3, α)ζ1(u4, α) dα, (46) perms 1 perms 1 where the sums run over permutations of (1, 2, 3, 4) so that the ﬁrst and third sums contain 4 terms, whilst the second sum contains 6 terms.

Proof. Employing the representation (38) for each Hurwitz function and integrating over (0, 1), we ﬁnd that the left-hand side of (46), which we denote by Q(u1, u2, u3, u4), is given by

4 ui−1 −(ρ +ρ +ρ +ρ ) 1 Z ∏ ρ dρi 1 − e 1 2 3 4 Q = i=1 i (47) 4 ( )4 + + + 4 ∏i=1 Γ(ui) 0,∞ ρ1 ρ2 ρ3 ρ4 ∏i=1 Ri where the functions {Ri} are deﬁned by

ρi Ri = e − 1, i = 1, 2, 3, 4. (48)

The following identity is valid:

1 − e−(ρ1+ρ2+ρ3+ρ4) 1 1 1 1 = e−(ρ1+ρ2+ρ3+ρ4) 1 + + + + 4 R R R R ∏i=1 Ri 1 2 3 4 (49) 1 1 1 1 1 1 1 1 1 1 + + + + + + + + + + . R1R2 R2R3 R3R4 R1R3 R2R4 R1R4 R1R2R3 R1R2R4 R1R3R4 R2R3R4

Using this in (47), we ﬁnd

4 ui−1 1 R ∏i=1 ρ dρi −( + + + ) 1 1 1 1 Q = i e ρ1 ρ2 ρ3 ρ4 1 + + + + 4 (0,∞)4 ρ + ρ + ρ + ρ R R R R ∏i=1 Γ(ui) 1 2 3 4 1 2 3 4 (50) 1 1 1 1 1 1 1 1 1 1 + + + + + + + + + + . R1R2 R2R3 R3R4 R1R3 R2R4 R1R4 R1R2R3 R1R2R4 R1R3R4 R2R3R4

In order to simplify the right-hand side of (50), we ﬁrst note the deﬁnition of the Gamma function—namely, the equation

Z ∞ Γ(u) = ru−1e−r dr, Re u > 0, (51) 0 Symmetry 2019, 11, 754 10 of 17 implies that 4 Z 4 ui−1 −x1−x2−x3−x4 Γ(ui) = x dxi e . ∏ 4 ∏ i i=1 (0,∞) i=1 Using in the right-hand side of this equation the transformations

xi = αρi, i = 1, 2, 3, 4, (52) dividing by the product of the four Gamma functions, and multiplying the resulting expression by α−u1−u2−u3−u4 , we ﬁnd the identity

1 Z 4 −u1−u2−u3−u4 ui−1 −α(ρ1+ρ2+ρ3+ρ4) α = ρ dρi e . 4 ( )4 ∏ i ∏i=1 Γ(ui) 0,∞ i=1

Integrating this equation over (1, ∞) with respect to α, we obtain 4 ui−1 1 1 Z ∏i=1 ρi dρi = e−ρ1−ρ2−ρ3−ρ4 . (53) u + u + u + u − 4 ( )4 + + + 1 2 3 4 1 ∏i=1 Γ(ui) 0,∞ ρ1 ρ2 ρ3 ρ4

Employing for ζ1(u, α) and Γ(u) Equations (38) and (51), respectively, we ﬁnd

u −1 1 Z ρ 1 1 u2−1 u3−1 u4−1 −αρ1−(x2+x3+x4) ζ1(u1, α)Γ(u2)Γ(u3)Γ(u4) = dρ1 dx2 dx3 dx4 x2 x3 x4 e . Γ(u1) (0,∞)4 R1

Using in the right-hand side of this equation the transformations (52) but restricted only to −(u +u +u ) i = 2, 3, 4, dividing by Γ(u2)Γ(u3)Γ(u4), multiplying by α 2 3 4 and integrating with respect to α over (1, ∞), we obtain

− 4 ui 1 −(ρ1+ρ2+ρ3+ρ4) Z ∞ 1 Z ∏i=1 ρi dρi e −(u2+u3+u4) α ζ1(u1, α) dα = . (54) 4 ( )4 R ( + + + ) 1 ∏i=1 Γ(ui) 0,∞ 1 ρ1 ρ2 ρ3 ρ4

