How to express the Riemann hypothesis in terms of the Gamma function?
Solution 1:
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.
Solution 2:
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?