Eigenfunction of the Fourier transform

Question

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.

Answer 1

Hint: let $z\in \mathbb C$. The zeroes of $f(z)=\exp(z)+\exp(-z)$ are all complex numbers $z_k=a+ib_k$ s.t.

$$a=0$$ $$b_k=\frac{\pi}{2}+k\pi,~~k\in\mathbb Z.$$

The integral $\int_{-\infty}^{\infty}\frac{\exp(-i\tilde{\omega}y)}{f(y)}dy,$ with $\tilde{\omega}:=\omega\sqrt{2}{\pi}$ can be computed using the residue theorem and an appropriate limiting procedure ($R\rightarrow \infty$, $K\rightarrow \infty$) on

$$\int_{\gamma}\frac{\exp(-i\tilde{\omega}z)}{f(z)}=Res(f(z),z_0)=\frac{\exp((-i\tilde{\omega}i\frac{\pi}{2})}{2exp(i\frac{\pi}{2})}=\frac{1}{2}\exp(\tilde{\omega}\frac{\pi}{2}),$$

with $\gamma=\gamma_1+\gamma_2$, and $\gamma_1(t)=-R(1-t)+tR$, $\gamma_2(t)=K\exp(i\pi t)$, $z_0=i\frac{\pi}{2}$. Here $K>1$ and $t\in[0,1]$.

Answer 2

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.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{{1 \over \root{2\pi}} \int_{-\infty}^{\infty}{\expo{-\ic\omega x} \over \cosh\pars{x\root{\pi/2}}}\,\dd x = {1 \over \cosh\pars{\omega\root{\pi/2}}}}:\ {\Large ?}}$.

---

Lets $\ds{\pars{~\root{\pi/2}x \mapsto x/4 \implies x \mapsto \root{2/\pi}x/4~}}$ and $\ds{\nu = \root{2/\pi}\omega/4}$: \begin{align} &\bbox[5px,#ffd]{{1 \over \root{2\pi}} \int_{-\infty}^{\infty}{\expo{-\ic\omega x} \over \cosh\pars{x\root{\pi/2}}}\,\dd x} = \root{2 \over \pi}\,\Re \int_{0}^{\infty}{\expo{-\ic\omega x} \over \cosh\pars{x\root{\pi/2}}}\,\dd x \\[5mm] = &\ {1 \over 2\pi}\,\Re \int_{0}^{\infty}{\expo{-\ic\nu x} \over \cosh\pars{x/4}}\,\dd x = {1 \over \pi}\,\Re\int_{0}^{\infty}{\sinh\pars{x/4} \expo{-\ic\nu x} \over \sinh\pars{x/2}}\,\dd x \\[5mm] = &\ {1 \over \pi}\,\Re\int_{0}^{\infty}{ \expo{-\pars{1/4 + \ic\nu}x} - \expo{-\pars{3/4 + \ic\nu}x}\over 1 - \expo{-x}}\,\dd x \\[5mm] = &\ {1 \over \pi}\,\Re\bracks{\int_{0}^{\infty}{% \expo{-x} - \expo{-\pars{3/4 + \ic\nu}x}\over 1 - \expo{-x}} \,\dd x - \int_{0}^{\infty}{% \expo{-x} - \expo{-\pars{1/4 + \ic\nu}x}\over 1 - \expo{-x}} \,\dd x} \\[5mm] = &\ {1 \over \pi}\,\Re\bracks{% \Psi\pars{{3 \over 4} + \ic\nu} - \Psi\pars{\bracks{{1 \over 4} + \ic\nu}}} \end{align}

where $\ds{\Psi}$ is the Digamma Function. See $\ds{\color{black}{\bf 6.3.22}}$ in A & S Table.

Then, \begin{align} &\bbox[5px,#ffd]{{1 \over \root{2\pi}} \int_{-\infty}^{\infty}{\expo{-\ic\omega x} \over \cosh\pars{x\root{\pi/2}}}\,\dd x} \\[5mm] = &\ {1 \over 2\pi}\bracks{\Psi\pars{{3 \over 4} + \ic\nu} - \Psi\pars{{1 \over 4} - \ic\nu}} \\[2mm] + &\ {1 \over 2\pi}\bracks{\Psi\pars{{3 \over 4} - \ic\nu} - \Psi\pars{{1 \over 4} + \ic\nu}} \\[5mm] = &\ {1 \over 2\pi}\bracks{% \pi\cot\pars{\pi\bracks{{1 \over 4} - \ic\nu}} + \pi\cot\pars{\pi\bracks{{1 \over 4} + \ic\nu}}} \end{align} In the last line I used the Euler Reflection Formula. See $\ds{\color{black}{\bf 6.3.7}}$ in A & S Table.

Therefore, \begin{align} &\bbox[5px,#ffd]{{1 \over \root{2\pi}} \int_{-\infty}^{\infty}{\expo{-\ic\omega x} \over \cosh\pars{x\root{\pi/2}}}\,\dd x} = \Re\cot\pars{\pi\bracks{{1 \over 4} - \ic\nu}} \\[5mm] = &\ \Re\bracks{% -\cot\pars{\ic\pi\nu} - 1 \over -\cot\pars{\ic\pi\nu} + 1} = \Re\bracks{% \ic\coth\pars{\pi\nu} - 1 \over \ic\coth\pars{\pi\nu} + 1} \\[5mm] = &\ -\,{1 - \coth^{2}\pars{\pi\nu} \over \coth^{2}\pars{\pi\nu} + 1} = \on{sech}\pars{2\pi\nu} = \on{sech}\pars{2\pi\root{2 \over \pi}{\omega \over 4}} \\[5mm] = &\ \bbx{1 \over \cosh\pars{\omega\root{\pi/2}}} \\ & \end{align}