Similarly, Equations (38) and (51) imply

ζ1(u1, α)ζ1(u2, α)Γ(u3)Γ(u4) u −1 u −1 Z x 3 x 4 e−α(ρ1+ρ2)−(x3+x4) 1 u1−1 u2−1 3 4 = dρ1 dρ2 dx3 dx4 ρ1 ρ2 . Γ(u1)Γ(u2) (0,∞)4 R1R2

Using in the right-hand side of this equation the transformations (52) but only for i = 3, 4, dividing −u −u by Γ(u3)Γ(u4), multiplying by α 3 4 and integrating the resulting expression with respect to α over (1, ∞), we obtain

− 4 ui 1 −ρ1−ρ2−ρ3−ρ4 Z ∞ 1 Z ∏i=1 ρi dρi e −u3−u4 α ζ1(u1, α)ζ1(u2, α)dα = . (55) 4 ( )4 R R ( + + + ) 1 ∏i=1 Γ(ui) 0,∞ 1 2 ρ1 ρ2 ρ3 ρ4

A similar procedure yields the identity

− 4 ui 1 −ρ1−ρ2−ρ3−ρ4 Z ∞ 1 Z ∏i=1 ρi dρi e −u4 α ζ1(u1, α)ζ1(u2, α)ζ1(u3, α) dα = . (56) 4 ( )4 R R R ( + + + ) 1 ∏i=1 Γ(ui) 0,∞ 1 2 3 ρ1 ρ2 ρ3 ρ4

Employing in Equation (50), Equations (53)–(56), and appropriate permutations of (1, 2, 3, 4), we ﬁnd Equation (46). Symmetry 2019, 11, 754 11 of 17

4.3. A New Derivation of Power Mean Estimates for the Hurwitz Zeta Function Here, we re-derive some of the results from [10]. The identity in (14) is a consequence of Equation (37) and of the following exact formula.

Lemma 6. Let ζ1(u, α), u ∈ C, and α > 0 denote the modiﬁed Hurwitz function, and let ζ(u), u ∈ C denote Riemann’s zeta function. Then,

Z ∞ −v Γ(1 − v) α ζ1(u, α) dα = Γ(u + v − 1)ζ(u + v − 1), Re (u + v) > 2, Re v > 1, (57) 0 Γ(u) and Z 1 Z 1 −v ζ(u) − 1 u 1−v α ζ1(u, α) dα = + α ζ1(u + 1, α) dα, Re v < 1, u ∈ C. (58) 0 1 − v 1 − v 0

Proof. In order to derive (57), we ﬁrst assume that Re u > 1, so that we can use the sum representation of ζ1(u, α). Furthermore, we assume that Re v < 1, so that the relevant integral converges at α = 0. Then,

Z ∞ ∞ Z ∞ ∞ Z ∞ −v −v −u 1−(u+v) −v −u α ζ1(u, α) dα = ∑ α (m + α) dα = ∑ m β (1 + β) dβ, 0 m=1 0 m=1 0 where we have used the change of variables α = βm in the second equation. Then, the deﬁnition of Riemann’s zeta function, together with the identity

Z ∞ β−v Γ(1 − v)Γ(u + v − 1) = u dβ , (59) 0 (1 + β) Γ(u) imply Equation (57). In order to derive Equation (58), we use integration by parts:

Z 1 1−v 1 Z 1 −v α 1 1−v ∂ α ζ1(u, α) dα = ζ1(u, α) − α ζ1(u, α) dα. 0 1 − v 0 1 − v 0 ∂α

Thus, (58) follows.

Proof of identity (14). Splitting the ﬁrst integral in the RHS of (37) and then using Equations (57) and (58), as well as using the analogous equations where u and v are interchanged, we ﬁnd Equation (14).

1 ε Remark 1. In order to derive Equation (15) we let σ = 2 + 2 in the LHS of (15), and employ the identities

1 Γ(ε) = − γ + O(ε), ε → 0, (60) ε 0 1 + Γ 2 it iπ 1 = ln t + + O , t → ∞, (61) 1 2 t2 Γ 2 + it as well as the identity 1 h i ζ(ε) = − 1 + ε ln(2π) + O(ε2) , ε → 0. (62) 2

Remark 2. By proving a simple estimate for the integrals in the RHS of (14), it is shown in [10] that these integrals do not contribute to the leading asymptotics of the LHS of Equation (14). Actually, it is straightforward to show that if v = σ1 − it, u = σ2 + it, σ1 < 2, σ1 > 0, Symmetry 2019, 11, 754 12 of 17 then Z 1 1 ∞ 1 1 α1−vζ (u + 1, α) dα = + O , t → ∞. (63) 1 ∑ ( + )u 2 0 it m=1 m m 1 t Indeed, Z 1 ∞ Z 1 1−v 1−σ1 −1−σ2 it[ln α−ln(α+m)] α ζ1(u + 1, α) = ∑ α (m + α) e dα. (64) 0 m=1 0 Noting that d [ln α − ln(α + m)] 6= 0, m ≥ 1, 0 < α < 1, dα it follows that integrals in the RHS of (64) do not possess any stationary points. Then, straightforward integration by parts yields (63).

