# Calculating the Fourier transform of $\frac{\sinh(kx)}{\sinh(x)}$

I'm trying to compute $$\int_{-\infty}^\infty \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx$$ i.e. the Fourier transform of $x\mapsto \frac{\sinh(kx)}{\sinh(x)}$, where $0<k<1$ is fixed.

But I'm having trouble with it.

Motivation: I'm trying to derive an expression for the solution of the Dirichlet problem $\Delta u = 0$ on the strip $[0,1]\times \mathbb R$ with values on the boundary $f_0, f_1$ (assuming necessary niceness conditions for all functions involved).

For this I took the Fourier transform of the solution $u$ and got a formula for $\hat u$ in terms of $\hat{ f_0}, \hat{ f_1}$ and now I want to transform it back. Doing this led (more or less) to the integral expression: $u(x,y) = \int \frac{\sinh(kx)}{\sinh(k)}e^{-iky} \hat f(k) \ dk$.

Now I'm trying to apply the product formula $\int f \hat g = \int \hat f g$ to get everything in terms of $f$. This is why I'm interested in computing the above integral.

My attempt: (which led nowhere, so you may actually ignore everything below)

I think it should be possible using residues. For this I thought of the path having the following components:

$$[-R,R], \ [R,R+i\pi], \ [R+i\pi, \delta + i\pi],$$ $$\ \text{semicircle from \delta + i\pi to -\delta + i\pi below i\pi},$$ $$[-\delta + i\pi, -R + i\pi], \ [-R+i\pi, -R]$$

with the intention of letting $\delta \to 0$ eventually.

The integrals over the vertical components will vanish for $R\to\infty$, so the integral over the path then becomes

\begin{align} 0 &= \int_{-\infty}^\infty \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx + \left(\int_{\infty}^{\delta} + \int_{-\delta}^{-\infty}\right) \frac{\sinh(k(x+i\pi))}{\sinh(x+i\pi)}e^{-i\omega (x+i\pi)} \ dx \\ & \qquad + \int_{\text{semicircle}} \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx \end{align}

Using $\sinh(a+ib) = \sinh(a)\cos(b) + i \cosh(a)\sin(b)$ for real $a,b$, we get

\begin{align} 0 &= \int_{-\infty}^\infty \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx \\ &\qquad + \left(\int_{\delta}^{\infty} + \int_{-\infty}^{-\delta}\right) \frac{\sinh(kx)\cos(k\pi) + i \cosh(kx)\sin(k\pi)}{\sinh(x)}e^{-i\omega x}e^{\omega \pi} \ dx \\ & \qquad + \int_{\text{semicircle}} \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx \end{align}

The integral over the semicircle should go to $$(-\pi i) \ \mathrm{Res}_{x = \pi i}\left(\frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x}\right) = -\pi \sin(k\pi)e^{\omega \pi}$$ as $\delta \to 0$. Therefore

\begin{align} \pi \sin(k\pi) e^{\omega \pi} &= \int_{-\infty}^\infty \frac{\sinh(kx)(1+\cos(k\pi)e^{\omega \pi}) + i \cosh(kx) \sin(x\pi)e^{\omega \pi}} {\sinh(x)} e^{-i\omega x} \ dx \\ &= (1+\cos(k\pi)e^{\omega \pi}) \int_{-\infty}^\infty \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx \\ & \qquad + i \sin(k\pi)e^{\omega \pi}\int_{-\infty}^\infty \frac{\cosh(kx)}{\sinh(x)}e^{-i\omega x} \ dx \end{align}

I don't see whether this has brought me any closer to my goal?

• I thought $(e^{kx} -e^{-kx})/(e^x - e^{-x})$ might help but nothing yet.
– user13838
Oct 30, 2011 at 13:55
• What happens if you now start over and try to compute the Fourier transform of $\cosh(kx)/\sinh(x)$ using the same contour? Maybe you will get a second relation between the two transforms, which together with the relation you already have will let you solve for both of them. (I haven't tried it, though.) Oct 30, 2011 at 15:10
• I got this horribly messy thing...
– user13838
Oct 30, 2011 at 15:25
• Out of curiosity: is this homework or something you bumped into in an application? Oct 30, 2011 at 15:58
• @J.M. I now have added some motivation above.
– Sam
Oct 30, 2011 at 16:26

The result is doable by method of residues. We complete the integration path by the arc crossing from $+\infty$ to $-\infty$ over the upper-half complex plane. Then $$\begin{eqnarray} \mathcal{F}(\omega, \kappa) &=& \int_{-\infty}^\infty \frac{\sinh(\kappa x)}{\sinh(x)} \mathrm{e}^{i \omega x} \mathrm{d} x = 2 \pi i \sum_{n=1}^\infty \operatorname{Res}_{x = i \pi n} \frac{\sinh(\kappa x)}{\sinh(x)} \mathrm{e}^{i \omega x} \\ &=& \sum_{n=1}^\infty 2 \pi (-1)^{n-1} \mathrm{e}^{-\omega \pi n} \sin(\pi \kappa n) = \frac{2 \pi e^{\pi \omega } \sin (\pi \kappa )}{2 e^{\pi \omega } \cos (\pi \kappa )+e^{2 \pi \omega }+1} \\ &=& \frac{\pi \sin (\pi \kappa )}{\cos (\pi \kappa )+\cosh\left( \pi \omega \right)} \end{eqnarray}$$

