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 $\vert z \vert \rightarrow \infty$. 



Asymptotic behavior of integral with gamma functions
If I just insert the large-$z$ asymptotics of $\Gamma(z)\rightarrow \sqrt{2 \pi } e^{-z} z^{z-\frac{1}{2}}$, and take $z>1/2$ real for simplicity, I find
$$5^z\,F(z)\rightarrow \int\_0^\infty \left(1+x^2/z^2\right)^{z-\frac{1}{2}} e^{\pi x/2-2 x \arctan \left(x/z\right)}\,dx$$
$$\qquad = z\int\_0^\infty(1+x^2)^{-1/2}\exp[z u(x)]\,dx$$
$$\text{with}\;\;u(x)=\pi x/2-2x\arctan x+\ln(1+x^2).$$
This is readily evaluated in the large-$z$ limit by steepest descent. Since $u'(x)=\frac{1}{2}(\pi-4\arctan x)$, the maximum of $u(x)$ is at $x=1$, so we expand 
$$u(x)=\ln 2-\tfrac{1}{2}(1-x)^2+\cdots,$$
and carry out the Gaussian integral, to arrive at 
$$F(z)\rightarrow (2/5)^{z}\sqrt{\pi z}.$$
Here is a numerical test, blue the exact $F(z)$ and gold the asymptote:


![](https://ilorentz.org/beenakker/MO/gamma_asymp2.png)


the agreement is very precise (the two curves are pretty much indistinguishable).


The steepest descent for complex $z$ can be carried out in the same way, but is a more tedious calculation.


