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We have $$\frac {\Gamma (a)}{2^a}=\int _{(c)}\Gamma (s)\Gamma (a-s)\,ds,$$ see e.g. Exercise C.23 of Montgomery and Vaughan's "Multiplicative Number Theory".

I would like a similar formula for $$\int_{(c)} \Gamma (s)\Gamma (1-s)\Gamma (a+2-s)\Gamma (s-a)\,ds.$$

I've looked at a few reference books for Mellin transforms but can't find this one. It seems to be a particular case of the Meijer $G$-function (https://en.wikipedia.org/wiki/Meijer_G-function), with arguments $G_{2,2}^{2,2}\left (\begin {array}{ll}1,&a+1\\ 1,&a+2\end {array}\right )$ however when I put this in WolframAlpha I don't get anything out.

I suspect that the integral is something like $p(a)/\sin (\pi a)$ with $p$ a quadratic polynomial, but it seems I'm not proficient enough in Mellin transforms to be sure I'm not misunderstanding or miscalculating something.

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    $\begingroup$ By using $\Gamma(s)\Gamma(1-s)=\pi / \sin(\pi s)$, $\Gamma(s)\Gamma(-s)=-\pi /(z \ \sin(\pi s) )$ and the well known $\Gamma(z+1)=z \ \Gamma(z)$ you can get rid of all $\Gamma$-functions. $\endgroup$
    – Johannes Trost
    Commented Jul 30 at 8:23
  • $\begingroup$ ye but i don't think that changes anything - still need to evaluate the contour integral and i don't think one involving sines is any easier. (for example it wouldn't help in the first example). (happy to be wrong though) $\endgroup$
    – tomos
    Commented Jul 30 at 9:17

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Using Mathematica 14 I got the following result $$ \lim_{z\rightarrow 1} G^{2,2}_{2,2}\left(\begin{matrix}1,&a+1\\ 1,&a+2\end{matrix}\middle|z\right)= \pi\ \frac{a(a+1)}{2\ \sin(a \ \pi)} $$

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  • $\begingroup$ thanks a lot! :) $\endgroup$
    – tomos
    Commented Jul 31 at 13:41

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