On a Formula of Mellin and Its Application to the Study of the Riemann Zeta-function

In this paper, we reconsider a formula of Mellin. We present a formula which relates the sum of two positive real numbers m, n to their product mn. We apply this formula to derivation of a relationship involving the Hurwitz zeta-function. Then we define a series function (stemming from the proved relationship) and discuss an analogy in regard to the Lindelöf hypothesis. Finally, a proof of the Lindelöf hypothesis of the Riemann zeta-function is deduced from this analogy.

By these, (2) is rewritten as This completes the proof of Theorem 1.
Regardless of the straightforward derivation of Theorem 1 from Mellin's formula, however, we decided to present it in a research paper because of its rare characteristic; that is, it relates the sum of two numbers m and n to their product mn.
Throughout the rest of the discussion, we choose −π ≤ arg z < π as the domain of complex logarithm log z.
An application of Theorem 1 is given in a proof of the following formula.

Notes:
1) The first integral on the right would be regarded as its analytic continuation to Re(δ) > −1 (see ( 15) for the existence of such an extension).
3) The conventional definition of ζ(s, a) takes the sum ∑ n≥0 .Because we often consider the case a = 0, however, we separate the n = 0-term.
4) The subscript ϵ for Z ϵ is to make clear the order of magnitude of the polynomially growing factor |α| ϵ in γ(ϵ, δ, α).
We note that the main term of the left side of Theorem 2 as α → ∞ is the function Z ϵ .This is by the appearance of the number (−1) in its terms; the other function E vanishes exponentially because it lacks a factor which should cancel out the exponentially decaying factor γ.
As another objective of this paper other than presentation of Theorem 1 and 2, we shall relate the latter theorem to the study of the Riemann zeta-function.
For instance, a property of the function Z analogous to the Riemann zeta-function is that when −1 < Re(δ) < 0, by (Whittaker & Watson, 2008) Γ the Equation ( 15) below, and by Theorem 2, it is easy to see that as α → ∞, Choosing ϵ = 1/2 and δ = 1 in Theorem 2 and using the well-known result (Whittaker & Watson, 2008) we obtain the following corollary to Theorem 2.
Note: The corollary does not include the case α → −∞.
The Lindelöf hypothesis states that for each ϵ > 0, the Riemann zeta-function would satisfy If we note that the function Z ϵ (1 − δ + ϵ − iα, a)|α| −ϵ is essentially reduced to the Riemann zeta-function by putting a = 0, that is, (by our convention on complex logarithm, (−1) = e −πi and (−1) iα = e απ ) we may conclude from the corollary that the Lindelöf hypothesis for the function Secondly, we will give a proof of the original Lindelöf hypothesis based on Theorem 2.

Method
All the techniques used in our analysis are in the realm of classics.
We mainly rely on the following well-established theoretical tools: 1) Complex analysis (especially analytic continuation); 2) Fourier transform; 3) Rigors by real analysis.
As for the rigors, we leave the reader for verifying any details, such as convergence in taking the summation ∑ m≥1 , etc.
When we use the theory of Fourier integrals, we discuss in the Schwartz space of rapidly decreasing functions.
Two important tools in our discussion, together with analytic continuation, are Theorem 1 and the following function where a and ω 1 are some complex numbers.It is easy to show that the series F s (z, a) has the period 2ω 1 as a function of z.The extensive use of these tools are to be seen as follows.

Proof of Theorem 2
Throughout the argument, δ is a complex variable, while ϵ is a positive real number.
First, for any small δ, ϵ satisfying 0 < Re(δ) < ϵ, we put Here, we note that if then with (4), it is easy to see that we could extend (7) in the variable n to a neighborhood of the number −2(m+a)ω 1 by analytic continuation.
One example of ω 1 and a for which arg(a) > 0, (8), and (10) are satisfied is Then since the right integral of ( 9) is it is easy to show that the summation Replacing z by ze α in (11), we have or multiplying by e δα , 2πΓ(1 Applying the Fourier inversion theorem to (13), namely, we obtain for 0 < Re(δ) < ϵ, Now, the right integral of ( 14) is a meaningful expression; in fact, in the same way as getting ( 12), an application of Theorem 1 gives for Re(δ) < Re(δ ′ ) < ϵ, from which it follows readily that With the change of variable u = e t , the right integral of ( 14) is rewritten as At this point, we break the right integral of (14) as and extend ( 14) in the variable δ to −1 < Re(δ) ≤ 0 by analytic continuation.But in this process, it is plain that the left expression of ( 14) keeps analytic in δ and valid for Re(δ) ≤ 0, and that f 2 (δ + iα, z) is also meaningful and analytic in δ for Re(δ) ≤ 0.
Furthermore, for Re(δ) > 0, we rewrite the integral for f 1 (δ + iα, z) with integration by parts as Then the integral on the right converges also for −1 < Re(δ) ≤ 0, and so this gives the extension of f 1 (δ + iα, z) into the the right half-plane Re(δ) > −1.
Thus, we have Next, we extend ( 16) in the variable z to some neighborhood of the point z = 2ω 1 by analytic continuation.To validate this procedure, we first note that the left member of ( 16) and (by ( 15)) f 1 (δ + iα, z) readily have their desired extensions.
Here we note that by (10) and arg(a) > 0, we could vary z (= a positive real number) continuously upward from some point in the positive real line to the point 2ω 1 , while any singularity is not brought out from F 1+ϵ (zu, a).
To give a new expression for f 2 (δ + iα, z) which is analytic in the variable δ + iα in C − {0} and for z = 2ω 1 , we analyze as follows.(With this step, we could remove the restriction Re(δ) < ϵ, and give a meaningful expression for the analytic continuation of the left members of (16) for Re(δ) ≥ ϵ.) For −1 < Re(δ) < 0, if z is chosen to be 2ω 1 , then by the 2ω 1 -periodicity of F 1+ϵ (z, a), we have (17) here, we used the change of variable u = n + r, and the interchanging of summation and integration symbols is verified easily if we note Re(δ) < 0.
But observing (3), it is easy to see that this restriction on a is improved (by analytic continuation in a) to arg(a) > 0.
(When a passes the real line, however, a singularity occurs.) This completes the proof of Theorem 2.