• Very nice, sir. =) This contour crossed my mind at first, but I didn't investigate it further, since I just thought "infinite sum" and went on to look for a different one. Thanks a bunch!
– Sam
Oct 30, 2011 at 17:17
• I just managed to derive the expression I wanted. =) Thanks to your help. Great!
– Sam
Oct 30, 2011 at 17:51
• I wonder whether somebody could also post the estimate necessary to show that the integral along the half-circle vanishes? thx, I've been trying for some hours, but I don't get a very clean estimate.
– user55315
Jan 16, 2013 at 23:09
• May I know how to get the closed form of the inifite series?
– John
Dec 15, 2014 at 16:18
• @JohnZHANG It is obtained by writing sine as a linear combination of exponents, summing two geometric series and simplifying the result. Dec 15, 2014 at 17:30

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \begin{align} &\bbox[5px,#ffd]{\left.\int_{-\infty}^{\infty}{\sinh\pars{kx}\over \sinh\pars{x}} \expo{-\ic\omega x} \,\dd x\,\right\vert_{0\ <\ k\ <\ 1}} \\[5mm] = &\ \int_{-\infty}^{\infty}{\sinh\pars{kx}\over \sinh\pars{x}}\cos\pars{\omega x}\,\dd x \\[5mm] = &\ 2\int_{0}^{\infty}{\sinh\pars{kx}\over \sinh\pars{x}}\cos\pars{\omega x}\,\dd x \\[5mm] = &\ 2\,\Re\int_{0}^{\infty}{\sinh\pars{kx}\over \sinh\pars{x}}\expo{-\ic\omega x}\,\dd x \\[5mm] = &\ 2\,\Re\int_{0}^{\infty} {\expo{-\pars{-k + 1 + \ic\omega}x}\,\,\, - \expo{-\pars{k + 1 + \ic\omega}x}\,\,\over 1 - \expo{-2x}}\,\dd x \\[5mm] \stackrel{2x\ \mapsto\ x}{=}\,\,\,&\ \Re\int_{0}^{\infty} {\expo{-\pars{-k/2 + 1/2 + \ic\omega/2}x}\,\,\,\,\,\, - \expo{-\pars{k/2 + 1/2 + \ic\omega/2}x}\,\,\, \over 1 - \expo{-x}}\,\dd x \\[5mm] = &\ \Re\left[\int_{0}^{\infty} {\expo{-x} - \expo{-\pars{k/2 + 1/2 + \ic\omega/2}x}\,\,\,\,\, \over 1 - \expo{-x}}\,\dd x -\right. \\[2mm] &\ \left.\phantom{\Re\left[\right.}\int_{0}^{\infty} {\expo{-x} - \expo{-\pars{-k/2 + 1/2 + \ic\omega/2}x}\,\,\,\,\, \over 1 - \expo{-x}}\,\dd x\right] \\[5mm] = &\ \Re\bracks{\Psi\pars{{1 \over 2} + {1 \over 2}\omega\ic + {1 \over 2}\,k} - \Psi\pars{{1 \over 2} + {1 \over 2}\omega\ic - {1 \over 2}\,k}} \\[5mm] = &\ {1 \over 2}\,\Psi\pars{{1 \over 2} + {1 \over 2}\omega\ic + {1 \over 2}\,k} - {1 \over 2}\,\Psi\pars{{1 \over 2} - {1 \over 2}\omega\ic - {1 \over 2}\,k} \\[2mm] + &\ {1 \over 2}\,\Psi\pars{{1 \over 2} - {1 \over 2}\omega\ic + {1 \over 2}\,k} - {1 \over 2}\,\Psi\pars{{1 \over 2} + {1 \over 2}\omega\ic - {1 \over 2}\,k} \\[5mm] = &\ {1 \over 2}\,\pi\cot\pars{\pi\bracks{{1 \over 2} - {1 \over 2}\omega\ic - {1 \over 2}\,k}} \\[2mm] + &\ {1 \over 2}\,\pi\cot\pars{\pi\bracks{{1 \over 2} + {1 \over 2}\omega\ic - {1 \over 2}\,k}} \\[5mm] = &\ {1 \over 2}\,\pi\tan\pars{{1 \over 2}\pi\,k + {1 \over 2}\pi\omega\ic} + {1 \over 2}\,\pi\tan\pars{{1 \over 2}\pi\,k - {1 \over 2}\pi\omega\ic} \\[5mm] = &\ \pi\,\Re\tan\pars{{1 \over 2}\pi\,k + {1 \over 2}\pi\omega\ic} \\[5mm] = &\ \bbx{\pi\sin\pars{\pi k} \over \cos\pars{\pi k} + \cosh\pars{\pi\omega}} \\ & \end{align}