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(1) The Riemann hypothesis (RH) states that all non-trivial zeros of the zeta function have real part 1/2.

(2) The zeta function is intimately connected with the Gamma function via the functional equation.

The second fact suggests that there is an equivalent form of RH which is expressed solely in terms of the Gamma function.

Question: What is the most natural form to translate RH as directly as possible (without mentioning the zeta function) into a hypothesis on the behaviour of the Gamma function?

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Based on the philosophy that the Gamma function is the term in the Euler product "at infinity" I would expect that this is not possible, since a local term shouldn't give us enough information about the whole thing, but I would be pleasantly surprised if I were wrong. –  Qiaochu Yuan Aug 29 '10 at 13:26
    
I agree with Qiaochu Yuan here. Trying to describe the Riemann hypothesis in terms of the gamma function alone is like trying to describe it with the function (1−2−s)−1(1-2^{-s})^{-1} alone, since it appears in the Euler product. All the gamma function does, afaict, is cancel out the trivial zeros at the negative even integers. –  George Lowther Aug 29 '10 at 18:23
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I meant $(1-2^{-s})^{-1}$ above (editing comments seems to cause formulas to get mangled, especially annoying since it won't let you re-edit to fix this). –  George Lowther Aug 29 '10 at 19:17
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Let me play devil's advocate to the OP by pushing the argument further until its dubiousness becomes clear. "(2) The zeta function is intimately connected with $\pi^s$ via the functional equation. The second fact suggests that there is an equivalent form of RH which is expressed solely in terms of the powers of $\pi$." Seems shakier, right? For more convincing, try replacing $\pi$ with $2$... –  Pete L. Clark Aug 29 '10 at 20:16
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...so, my point was that concentrating on the gamma function is like concentrating on one of the $(1-p^{-s})^{-1}$ terms. Actually, if you wanted, you could redefine the zeta function as a sum over the odd positive integers which would have the effect of shifting the $(1-2^{-s})^{-1}$ term explicitly into the functional equation alongside the gamma function. –  George Lowther Aug 29 '10 at 21:36

2 Answers 2

up vote 4 down vote accepted

Since

$$\zeta(z)=\frac{\Gamma (1-z) \left(2^{-z} \left(\psi \left(z-1,1\right)+\psi \left(z-1,\frac{1}{2}\right)\right)-\psi(z-1,1)\right)}{\ln(2)}$$

where $\psi(x,z)$ is the generalized polygamma following Espinosa's generalization, whatever we say about Zeta function we can also say about the right hand part of this identity. It consists only of Gamma function, its (fractional) derivatives and integrals.

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But it seems that Espinosa and Moll define their generalized polygamma function in terms of the Hurwitz Zeta function. Is there some equivalent characterization of their polygamma function that uses only Gamma? –  castal Dec 5 '10 at 19:02
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The formula by Espinosa is obtained from fractional integration of polygamma by Adamchik (Adamchik, V. (1998). Polygamma functions of negative order. Jour. Comp. Appl. Math., 100, 191–199.) whose modified (by introducing proper integration constants for balancing) definition is used by Espinosa. –  Anixx Dec 5 '10 at 21:22

The Wikipedia article gives a Mellin transform

$$\Gamma(s)\zeta(s) =\int_0^\infty\frac{x^{s-1}}{e^x-1} dx.$$

The Dirichlet series over the Möbius function gives the reciprocal

$$ \frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} .$$

Thus we may write

$$\Gamma(s) = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} \int_0^\infty\frac{x^{s-1}}{e^x-1} dx .$$

This holds true for every complex number s with real part > 1. Now let's try to enlarge the domain of validity of this representation. Riemann showed (see the book of H. M. Edwards, Riemann Zeta Function, for the details) that modifying the contour gives a formula valid for all complex s.

$$ 2\sin(\pi s)\Gamma(s)\zeta(s) = i \oint_C \frac{(-x)^{s-1}}{e^x-1}dx .$$

This leads to

$$ \sin(\pi s) \Gamma(s) = \frac{i}{2} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} \oint_C \frac{(-x)^{s-1}}{e^x-1} dx . \qquad (*) $$

However, this formula is again only valid for s with real part > 1 because of the use of the Dirichlet series. Wikipedia remarks: "The Riemann hypothesis is equivalent to the claim that [this representation of the reciprocal of the zeta function] is valid when the real part of s is greater than 1/2." Thus a possible answer to my question is: "The representation (*) is valid for all s with real part > 1/2 if and only if the RH holds."

Perhaps someone can elaborate further to give this relation a more geometric meaning? Where are the non-trivial zeros of the zeta function to be spotted in this setup?

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this is sort of cheating. You certainly haven't gotten rid of the zeta function; you've just moved it to the other side. –  Qiaochu Yuan Aug 31 '10 at 14:40
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Indeed, this is like claiming that $x=\tfrac12y$ is the result of getting rid of the $2$ in the equation $2x=y$ :) –  Mariano Suárez-Alvarez Oct 28 '10 at 22:37
    
The remarks are delightful, but do they aid to answer the question which was: "What is the most natural form to translate RH as directly as possible (without mentioning the zeta function) into a hypothesis on the behaviour of the Gamma function?" Qiaochu Yuan's remark "I would expect that this is not possible" is the remark most agreed upon; however it does not help me to see the reason, as "the philosophy that the Gamma function is the term in the Euler product "at infinity"" is above my head and perhaps better suited for MO than for SE. –  Bruce Arnold Oct 31 '10 at 16:56

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