Proof of the Lindelöf Hypothesis
The main difficulty is to treat several singularities which arise when we let a → 0 in Theorem 2. Throughout this section, a 0 is fixed, unless otherwise a is chosen to be some number.
To resolve the aforementioned problem, we first rewrite and sort out the m = 0-term of the left side of Theorem 2, namely, (−a) and the following terms of the right side, and Referring to ( 14), we have (considering only the m = 0-term in ( 14)) or with the change of variable u = e t on the right, multiplying by (2ω 1 ) 1+ϵ , and choosing z = 2ω 1 , If we write then it is easy to see that ( 18) is equal to We note that B 1 is a meaningful expression for a = 0 and Re(δ) ≤ 1.
Next, we rewrite the integrals in ( 19) and ( 20) which contain the factor (u − 1 − a) −1−ϵ by the change of variable Summing up ( 22) and ( 23), if we recall the series definition of ζ(s, a) and note that the integral has an analytic continuation to the region −1 < Re(δ) ≤ 0 (easily shown by integration by parts as we saw in ( 15)), we have Hence, the sum of two integrals in ( 19) and the second integral in ( 20) is Using all the results above, Theorem 2 is rewritten as, for Re(δ) > 0, where Integrating (25) with respect to a over any path [a 1 , a 2 ] on the upper half-plane, we obtain (recall the series definition of Z ϵ and E) The final step of the proof is to estimate all the terms in ( 26) except for Z ϵ (−δ + ϵ − iα, a 1 ) one by one under the conditions where η > 0 is arbitrarily small.
Here, with Theorem 2 (choose ϵ = 1/2 − η and δ = 1) and ( 5), it is plain that Besides, with integration by parts and ( 5), it is easy to show that where Thus, we are left with the integral In order to rewrite (29) in a form to which estimates from Theorem 2 are applicable, we apply Cauchy's theorem in the variable L as follows.
We choose a 1 to be arbitrarily close to 0 (with arg(a 1 ) > 0), and consider ∫ where here, Γ g 1 is the incomplete semicircle Γ g 1 := {z : z = e it , 0 < t < π − g 1 }, g 1 > 0 is arbitrarily small, and R g 1 , R g 2 are any path from the point z = e i(π−g 1 ) to z = −1 + g 1 and from z = −g 2 to z = g 2 , respectively, such that the contour I g 1 ,g 2 ∪ Γ g 1 ∪ R g 1 does not contain the point z = −1 and z = a 1 .The analyticity of ζ(1 − δ − iα, L) with respect to the variable L and with Re(δ) < 1 on the given contour (necessary in using Cauchy's theorem) follows from which is shown with integration by parts in the sense of Riemann-Stieltjes integration.
By the uniform convergence of the integral throughout a 1 in any neighborhood of a 1 = 0 and lim we have lim Now, if we let g 1 → 0 + in (30), then integrals with diverging integrands on the right side are bounded by, using arg(1 + L) > 0, ∫ Next, we shall estimate the remaining integrals with analytic integrands on the right side of (30), namely, as follows.A difficulty at this step is to give an estimate for That is, one might like to obtain a bound for this function which is valid for any closed segment t ∈ [π − h, π] (or t ∈ [0, h]) with h > 0 arbitrarily small and fixed, while (25) alone is not sufficient for this purpose.(For instance, when a → −1 in (25), we need to handle the singularity (L − a) −1−ϵ inside the integral ∫ 1 −1 .)In order to resolve the issue, we consider another contour K h such that where l h is any small segment which goes from z = −1 + h to z = e i(π−h) , and directly calculate the part lim With the contour K h , we have ∫ Inserting each l ∈ l h (so that l = −1 + µ with µ 0 and µ > 0 as l → −1 + h) for a in (25), and choosing ϵ = 1/2, δ = 1, we get, as α → ∞, (42)