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

Question

Is it possible to compute the following Fourier transform analytically? $$\psi(x) = \frac{1}{\sqrt{4 \pi}}\int \Gamma \left (\frac{1}{2}-i \frac{p}{2 \pi} \right) \frac{e^{i p x}}{\sqrt{ \cosh(p/2)}} dp $$

I am aware that $$ \Gamma \left (\frac{1}{2}-i \frac{p}{2 \pi} \right) \Gamma \left (\frac{1}{2}+i \frac{p}{2 \pi} \right)= \frac{\pi}{\cosh(p/2)}$$ and both sides of the identity have poles only at $p=(\text{odd integer}) \times 2 \pi i$, but can not figure out how this can help with the integral.

Mathematica v10 returns the integral unevaluated, both with Integrate[] and InverseFourierTransform[].

The question comes from a research problem in quantum physics.

Edit-3: (makes Edit-1 obsolete). The function is real-valued and positive, see continuous line on the numerically produced graph. Dashed lines are $\sqrt{2} \, e^{-\pi x}/\sqrt{x}$ for $x \to +\infty$ and $20 \, e^{+\pi x} \sqrt{-x}$ for $x \to -\infty$, confirming the asymptotics derived by @tired.

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Edit-2: Here is what is relevant for the publication on the physics side. We are are proposing a method to generate quantum particles described by the wave-function $\psi(x)$. The unprecedented feature of the method (in the particular physics context) is that the Heisenberg product of coordinate-momentum uncertainty $\Delta x \, \Delta p \approx 0.537541$ is close to the so-called Kennard bound of $1/2$ (known to be satisfied by Gaussian wave-packet only).

For the momentum the result is neat: $$(\Delta p)^2 = \int \frac{p^2}{4 \cosh^2(p/2)} dp= \frac{\pi^2}{3}$$

For the coordinate I need to compute $\Delta x \equiv \langle x^2 \rangle-\langle x \rangle^2$ where

$$ \langle x^n \rangle =\frac{1}{2\pi} \int x^n \, |\psi(x)|^2 \, dx$$

Numerically I have $ \langle x \rangle \approx -0.251022$ and $\langle x^2 \rangle \approx 0.150842$ which gives the above quoted uncertainty product.

So it would be very nice to find an analytic form for the first two moments of the probability distribution corresponding to $| \psi(x)|^2$ and also to prove analytically the exponential asymptotics (which distinguishes these wave-packets from other types known in the literature in my field of physics).

Answer 1

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:

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

Answer 2

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.

Answer 3

I was able to answer to the second of the question (about the moments), using the ideas form @tired's answer and the answer to this question. The first two moments evaluate to $$ \langle x \rangle =-\frac{1+\gamma}{ 2\pi}$$ and $$ \langle x^2 \rangle=\frac{1}{2 \pi^2} \left ( 1+ \gamma+ \frac{\gamma^2}{2}\right )+\frac{1}{16}$$ where $\gamma$ is Euler constant.