How to calculate $\int_{-\infty}^{+\infty}\Gamma(x+yi)\Gamma(x-yi) \, dy$?

Question

I tried. Kilbas says that $\int_{-\infty}^{+\infty} \Gamma(x+yi) \Gamma(x-yi) \, dy=(2\pi)^{3/2}$$_2F_1(1/2,1/2,1/2;-1)$. In this case, to the function $_2F(a,b.c;z)$, we have $z=-1 \ (|z|=1)$ and a conditionally convergence if $-1<\Re(c-a-b)\leq 0$.

There is a better way to solve this integral? Thanks for helping!

Answer 1

First we evaluate a symmetric integral: $$\int_0^{\infty } \frac{\cosh (a x)}{\cosh ^v(b x)} \, dx=2^{v-1} \int_0^1 \frac{t^{-a}+t^a}{t \left(t^{-b}+t^b\right)^v} \, dt=2^{v-1} \int_1^\infty \frac{t^{-a}+t^a}{t \left(t^{-b}+t^b\right)^v} \, dt\\=2^{v-2} \int_0^\infty\frac{t^{-a}+t^a}{t \left(t^{-b}+t^b\right)^v} \, dt=\frac{2^{v-2} \Gamma \left(\frac{v}{2}-\frac{a}{2 b}\right) \Gamma \left(\frac{a}{2 b}+\frac{v}{2}\right)}{b \Gamma (v)}$$ Where the first equality is given by $e^{-x}\to t$, the second by $t\to\frac1t$, the third by taking averages of above two, the last by recalling Beta integral $\int_0^\infty \frac{t^{s-1}}{(1+t)^{s+t}}=B(s,t)$. Since both sides are analytic w.r.t $a$, one may let $a\to i a$ to arrive at $$\int_0^{\infty } \frac{\cos (a x)}{\cosh ^v(b x)} \, dx=\frac{2^{v-2} \Gamma \left(\frac{v}{2}-\frac{i a}{2 b}\right) \Gamma \left(\frac{a i}{2 b}+\frac{v}{2}\right)}{b \Gamma (v)}$$ Therefore, based on suitable change of variables and Fourier inversion $$\int_{-\infty}^{\infty } \Gamma (x+i y) \Gamma (x-i y) e^{2 \pi i b y} \, dy= \sqrt{\pi } \Gamma (x) \Gamma \left(x+\frac{1}{2}\right) \text{sech}^{2 x}(\pi b)$$ Finally, letting $b\to0$ gives

$$\int_{-\infty }^{\infty } \Gamma (x+i y) \Gamma (x-i y) \, dy=\sqrt{\pi } \Gamma (x) \Gamma \left(x+\frac{1}{2}\right)$$

Bonus: By Parseval one arrive at Ramanujan's celebrated $$\int_{-\infty }^{\infty } \Gamma (x+i y) \Gamma (x-i y) \Gamma (z+i y) \Gamma (z-i y) \, dy=\frac{\sqrt{\pi } \Gamma (x) \Gamma \left(x+\frac{1}{2}\right) \Gamma (z) \Gamma \left(z+\frac{1}{2}\right) \Gamma (x+z)}{\Gamma \left(x+z+\frac{1}{2}\right)}$$ Which is a special case of Barnes integral.

Answer 2

By the beta function identity, we may write

\begin{align*} \Gamma(x+iy)\Gamma(x-iy) &= \Gamma(2x) \int_{0}^{\infty} \frac{t^{x+iy-1}}{(1+t)^{2x}} \, \mathrm{d}t \\ &= 2 \Gamma(2x) \int_{-\infty}^{\infty} \frac{e^{2isy}}{(e^{s} + e^{-s})^{2x}} \, \mathrm{d}s \tag{$t=e^{2s}$}. \end{align*}

Now let $\varepsilon > 0$ and consider the following regularized integral:

$$ I(\varepsilon) := \int_{-\infty}^{\infty} \Gamma(x+iy)\Gamma(x-iy)e^{-\varepsilon y^2} \, \mathrm{d}y. $$

Then the original integral is obtained by computing $\lim_{\varepsilon \to 0^+} I(\varepsilon)$. By the above identity, we find that

\begin{align*} I(\varepsilon) &= 2 \Gamma(2x) \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} e^{2isy}e^{-\varepsilon y^2} \, \mathrm{d}y \right) \, \frac{\mathrm{d}s }{(e^{s} + e^{-s})^{2x}} \\ &= 2 \Gamma(2x) \int_{-\infty}^{\infty} \sqrt{\frac{\pi}{\varepsilon}} e^{-s^2/\varepsilon} \frac{\mathrm{d}s }{(e^{s} + e^{-s})^{2x}} \\ &= 2 \Gamma(2x) \int_{-\infty}^{\infty} \sqrt{\pi} e^{-r^2} \frac{\mathrm{d}s }{(e^{\sqrt{\varepsilon}r} + e^{-\sqrt{\varepsilon}r})^{2x}} \tag{$s=\sqrt{\varepsilon}r$}. \end{align*}

So, as $\varepsilon \to 0^+$, this converges to

$$ I(0) = 2 \Gamma(2x) \int_{-\infty}^{\infty} \sqrt{\pi} e^{-r^2} \frac{\mathrm{d}s}{2^{2x}} = 2^{1-2x}\pi \Gamma(2x). $$

This also matches @User's answer via the Legendre's duplication formula.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ I'll use an Identity related to Gamma Function Integration: \begin{align} &\bbox[5px,#ffd]{\int_{-\infty}^{\infty} \Gamma\pars{x + y\ic}\Gamma\pars{x - y\ic}\,\dd y} \\[5mm] = & 2\pi\bracks{{1 \over 2\pi}\int_{-\infty}^{\infty} \verts{\Gamma\pars{x + y\ic}}^{2} \expo{\pars{2b - \pi}y}\,\dd y} _{\ b\ =\ \color{red}{\pi/2}} \\[5mm] = &\ 2\pi\braces{\Gamma\pars{2x} \over \bracks{2\sin\pars{\color{red}{\pi/2}}}^{\,2x}} = \bbx{2^{1 - 2x}\,\,\pi\,\Gamma\pars{2x}} \\ & \end{align} The above link enforces the conditions $\ds{x > 0}$ and $\ds{b \in \pars{0,\pi}}$.