How to calculate $\int_{-\infty}^{+\infty}\Gamma(x+yi)\Gamma(x-yi) \, dy$? I tried. Kilbas says that $\int_{-\infty}^{+\infty} \Gamma(x+yi) \Gamma(x-yi) \, dy=(2\pi)^{3/2}$$_2F_1(1/2,1/2,1/2;-1)$. In this case, to the function $_2F(a,b.c;z)$, we have $z=-1 \ (|z|=1)$ and a conditionally convergence if $-1<\Re(c-a-b)\leq 0$.
There is a better way to solve this integral? Thanks for helping!
 A: First we evaluate a symmetric integral:
$$\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.
A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
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
$\ds{b \in \pars{0,\pi}}$.
