Fourier transform of $\Gamma \left (\frac{1}{2}-i \frac{p}{2 \pi} \right) /\sqrt{ \cosh(p/2)}$

Ok, i will give a derivation of the leading order asymptotics for $x\rightarrow\infty$. For now, i also concentrate on the case where $x>0$, the opposite possibility should be doable with quite similiar techniques.

We start from the complex function given by @Leucippus.

$$ f(z,x)=\frac{e^{2ixz}}{\Gamma(\frac12+\frac{i}{\pi}z)\cosh^{3/2}(z)} $$

We observe the following four facts:

1.) For $x>0$ the function is convergent for $|z|\rightarrow\infty$ so will close our contour of integration in the upper halpf plane.

2.) $f(x,z)$ has branch points at $z_n=\frac{i \pi}{2}+i n\pi$

3.) The corresponding singularities are integrable which may be checked by using the expansion of $\frac{1}{\Gamma(\frac12+\frac{i}{\pi}z)}$ near $z_n$

4.) The integral we are looking for is given by ($z=\sigma+i s$) $$ I(x)=\int_{-\infty}^{\infty}d\sigma f(\sigma,x) $$

It follows that

$$ I(x)=-\oint_{C_{cut}}dzf(z,x) $$

Here $\oint_{C_{cut}}$ denotes an integral around the branch cut which connects $z_0$ and $i\infty$ along the imaginary axis as @RandomVariable pointed out.

So far everything was exact, but now we need to start to make some approximations.

Let's write this integral a little bit more explicit.For more details how to correctly choose all the phase factors, see here and noting that the integral around the small edge of the cut vanishs (due to the integrable singularity) we obtain:

$$ I(x)=-2i\sum_{n=1}^{\infty}\sin(\frac{3}{2}\pi n)\int_{n\pi-\pi/2}^{n\pi+\pi/2}ds\frac{e^{-2sx}}{\Gamma(\frac12-\frac{s}{\pi})|\cos(s)|^{3/2}} $$

It is clear that this integral is strongly dominated from the region of $s\in[\frac{\pi}{2},\frac{\pi}{2}+\epsilon]$ (everything else is strongly surpressed by the exponential) which means that only a small fraction the first part of the sum will be sufficent to get the leading order asymptotics.

$$ I(x)\sim2i\int_{\pi/2}^{\pi/2+\epsilon}ds\frac{e^{-2sx}}{\Gamma(\frac12-\frac{s}{\pi})|\cos(s)|^{3/2}} $$

Using again the expansion of $\frac{1}{\Gamma(\frac12-\frac{x}{\pi})}$ (or asking mathematica) we therefore end up with

$$ I(x)\sim\frac{2}{\pi}\int_{\pi/2}^{\pi/2+\epsilon}ds\frac{e^{-2sx}}{\sqrt{s-\frac{\pi}{2}}} $$

Because of the exponential we only make an exponentially small error if we blow up our upper limit of integration to infinity so we are fine to calculate

$$ I(x)\sim\frac{2}{\pi}\int_{\pi/2}^{\infty}ds\frac{e^{-2sx}}{\sqrt{s-\frac{\pi}{2}}} $$

This last integral can be done in closed form (it's not too difficult) and we end up with

$$ I(x)\sim\sqrt{2}\frac{e^{-\pi x}}{\sqrt{\pi x}} $$

Comparing with numerical calculation we find an error of just $0,8 \%$ for $x=5$ and $1,2\%$ for $x=3$ which seems incredibly good for me keeping in mind all the approximations we did on our journey!

Edit:

I think it's not too difficult to obtain corrections to this results by taken into account 1.) the second branch point 2.) more branch cuts. i also think that the corrections will be of order $\mathcal{O}(\frac{e^{-(2n+1)\pi x}}{\sqrt{x}})$

Edit2:

I corrected my answer using a correct choice of branches.

Appendix

Derivation of the last integral:

$$ \int_{\pi/2}^{\infty}ds\frac{e^{-2sx}}{\sqrt{s-\frac{\pi}{2}}}=e^{-\pi x}\int_{0}^{\infty}dy\frac{e^{-2xy}}{\sqrt{y}}=e^{-\pi x}\int_{0}^{\infty}dqe^{-2xq^2}=e^{-\pi x}\frac{\sqrt{\pi}}{\sqrt{2 x}} $$

Given the two conditions listed in the proposed problem let $p = 2 z$ to obtain \begin{align} \frac{\psi(x)}{\sqrt{\pi}} = \int_{\Gamma} \, \frac{e^{2 i x \, z} \, dz}{\Gamma\left(\frac{1}{2} + \frac{i \, z}{\pi}\right) \, \left(\cosh(z)\right)^{\frac{3}{2}}}. \end{align} There is a branch cut due to the $(\cosh(z))^{3/2}$ term. The poles are determined by $\cosh(z_{n}) = 0$. This lead to poles of the form $$ z_{\text{poles}} = i \, \pi \, \left(m + \frac{1}{2}\right) \hspace{10mm} m \geq 0$$ From here it is a matter of finding the right contour to evaluate the integral and work around the branch cut(s) to find the integral's value.