5. A Relation between Quadratic Products of Hurwitz Zeta Functions and Their Fourier Series Theorem2 will be proved by examining the Fourier series for the function

ζ1(u, α)ζ1(v, α) for Re u, Re v > 0. Following Rane [15], we ﬁrst construct the Fourier series for ζ1(s, α).

Lemma 7. Let σ ∈ (0, 1). Then, the Fourier series

1 Z ∞ + a (s)e2πinα, a (s) = α−se−2πinα dα − ∑ n n s 1 n6=0 1 converges pointwise to ζ1(s, α) for each α ∈ (0, 1).

Proof. Since ζ1(s, α) is a smooth function of α (for fixed s), its Fourier series converges pointwise for α ∈ (0, 1). The Fourier coefficients are defined by

Z 1 −2πinα an(s) = e ζ1(s, α) dα. 0

−s Note that the Fourier series for ζ(α, s) = α + ζ1(s, α) is well-known, and has Fourier coefﬁcients Γ(1 − s)(2πin)s−1. Hence,

Z 1 s−1 −s −2πinα an(s) = Γ(1 − s)(2πin) − α e dα. 0

Using Euler’s integral representation of the Gamma function, we arrive at the desired result.

Remark 3. Since an(s) is expressible in terms of the incomplete Gamma function, we conclude that it has an analytic extension to all complex s 6= 1.

Note that for σ > 1, we have 1 Z ∞ = α−s dα, s − 1 1 so we may write Z ∞ −s a0(s) = α dα, 1 for σ > 1, and by analytic continuation elsewhere. Now, we write

2πinα ζ1(s, α) = ∑ an(s)e , σ > 0 n Symmetry 2019, 11, 754 13 of 17

where the an(s) are deﬁned accordingly.

Lemma 8. Let Re u, Re v > 1 and deﬁne the functions

Z ∞ Z ∞ −v −2πinα −u −2πinα qn(u, v) = an(u + v) + ζ1(u, α)α e dα + ζ1(v, α)α e dα n ∈ Z. 1 1

2πinα Then, the Fourier series ∑n qn(u, v)e converges pointwise to ζ1(u, α)ζ1(v, α) for α ∈ (0, 1).

Proof. Since the Fourier coefﬁcients for ζ1(u, α) are an(u), the Fourier coefﬁcients of the product ζ1(u, α)ζ1(v, α) are given by the convolution

qn(u, v) = ∑ am(u)an−m(v). m

For Re u, Re v > 1, we have

Z ∞ Z ∞ −u −v −2πim(α−β) −2πinβ ∑ am(u)an−m(v) = ∑ dα dβ α β e e , m m 1 1 the double integral being absolutely convergent. Now, we recall the distributional result

∑ e−2πim(α−β) = ∑ δ(β − α − m). m m

Using this in the above, we ﬁnd

Z ∞ −u −v −2πinα qn(u, v) = ∑ α (m + α) e dα m≥1 1 Z ∞ Z ∞ + α−u−ve−2πinα dα + ∑ α−v(m + α)−ue−2πinα dα. (65) 1 m≥1 1

Since Re u, Re v > 1, the integrands of the ﬁrst and third terms can be dominated by the integrable functions α−Re u and α−Re v, respectively, allowing us to pass the sum inside the integral,

Z ∞ Z ∞ −v −2πinα −u −2πinα qn(u, v) = an(u + v) + ζ1(u, α)α e dα + ζ1(v, α)α e dα. 1 1

We note that both the integrals are absolutely convergent for Re u, Re v > 1.

To establish the main result in this section, we must ﬁrst perform an analytic continuation of the functions qn(u, v) valid for Re u, Re v > 0. To this end, we recall the following result [15]:

1−s −s Z ∞ α α −s 2πimx −2πimα ζ1(s, α) = − + lim x e dx e , (66) s − 1 2 N→∞ ∑ 0<|m| 0. This result can be derived using the Euler-Maclaurin formula. We will need the following lemma to control the ﬁnal term.

