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  1. The Riemann hypothesis (RH) states that all non-trivial zeros of the zeta function have real part $\frac{1}{2}$.

  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.2010-08-29
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    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.2010-08-29
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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).2010-08-29
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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 $2$...2010-08-29
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    Pete - I think that's taking it a bit far. I guess that where the OP is coming from is supposing that the gamma and zeta function play more or less equal roles in the functional equation, and might not be convinced by suggesting looking at the $\pi^s$ term instead. Actually, using the product expansion, the functional equation involves a product of infinitely many terms, one $(1-p^{-s})^{-1}$ for each prime p and one $\Gamma(s/2)$ for the "prime at infinity" (aka, the Euclidean valuation). The zeta function encodes all but one of these terms, and the gamma function supplies the missing term.2010-08-29
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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.2010-08-29
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    As I see that refering to the functional equation is misleading I suggest to ignore (2) as an unfortunate choice of motivation. The question was: "What is the most natural form to translate RH into a hypothesis on the behaviour of the Gamma function?" Riemann defined $\zeta(s)=\Gamma(1-s)I(s)$ where $I(s)$ is a certain integral divided by a constant (H. M. Edwards, Riemann zeta function, 1.4(3)). Thus a way to put the question is: How to restate the RH for $\zeta(s)/I(s)$; however this might be futile as George points out.2010-08-30
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    Or Can there be an integral representation of the $\zeta$ function? Because for example there are certain integrals, whose values equal $\frac{\pi^{2}}{6} = \zeta(2)$, may be we should and can think of this question in that manner.2010-08-31

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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?2010-12-05
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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.2010-12-05
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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 greater than $1$. 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 greater than $1$ 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 $\frac{1}{2}$."

Thus a possible answer to my question is:

The representation $\,*\,$ is valid for all $s$ with real part greater than $\frac{1}{2}$ 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.2010-08-31
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    Indeed, this is like claiming that $x=\tfrac12y$ is the result of getting rid of the $2$ in the equation $2x=y$ :)2010-10-28
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    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.2010-10-31