Fourier transform of 1/cosh

How do you take the Fourier transform of $$ f(x) = \frac{1}{\cosh x} $$ This is for a complex class so I tried expanding the denominator and calculating a residue by using the rectangular contour that goes from $-\infty$ to $\infty$ along the real axis and $i \pi +\infty$ to $i \pi - \infty$ to close the contour (with vertical sides that go to 0). Therefore, I tried to calculate the residue at $\frac{i \pi}{2}$ of $$ \frac{e^{-ikx}}{e^x + e^{-x}} $$ which will be give me the answer, but I don't know how to do this. Thanks for the help!

Solution 1:

First, let's compute the FT of $\text{sech}{(\pi x)}$, which may be derived using the residue theorem. We simply set up the Fourier integral as usual and comvert it into a sum as follows:

$$\begin{align}\int_{-\infty}^{\infty} dx \, \text{sech}{(\pi x)} \, e^{i k x} &= 2 \int_{-\infty}^{\infty} dx \frac{e^{i k x}}{e^{\pi x}+e^{-\pi x}}\\ &= 2 \int_{-\infty}^0 dx \frac{e^{i k x}}{e^{\pi x}+e^{-\pi x}} + 2 \int_0^{\infty}dx \frac{e^{i k x}}{e^{\pi x}+e^{-\pi x}}\\ &= 2 \sum_{m=0}^{\infty} (-1)^m \left [\int_0^{\infty}dx \, e^{-[(2 m+1) \pi+i k] x} +\int_0^{\infty}dx \, e^{-[(2 m+1) \pi-i k] x} \right ] \\ &= 2 \sum_{m=0}^{\infty} (-1)^m \left [\frac{1}{(2 m+1) \pi-i k} + \frac{1}{(2 m+1) \pi+i k} \right ]\\ &= 4\pi \sum_{m=0}^{\infty} \frac{(-1)^m (2 m+1)}{(2 m+1)^2 \pi^2+k^2}\\ &= \frac{1}{2 \pi}\sum_{m=-\infty}^{\infty} \frac{(-1)^m (2 m+1)}{\left (m+\frac12\right)^2+\left(\frac{k}{2 \pi}\right)^2} \end{align}$$

By the residue theorem, the sum is equal to the negative sum of the residues at the non-integer poles of

$$\pi \csc{(\pi z)} \frac{1}{2 \pi}\frac{2 z+1}{\left ( z+\frac12\right)^2+\left (\frac{k}{2 \pi}\right)^2}$$

which are at $z_{\pm}=-\frac12 \pm i \frac{k}{2 \pi}$. The sum is therefore

$$-\frac12\csc{(\pi z_+)} - \frac12 \csc{(\pi z_-)} = -\Re{\left [\frac{1}{\sin{\pi \left (-\frac12+i \frac{k}{2 \pi}\right )}}\right ]} = \text{sech}{\left ( \frac{k}{2}\right)}$$

By this reasoning, the FT of $\operatorname{sech}{x}$ is $\pi\, \text{sech}{\left ( \frac{\pi k}{2}\right)}$.

Solution 2:

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{{1 \over \cosh\pars{x}} =\int_{-\infty}^{\infty}\tilde{\fermi}\pars{k}\expo{\ic kx} \,{\dd k \over 2\pi}\quad\imp\quad\tilde{\fermi}\pars{k} =\int_{-\infty}^{\infty}{\expo{-\ic k x} \over \cosh\pars{x}}\,\dd x:\ {\large ?}}$

In order to avoid the infinite poles of $\ds{\cosh\pars{x}}$, we can make a suitable change of variables which 'leave us' with just ${\large\tt\ul{one}}$ pole. As an extra bonus, we don't have to sum a serie: \begin{align} \tilde{\fermi}\pars{k}&=2\ \overbrace{% \int_{-\infty}^{\infty}{\expo{-\ic k x} \over \expo{x} + \expo{-x}}\,\dd x} ^{\ds{\mbox{Set}\ t = \expo{x}\ \imp\ x = \ln\pars{t}}} =2\int_{0}^{\infty}{t^{-\ic k} \over t + 1/t}\,{\dd t \over t} =2\int_{0}^{\infty}{t^{-\ic k} \over t^{2} + 1}\,\dd t \\[3mm]&=\ \overbrace{\color{#c00000}{% \int_{0}^{\infty}{t^{-1/2 - \ic k/2} \over t + 1}\,\dd t}} ^{\ds{=\ \tilde{\fermi}\pars{k}}}\ =\ 2\pi\ic\pars{\expo{\ic\pi}}^{-1/2 - \ic k/2} -\int^{0}_{\infty} {t^{-1/2 - \ic k}\pars{\expo{2\pi\ic}}^{-1/2 - \ic k/2} \over t + 1}\,\dd t \\[3mm]&=2\pi\expo{\pi k/2} -\expo{\pi k}\ \overbrace{\color{#c00000}{\int_{0}^{\infty}{t^{-1/2 - \ic k/2} \over t + 1}\,\dd t}} ^{\ds{=\ \tilde{\fermi}\pars{k}}} \ \imp\ \tilde{\fermi}\pars{k}=\pi\,{2\expo{\pi k/2} \over 1 + \expo{\pi k}} =\pi\,{2 \over \expo{-\pi k/2} + \expo{\pi k/2}} \end{align}

$$\color{#00f}{\large% \int_{-\infty}^{\infty}{\expo{-\ic k x} \over \cosh\pars{x}}\,\dd x} = \color{#00f}{\large\pi\sech\pars{{\pi \over 2}\,k}} $$

Solution 3:

I know that this question is rather old, but here is an answer which does not rely on the residue theorem:

Letting $t = (1 + x^\beta)^{-1}$ we find for $0 < \operatorname{Re} \alpha < \beta$ \begin{align} \int \limits_0^\infty \frac{x^{\alpha -1}}{1+x^\beta} \, \mathrm{d} x &= \frac{1}{\beta} \int \limits_0^1 t^{-\frac{\alpha}{\beta}} (1-t)^{\frac{\alpha}{\beta} - 1} \, \mathrm{d} t = \frac{1}{\beta} \operatorname{B} \left(1-\frac{\alpha}{\beta},\frac{\alpha}{\beta}\right) \\ &= \frac{1}{\beta} \Gamma \left(1-\frac{\alpha}{\beta}\right) \Gamma \left(\frac{\alpha}{\beta}\right) = \frac{\pi}{\beta} \csc\left(\frac{\alpha}{\beta} \pi \right) \, . \end{align} Now using Felix Marin's substitution $x = \ln(t)$ we obtain for $k \in \mathbb{R}$ \begin{align} \int \limits_{-\infty}^\infty \frac{\mathrm{e}^{-\mathrm{i} k x}}{\cosh(x)} \, \mathrm{d} x &= 2 \int \limits_0^\infty \frac{t^{-\mathrm{i} k}}{1+t^2} \, \mathrm{d} t = 2 \frac{\pi}{2} \csc \left(\frac{1 - \mathrm{i} k}{2} \pi \right) = \pi \sec \left(\mathrm{i} \frac{\pi}{2} k\right) \\ &= \pi \operatorname{sech} \left( \frac{\pi}{2} k\right) \, . \end{align}