The Mathematica 11.1.0.0 command

```
Series[Abs[(E^((\[Pi] x)/2) Gamma[-I x + z] Gamma[I x + z])/
Gamma[z]^2], {x, Infinity, 3}, Assumptions -> x > 0]
```

outputs
$$\frac{\exp \left(\Re\left(-\pi x+(\log (2 \pi )+2 z \log (x)-\log (x))+\frac{2 z^3-3 z^2+z}{6 x^2}+O\left(\left(\frac{1}{x}\right)^3\right)\right)+\frac{\pi x}{2}\right)}{\left| \Gamma (z)\right| ^2} . $$

Hope this would be useful.

Addition. The asymptotics of the integrand as $z \to \infty$ can be found as follows.

```
Series[Abs[(E^((\[Pi] x)/2) Gamma[-I x + z]*Gamma[I x + z])/
Gamma[z]^2], {z, ComplexInfinity, 2}, Assumptions -> x > 0]
```

Because I have a problem with inserting its output in $\LaTeX$ form, the one may be submitted through Dropbox on demand. The output is neither simple nor nice.

Addition 2. The math experiment

```
Table[Log[1/5^Re[z]* NIntegrate[Abs[(E^((\[Pi] x)/2)* Gamma[-I x + z]*Gamma[I x + z])/
Gamma[z]^2], {x, 0, Infinity}]], {z, 100, 400, 100}]
```

performs a warning and $ \{-88.7554,-180.037,-271.463,-362.949\}$ and

```
Table[Log[1/5^Re[z]* NIntegrate[Abs[(E^((\[Pi] x)/2)* Gamma[-I x + z]*Gamma[I x + z])/
Gamma[z]^2], {x, 0, Infinity}]], {z, 20 + 100*I, 20 + 400*I, 100*I}
```

produces a warning and $\{89.3217,232.976,382.167,533.642\} $.

Addition 3. The plot produced by

```
Plot3D[Log[(1/5^Re[z] /. z -> u +I*v)*NIntegrate[ Abs[(E^((Pi*x)/2)*
Gamma[-I *x + z]*Gamma[I* x + z])/Gamma[z]^2] /. z -> u + I*v, {x, 0, Infinity},
Method -> "GlobalAdaptive", AccuracyGoal -> 4, WorkingPrecision -> 15]], {u, -50, 50},
{v, -50, 50},AxesLabel -> Automatic]
```

suggests the dependence of the integral under consideration on $u=\Re z$ mostly.