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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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    $\begingroup$ 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. $\endgroup$ Aug 29, 2010 at 13:26
  • $\begingroup$ 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. $\endgroup$ Aug 29, 2010 at 18:23
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    $\begingroup$ 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). $\endgroup$ Aug 29, 2010 at 19:17
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    $\begingroup$ 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 $2$... $\endgroup$ Aug 29, 2010 at 20:16
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    $\begingroup$ ...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. $\endgroup$ Aug 29, 2010 at 21:36

3 Answers 3

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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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    $\begingroup$ 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? $\endgroup$
    – castal
    Dec 5, 2010 at 19:02
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    $\begingroup$ 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. $\endgroup$
    – Anixx
    Dec 5, 2010 at 21:22
  • $\begingroup$ @Anixx This is interesting. How did you derive it? $\endgroup$
    – DUO Labs
    May 4, 2020 at 20:31
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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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    $\begingroup$ this is sort of cheating. You certainly haven't gotten rid of the zeta function; you've just moved it to the other side. $\endgroup$ Aug 31, 2010 at 14:40
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    $\begingroup$ Indeed, this is like claiming that $x=\tfrac12y$ is the result of getting rid of the $2$ in the equation $2x=y$ :) $\endgroup$ Oct 28, 2010 at 22:37
  • $\begingroup$ 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. $\endgroup$ Oct 31, 2010 at 16:56
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I don't think it's possible. There is a paper that expresses very good approximations of zeta using truncated euler products and gamma functions.

The author in this paper, uses finite euler products, gamma functions and forces functional equation on the approximation, subtracts the principal component part from the approximation, and some other tricks. The approximation is very efficient to compute the zeros of Zeta. However, the principal part on the imaginary line is an infinite sum of of reciprocal functions with coefficients involving gamma values, and a finite number of primes. The regularization is based on the observation that the euler product should have a natural boundary at sigma=0, but the analytical continuation of zeta has none on the imaginary axis. Hence the principal part of the euler product must vanish somehow, and the author simply removed it cleverly. However, I don't think you can express the principal part as finite sum of gamma functions, but as a contour in integral yes of course.

But why even bother, when you already have a hankel contour integral and mellin integral representation for the zeta function in terms of the exponential.

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