Lemma 9. Let s = σ + it with σ > 0. Then if α ≥ η > t/2π we have

Z ∞ −s 2πimx −2πimα −σ−1 lim x e dx e η tα . N→∞ ∑ 0<|m|

Proof. For α in the stated range, we can integrate by parts using

Z ∞ Z ∞ x−σ d x−se2πim(x−α) dx = x−ite2πim(x−α) dx α α i(2πm − t/x) dx α−s Z ∞ d x−σ = − + i x−ite2πim(x−α) dx. i(2πm − t/α) α dx 2πm − t/x

We can estimate the sum arising from the ﬁrst term

−s α −σ 2t/α −σ−1 = α η tα . ∑ i(2πm − t/α) ∑ (4π2m2 − t2/α2) 0<|m|

Computing the derivative, the second term becomes

Z ∞ x−σ−1(2πσm + (1 − σ)t/x)) d x−ite2πim(x−α) dx. α i(2πm − t/x)3 dx

An application of the second mean value theorem for integrals on the real and imaginary parts of −σ−1 −2 −σ−2 −3 this term show it to be Oη(α m ) + Oη(tα m ). In particular

Z ∞ −σ d x − ( − ) − − x ite2πim x α dx α σ 1 for α ≥ η > t/2π ∑ dx 2πm − t/x η 0<|m|

Now, we return to the analytic continuation of qn(u, v) for Re u, Re v > 0. The previous lemma establishes that

1−u−v −u−v − α α −Re −Re − α vζ (u, α) − + ≤ tα u v 1, for α ≥ η > t/2π. (67) 1 u − 1 2 η

In particular, the left-hand side is an absolutely integrable function of α on (1, ∞), provided that Re u, Re v > 0. This suggests the splitting

Z ∞ Z ∞ 1−u−v −u−v −v −2πinα −v α α −2πinα ζ1(u, α)α e dα = α ζ1(u, α) − + e dα 1 1 u − 1 2 a (u + v − 1) a (u + v) + n − n , u − 1 2 which is valid for Re u, Re v > 1. This gives rise to the representation

Z ∞ 1−u−v −u−v 1 1 −v α α −2πinα qn(u, v) = + an(u + v − 1) + α ζ1(u, α) − + e dα u − 1 v − 1 1 u − 1 2 Z ∞ 1−u−v −u−v −u α α −2πinα + α ζ1(v, α) − + e dα, 1 v − 1 2 which provides an analytic continuation of qn(u, v) for Re u, Re v > 0.

Remark 4. Using ∂αζ1(s, α) = −sζ1(s + 1, α) in (66), we see that

∂ζ sα−s−1 Z ∞ 1 (u, α) = −α−s + − lim s x−s−1e2πimx dx e−2πimα. ∂α 2 N→∞ ∑ 0<|m|

Using the previous lemma, this then implies that for α ≥ η > t/2π

1−s −s ∂ α α − − ζ (s, α) − + t2α σ 2. (68) ∂α 1 s − 1 2 η

Lemma 10. If u = σ + it and v = σ − it, then for each η > 0

Z ∞ 1−u−v −u−v 1−2σ −v α α −2πinα t α ζ1(u, α) − + e dα η t/2π+η u − 1 2 n where the implied constant is independent of t.

Proof. Integrating by parts, we ﬁnd the above integral can be rewritten as

1−u−v −u−v 1 −v α α −2πinα α ζ1(u, α) − + e 2πin u − 1 2 α=t/2π+η Z ∞ 1−u−v −u−v 1 ∂ −v α α −2πinα + α ζ1(u, α) − + e dα. 2πin t/2π+η ∂α u − 1 2

The ﬁrst term can be estimated using (67), giving

1−u−v −u−v −2σ 1 −v α α −2πinα t α ζ1(u, α) − + e η . 2πin u − 1 2 α=t/2π+η n

And the second term can be estimated using (68)

Z ∞ 1−u−v −u−v 2 Z 1 ∂ −v α α −2πinα t −2σ−2 α ζ1(u, α) − + e dα η α dα. 2πin t/2π+η ∂α u − 1 2 n t/2π+η

1−2σ Performing the ﬁnal integration shows that this term is Oη(t /n).

