TITLE: Is $\frac{1}{\pi}\int_{0}^{\infty} \Gamma(\sigma +xi)^2\,\Gamma(\sigma-xi)^2 \,dx = \frac{\Gamma(2\,\sigma)^4}{\Gamma(4\,\sigma)}$?
QUESTION [12 upvotes]: Using the approach from the answer to this question, it can be shown that for $\sigma \in \mathbb{C}, x\in \mathbb{R}$:
$$\frac{1}{\pi}\int_{0}^{\infty} \Gamma(\sigma +xi)\,\Gamma(\sigma-xi) \,dx =\frac{\Gamma(2\sigma)}{2^{2\sigma}} \qquad \Re(\sigma)>0$$

In a nutshell:
$$\Gamma(s) = \int_0^\infty \frac{x^{s-1}}{e^{x}}dx \overset{(x \, =
 \,e^u)}= \int_{-\infty}^\infty \frac{e^{su}}{e^{e^u}}du =
 \int_{-\infty}^\infty \frac{e^{-su}}{e^{e^{-u}}}du$$
$\Gamma(s)$ is the Laplace transform of $\dfrac{1}{e^{e^{-u}}}$ and 
   $F_\sigma(\xi) = \Gamma(\sigma+2i \pi \xi)$ is the Fourier transform
   of $f_\sigma(u) = \dfrac{e^{-\sigma u}}{e^{e^{-u}}}$. Apply the
   Parseval theorem
   and get: $$\int_{-\infty}^\infty |F_\sigma(\xi)|^2d\xi =
 \int_{-\infty}^\infty |f_\sigma(u)|^2du = \int_{-\infty}^\infty
 \frac{e^{-2\sigma u}}{(e^{e^{-u}})^2}du $$
$$=\int_{-\infty}^\infty \frac{e^{2\sigma u}}{(e^{e^{u}})^2}du
 \overset{(e^u \, = \,x)}= \int_0^\infty \frac{x^{2\sigma-1}}{e^{2x}}dx
 =  \frac{\Gamma(2\sigma)}{2^{2\sigma}}$$

For the slightly altered function, I derived from numerical testing that:
$$\frac{1}{\pi}\int_{0}^{\infty} \Gamma(\sigma +xi)^2\,\Gamma(\sigma-xi)^2 \,dx = \frac{\Gamma(2\,\sigma)^4}{\Gamma(4\,\sigma)}\ \qquad \Re(\sigma)>0$$
Could this be proven? (I tried a similar approach as above, however the squares got me stuck).
Thanks!

REPLY [5 votes]: If you're still interested in evaluating the integral using Parseval's/Plancherel's theorem for the Mellin transform, we just have to show that $$2\int_{0}^{\infty} x^{s-1} K_{0}(2 \sqrt{x}) \, \mathrm dx = \Gamma^{2}(s) \; , \quad \Re(s) > 0\, ,  \tag{1}$$
where $K_{0}(z)$ is the modified Bessel function of the second kind of order zero.
Then for $\sigma >0$,
$$ \int_{-\infty}^{\infty} \Gamma^{2} (\sigma +it) \Gamma^{2} (\sigma -it) \, \mathrm dt = \int_{-\infty}^{\infty} \lvert \Gamma^{2}(\sigma +it) \rvert ^{2} \, \mathrm dt = 8 \pi\int_{0}^{\infty} x^{2\sigma -1} \, K_{0}^{2}(2 \sqrt{x}) \, \mathrm dx .$$
To show $(1)$, we can use the integral representation $$K_{0}(z) = \int_{0}^{\infty} e^{-z \cosh t} \, \mathrm dt \, , \quad \operatorname{Re}(z) >0, $$
and then switch the order of integration (which is justified since the integrand is nonnegative).
$$ \begin{align} 2\int_{0}^{\infty} x^{s-1} K_{0}(2 \sqrt{x}) \, \mathrm dx &= 4 \int_{0}^{\infty} u^{2s-1} K_{0} (2u) \, \mathrm du \\ &= 4 \int_{0}^{\infty} u^{2s-1} \int_{0}^{\infty} e^{-2 u \cosh t} \, \mathrm dt \,\mathrm du \\ &=4 \int_{0}^{\infty} \int_{0}^{\infty} u^{2s-1} e^{-2u \cosh t} \, \mathrm du \, \mathrm dt \\ &=\frac{4 \, \Gamma(2s)}{2^{2s}}\int_{0}^{\infty} (\cosh t)^{-2s} \,   \, \mathrm dt  \\ &= \frac{2 \, \Gamma(2s)}{2^{2s-1}}  \frac{\sqrt{\pi}}{2} \frac{\Gamma(s)}{\Gamma(s+ \frac{1}{2})} \tag{2} \\ &= \Gamma^{2}(s) \tag{3} \end{align}$$

