TITLE: Asymptotic behavior of integral with gamma functions
QUESTION [5 upvotes]: Consider the following function defined for complex numbers $z\in\mathbb{C}$ with $\Re(z)\geq \frac{1}{2}$:
$$F(z)=\frac{1}{5^{\Re(z)}}\int_0^\infty \left| \frac{\Gamma(z+ix)\Gamma(z-ix)}{\Gamma(z)^2} \cdot \exp\left( \frac{\pi}{2} x\right)\right| dx.$$
I am wondering about the behavior of $F(z)$ for $\vert z \vert \rightarrow \infty$.

REPLY [2 votes]: Too long for a comment:
By using the approach of @Carlo Beenakker I was able to get an asymptotic formula for $z=\frac{1}{2} + z_{0} e^{i \phi}$ with $z_{0}\rightarrow\infty$ and $-\frac{\pi}{2}<\phi<\frac{\pi}{2}$. Here it is:
$$
F(z) \sim \left(\frac{2}{5}\right)^{\Re (z)} \sqrt{\pi z_0 \cos \phi} \exp\left\{z_0 \left[\phi \ \sin{\phi}+\cos{\phi} \ln{(\cos{\phi})}\right]\right\}.
$$
Near the limits of the validity interval for $\phi$ the approximation breaks down. For $\phi = \pm\frac{\pi}{2}$ and $z_0 \rightarrow \infty$ I found
$$
F(\frac{1}{2} \pm i z_{0})\sim \frac{2}{\pi\sqrt{5}}\left(2\  e^{z_{0}\frac{\pi}{2}}-1\right). 
$$
Edit: However, in the last formula a factor is missing. But for the asymptotic case $z=\frac{1}{2} + i z_{0}$ with $z_{0}\rightarrow \infty$ it is anyway easier to exploit the identity
$$
\left|\Gamma\left(\frac{1}{2} + i y\right)\right|^2 = \frac{\pi}{\cosh (\pi y)}.
$$
After inserting in the definition of $F(z)$ and some variable transformations on gets
$$
\sqrt{\frac{2}{5}}\frac{\beta+\beta^{-1}}{2^{3/2}\pi}\int_{1}^{\infty} dz \ z^{-1/4} \left(z^2 +(\beta+\beta^{-1})z+1\right)^{-1/2}
$$
with the abbreviation $\beta = e^{\pi z_{0}}$. Mathematica can solve this integral in terms of Gaussian Hypergeometric functions and an Appell function, $F_{1}$ (see for example here). This can be expanded for large $z_{0}$
$$
F\left(\frac{1}{2}+ i z_{0}\right) = \sqrt{\frac{2}{5}} \frac{\Gamma\left(\frac{1}{4}\right)^2}{(2 \pi)^{3/2}} e^{\frac{\pi}{2}z_{0}}+O(e^{-\frac{\pi}{2}z_{0}})
$$