The previous Lemma gives the following

Z t/2π+η 1−u−v −u−v 1 1 −v α α −2πinα qn(u, v) = + an(u + v − 1) + α ζ1(u, α) − + e dα u − 1 v − 1 1 u − 1 2 Z t/2π+η 1−u−v −u−v 1−2σ −u α α −2πinα t + α ζ1(v, α) − + e dα + Oη , 1 v − 1 2 n valid for u = σ + it, v = σ − it and σ > 0. We note that for n 6= 0, a simple integration-by-parts argument provides an analytic continuation for an(s) into Re s > −1. That is,

Z ∞ −s −2πinα an(s) = α e dα 1 1 s Z ∞ = − α−s−1e−2πinα dα. 2πin 2πin 1

In particular,

Z ∞ 1 (2σ − 1) −2σ −2πinα 1 an(2σ − 1) = − α e dα = O . 2πin 2πin 1 n

Using integration by parts, we also have for 0 < σ ≤ 1/2,

1 Z t/2π+η t1−2σ Z t/2π+η 1 α1−u−ve−2πinα dα = O , α−u−ve−2πinα = O . u − 1 1 n 1 n Symmetry 2019, 11, 754 16 of 17

So for 0 < σ ≤ 1/2 and (u, v) = (σ + it, σ − it), we have

Z t/2π+η Z t/2π+η 1−2σ −v −2πinα −v −2πinα t qn(u, v) = α ζ1(u, α)e dα + α ζ1(u, α)e dα + Oη . 1 1 n

Finally, we show that terms with |n| > t/2π are easily controllable. For this, we once again use the approximate functional equation for the Hurwitz zeta function in the form

(M + α)1−s 1 Z ∞ ζ (s, α) = (α + m)−s + − (M + α)−s − s (α + x)−s−1(x − [x] − 1 ) dx 1 ∑ − 2 1≤m≤M s 1 2 M which holds for M > 1. This follows directly from the Euler-Maclaurin formula when Re s > 1, and then by analytic continuation for Re s > 0. We will require the following lemma:

Lemma 11. For y > 1, σ1, σ2 > 0, and |n| > t/2π, we have

Z t/2π+η −σ2 − + − − − y α σ1 it(α + y) σ2 ite 2πinα dα , 1 |n − t/2π|

Z t/2π+η −σ2 − − − + − y α σ1 it(α + y) σ2 ite 2πinα dα . 1 |n + t/2π|

Proof. The proof is essentially the same as that used for Lemma9. The oscillatory term does not have stationary points if |n| > t/2π, so integrating by parts yields the desired estimate.

Applying this lemma and using the approximate functional equation, we ﬁnd that for |n| > t/2π, the following estimate is valid for each M > 1:

Z t/2π+η −v −2πinα α ζ1(u, α)e dα 1 M1−σ M1−σ M−σ tM−σ = O + O + O + O . |n − t/2π| t|n − t/2π| |n − t/2π| |n − t/2π|

By choosing M = O(t), we ﬁnd

Z t/2π+η 1−σ −v −2πinα t α ζ1(u, α)e dα . 1 |n − t/2π|

Similarly, Z t/2π+η 1−σ −u −2πinα t α ζ1(v, α)e dα . 1 |n − t/2π| So for u = σ + it and v = u¯,

2−2σ 2−2σ 2−4σ 2 t t t 1−2σ |qn(u, v)| + + = O t . ∑ η ∑ |n − t π|2 |n + t π|2 n2 n>t/π |n|>t/π /2 /2 Symmetry 2019, 11, 754 17 of 17

Now, by Parseval’s theorem,

Z 1 4 2 |ζ1(u, α)| dα = ∑ |qn(u, u¯)| 0 n Z t/2π+η 1−2σ 2 − − t − α u¯ ζ (u, α)e 2πinα dα + O + O t1 2σ η ∑ 1 η n |n|≤t/π 1 Z t/2π+η 2 − − − − α u¯ ζ (u, α)e 2πinα dα + O t2 4σ + O t1 2σ η ∑ 1 |n|≤t/π 1

1 Taking σ = 2 and using this in Theorem1 gives rise to the estimate in Theorem2.

Author Contributions: A.C.L.A. was responsible for: formal analysis; investigation, writing–original draft preparation; and A.S.F. was responsible for: formal analysis; investigation; writing–review and editing. Funding: The first author is funded by Homerton College, University of Cambridge. The second author acknowlegdes support from EPSRC via a Senior Fellowship Award (Engineering and Physical Sciences Research Council grant number EP/N006593/1) Acknowledgments: The first author is supported by Homerton College, University of Cambridge. The second author acknowledges support from EPSRC via a Senior Fellowship award. Conflicts of Interest: The authors declare no conflict of interest.

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