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.

graph of the function + asymptotics

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).

  • $\begingroup$ What type of problem doth this integral begin its journey in physics? $\endgroup$ – Leucippus Aug 11 '15 at 16:08
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    $\begingroup$ Shaping of on-demand electron wave-packets, here is an abstract I have presented on a conference few weeks ago: fqmt.fzu.cz/15/func/viewpdf.php?reg=650&num=1 $\endgroup$ – Slava Kashcheyevs Aug 11 '15 at 16:10
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    $\begingroup$ @Slaviks OK that is good. Sometimes these kinds of expressions can be evaluated. I think it would be good to note in the question that CAS didn't return any results. $\endgroup$ – Ali Caglayan Aug 11 '15 at 16:16
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    $\begingroup$ @Slaviks I do not think this can be evaluated. Should someone succeed please give me a sign, I would be glad to award a bounty to his answer. $\endgroup$ – Start wearing purple Aug 11 '15 at 17:43
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    $\begingroup$ @tired Nope. The asymptotics of course deserves an upvote but is relatively straightforward to obtain. Finding explicit formula seems to be much more difficult. (+1) $\endgroup$ – Start wearing purple Aug 13 '15 at 16:31

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!


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 $\mathcal{O}(\frac{e^{-(2n+1)\pi x}}{\sqrt{x}})$


I corrected my answer using a correct choice of branches.


Derivation of the last integral:

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

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    $\begingroup$ I see what you did, excellent!! Simultaneously with your post, I have update the plot which confirms your asymptotics beautifully ($I(x)$ defined by @Leucippus equals $\psi(x)/\sqrt{\pi}$). $\endgroup$ – Slava Kashcheyevs Aug 13 '15 at 14:37
  • $\begingroup$ @Slaviks thanks , i recognized the $\sqrt{\pi}$ a little bit too late :) . But this was fun, thanks for the nice question! $\endgroup$ – tired Aug 13 '15 at 14:40
  • $\begingroup$ I could be mistaken, but since $\log (\cosh)$ can be defined by integrating $\tanh z$ on the complex plane from the origin to $z$, I don't think $\frac{1}{\cosh^{3/2}(z)}$ (and thus $f(z,x)$) is well-defined if just the line segments between the branch points are omitted. $\endgroup$ – Random Variable Aug 13 '15 at 16:18
  • $\begingroup$ @RandomVariable your interpretation of the cuts is correct. i learned this technique some years ago from some very thrustworthy people and it worked in every case i tried. but i will rethink this part and give some additional information as soon as i can! $\endgroup$ – tired Aug 13 '15 at 17:13
  • $\begingroup$ @RandomVariable i may be wrong, but is this not the same logic then in this answer? math.stackexchange.com/questions/657668/… $\endgroup$ – tired Aug 13 '15 at 17:46

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.

  • $\begingroup$ Where does your second condition for the poles (with $z_m$) come from? $\endgroup$ – Slava Kashcheyevs Aug 11 '15 at 16:55
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    $\begingroup$ @Slaviks I removed the second condition. In general $\Gamma(z) = \infty$ for $z = 0, -1, -2, \cdots$. While typing I had inverted the values which led to an error. $\endgroup$ – Leucippus Aug 11 '15 at 17:01
  • $\begingroup$ Aha, and since the Gamma function is never zero, I can close the contour either in the upper or the lower half plane of $z$, depending on the sign of $x$, right? But then I think the poles are not in general limited to $m>0$ $\endgroup$ – Slava Kashcheyevs Aug 11 '15 at 17:09
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    $\begingroup$ @Slaviks The branch cut will be the ultimate issue. For some examples of how to deal with branch cuts see Dean G. Duff's 2nd edition of 'Transform methods for solving partial differential equations'. $\endgroup$ – Leucippus Aug 11 '15 at 18:49
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    $\begingroup$ @Leucippus at least this approach is well suited for an asymptotic analysis of the problem :) $\endgroup$ – tired Aug 13 '15 at 13:09

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.


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