For fixed $h>0$, consider the function $$F(x) = e^{-x\pi/2} \left| \Gamma\left(h-i\frac x2\right)\right|^2.$$ What is its asymptotic expansion as $x\to 0$ or $x\to \pm\infty$?
For $x\to \pm\infty$, I only need the leading behavior, while for $x\to 0$, I need the first subleading term that depends on $x$.
I tried Mathematica, but it does not provide a useful answer.
Mathematica provides a useful answer.
Hint
Consider $$\Gamma(h-\epsilon)=\Gamma (h)\left(1- \psi ^{(0)}(h)\,\epsilon+\frac{1}{2} \left(\psi ^{(0)}(h)^2+\psi ^{(1)}(h)\right)\,\epsilon ^2+O\left(\epsilon ^3\right) \right)$$ Square it $$\Gamma (h-\epsilon )^2=\Gamma (h)^2\left( 1-2 \psi ^{(0)}(h)\,\epsilon+\left(2 \psi ^{(0)}(h)^2+\psi ^{(1)}(h)\right)\,\epsilon ^2 +O\left(\epsilon ^3\right)\right)$$
By this known result $$ \mathrm{e}^{ - x\pi /2} \left| {\Gamma \!\left( {h \pm \mathrm{i}\frac{x}{2} } \right)} \right|^2 \sim 2\pi \left( {\frac{x}{2}} \right)^{2h - 1} \mathrm{e}^{ - x\pi} $$ as $x\to +\infty$ with any fixed $h>0$. As $x\to 0$, we have $$ \mathrm{e}^{ - x\pi /2} \left| {\Gamma \!\left( { h\pm \mathrm{i}\frac{x}{2} } \right)} \right|^2 = \Gamma ^2 (h) - \frac{\pi }{2}\Gamma ^2 (h)x + \mathcal{O}_h(x^2 ). $$ The absolute value of the gamma function is an even function, thus the more complicated behaviour is hidden in the $\mathcal{O}$-term. Only the exponential function contributes to the linear term.