$(2)$ https://math.stackexchange.com/a/1379526
$(3)$ Legendre Duplication Forumula

If we then use the integral representation$$K_{0}^{2}(z) = 2 \int_{0}^{\infty} K_{0}(2z \cosh t) \, \mathrm dt \, , \quad \operatorname{Re}(z) >0, \tag{4}$$ (which I'll talk about at the end), we get
$$\begin{align}\int_{-\infty}^{\infty} \Gamma^{2} (\sigma +it) \Gamma^{2} (\sigma -it) \, \mathrm dt &=8 \pi\int_{0}^{\infty} x^{2\sigma -1} \, K_{0}^{2}(2 \sqrt{x}) \, \mathrm dx  \\ &= 16 \pi\int_{0}^{\infty} u^{4\sigma -1} \, K_{0}^{2}(2 u) \,  \mathrm du  \\ & =32 \pi \int_{0}^{\infty} u^{4 \sigma-1} \int_{0}^{\infty} K_{0}(4u \cosh v) \, \mathrm dv \, \mathrm du \\ &= 32 \pi \int_{0}^{\infty} u^{4 \sigma -1} \int_{0}^{\infty} \int_{0}^{\infty} e^{-4u \cosh v \cosh w} \, \mathrm dw \, \mathrm dv \, \mathrm du  \\ &= 32 \pi \int_{0}^{\infty} \int_{0}^{\infty} \int_{0}^{\infty} u ^{4 \sigma-1} e^{-4u \cosh v \cosh w} \, \mathrm du \, \mathrm dw \, \mathrm dv \\ &= 32 \pi \int_{0}^{\infty} \int_{0}^{\infty} (4 \cosh v \cosh w)^{-4 \sigma} \, \Gamma(4 \sigma) \, \mathrm dw \, \mathrm dv \\ &=  \frac{32 \pi \, \Gamma(4 \sigma)}{2^{8 \sigma}} \left(\int_{0}^{\infty} (\cosh v)^{- 4 \sigma}  \, \mathrm dv \right)^{2} \\ &= \frac{32 \pi \, \Gamma(4 \sigma)}{2^{8 \sigma}} \frac{\pi}{4} \frac{\Gamma^{2}(2 \sigma)}{\Gamma^{2}(2 \sigma + \frac{1}{2})} \\ &= \frac{32 \pi \, \Gamma(4 \sigma)}{2^{8 \sigma}} \frac{\pi}{4} \,  \Gamma^{2}(2 \sigma) \,  \frac{2^{8 \sigma-2} \,   \Gamma^{2}(2 \sigma)}{\pi \, \Gamma^{2}(4 \sigma)}  \\ &= \frac{2 \pi \, \Gamma^{4}(2 \sigma)}{\Gamma(4 \sigma)}. \end{align}$$

Integral $(4)$ can be derived by first expressing $K_{0}^{2}(z)$ as the iterated integral $$\frac{1}{4} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{- z \cosh t} e^{-z \cosh u} \, \mathrm dt \, \mathrm du.    $$ Then interpret the iterated integral as a double integral and make the transformation $ v = \frac{t+u}{2}$, $w= \frac{t-u}{2}$ (which does not change the limits of integration).
Finally use the identity $\cosh (v+w) + \cosh(v-w) = 2 \cosh(v) \cosh(w) $ and then integrate.
G.N. Watson uses this approach in A Treatise on the Theory of Bessel Functions to derive the generalization of $ (4)$ that I linked to above.