<s>[INST] Eigenfunction of the Fourier transform I want to show that $$ \frac{1}{ \sqrt{ 2 \pi}} \int_{-\infty}^{\infty} \frac{e^{-iwx}}{\cosh{ (x  \sqrt{\frac{\pi}{2}}} ) } = \frac{1}{\cosh{ (w  \sqrt{\frac{\pi}{2}}} ) }  .$$
My attempt is to first make a substitution $y = x \sqrt{ \frac{\pi}{2} }$, and write $\cosh$ as exponentials which yeilds
$$ \frac{1}{ \sqrt{ 2 \pi}} \int_{-\infty}^{\infty} \frac{e^{-iwx}}{\cosh{ (x  \sqrt{\frac{\pi}{2}}} ) }dx =  \frac{2}{  \pi} \int_{-\infty}^{\infty} \frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{e^{y} + e^{-y} }dy =  \frac{2}{  \pi} \int_{-\infty}^{\infty} e^{-y}\frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{1 - (-e^{-2y}) }dy$$
Now I can write it as a convergent geometric series
$$\frac{2}{  \pi} \int_{-\infty}^{\infty} e^{-y}\frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{1 - (-e^{-2y}) }dy = \frac{2}{  \pi} \int_{-\infty}^{\infty} \sum_{k = 0}^{\infty} -e^{-2yk}  e^{-y}e^{-iwy\sqrt{\frac{2}{\pi}}}dy$$ $$ = \frac{2}{  \pi} \sum_{k = 0}^{\infty}\int_{-\infty}^{\infty}  -e^{-y(2k+1+iw\sqrt{\frac{2}{\pi}})}dy$$
However this integral diverges. Does anyone have any hint on what to do next, or another strategy to show this?
Since this would mean that $\frac{1}{\cosh( x \sqrt{\frac{\pi}{2}})}$ would be an eigenfunctions of the Fourier transforms, should you be able to write it as a linear combination of gaussians?
Thanks!
$\textbf{Added:}$
So before writing it as a geometric series, I split the integration in two parts
$$\frac{2}{  \pi} \int_{-\infty}^{\infty} \frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{e^{y} + e^{-y} }dy =  \frac{2}{  \pi} \int_{0}^{\infty} e^{-y}\frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{1 - (-e^{-2y}) }dy +  \frac{2}{  \pi} \int_{-\infty}^{0} e^{y}\frac{e^{-iwy\sqrt{\frac{2}{\pi}}}}{1 - (-e^{2y}) }dy$$
$$= \frac{2}{  \pi} \sum_{k = 0}^{\infty}\int_{0}^{\infty} (-1)^k e^{-y(2k+1+iw\sqrt{\frac{2}{\pi}})}dy + \frac{2}{  \pi} \sum_{k = 0}^{\infty}\int_{-\infty}^{0} (-1)^k  e^{y(2k+1-iw\sqrt{\frac{2}{\pi}})}dy$$
$$ =  \frac{4}{\pi} \sum_{k=0}^{\infty} (-1)^k \frac{ 2k+1 }{ (2k+1)^2 + (w \sqrt{\frac{2}{\pi}} )^2} $$
Now in a previous exercise I proved this identity:
$$ \frac{\cosh(a( \pi -x))}{\sinh(a \pi)} = \frac{1}{a \pi} + \frac{2}{\pi} \sum_{n = 1}^{\infty}\frac{a}{a^2+n^2}\cos(nx). $$ for $ 0 \leq x \leq \pi$, which look very similar with $x = \pi$. However I'm not sure how to proceed.
 [/INST] You are just about there.  One trick that I find helps over this hump is to recognize that your sum may be rewritten as
$$ \frac{2}{\pi} \sum_{k=-\infty}^{\infty} (-1)^k \frac{ 2k+1 }{ (2k+1)^2 + (w \sqrt{\frac{2}{\pi}} )^2}$$
Now you can apply the residue theorem to this infinite sum.  I will state the following result without (much) proof: the following convergent sum satisfies
$$\sum_{k=-\infty}^{\infty} (-1)^k \, f(k) = -\sum_m \text{Res}_{z=z_m} \pi \csc{(\pi z)} f(z)$$
where $z_m$ are the poles of $f$ that are not real integers.  You may prove this by integrating the function $\pi \csc{(\pi z)} f(z)$ along a rectangular contour about the real line interval $x \in [-N,N]$, and taking the limits as $N \to \infty$.  Now in this case, 
$$f(z) = \frac{2}{\pi}\frac{ 2k+1 }{ (2k+1)^2 + (w \sqrt{\frac{2}{\pi}} )^2}$$
Your poles are at $z_{\pm} = -(1/2) \pm i w \sqrt{\frac{1}{2 \pi}} $.  Now simply take these poles and plug into the above residue formula:
$$\frac{2}{\pi}\sum_{k=-\infty}^{\infty} (-1)^k \frac{ 2k+1 }{ (2k+1)^2 + (w \sqrt{\frac{2}{\pi}} )^2} = - \left [\csc{\left(-\frac{\pi}{2}+i w \sqrt{\frac{\pi}{2}}\right)} + \csc{\left(-\frac{\pi}{2}-i w \sqrt{\frac{\pi}{2}}\right)} \right ]$$
After simplification, I get as your FT:
$$\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} dx \, \frac{e^{-i w x}}{\cosh{[x \sqrt{(\pi/2)}]}} = \frac{2}{\cosh{[w \sqrt{(\pi/2)}]}}$$
which would certainly serve as an eigenfunction of the FT.
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