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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?

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  • $\begingroup$ I thought $(e^{kx} -e^{-kx})/(e^x - e^{-x})$ might help but nothing yet. $\endgroup$
    – user13838
    Oct 30, 2011 at 13:55
  • $\begingroup$ 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.) $\endgroup$ Oct 30, 2011 at 15:10
  • $\begingroup$ I got this horribly messy thing... $\endgroup$
    – user13838
    Oct 30, 2011 at 15:25
  • $\begingroup$ Out of curiosity: is this homework or something you bumped into in an application? $\endgroup$ Oct 30, 2011 at 15:58
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    $\begingroup$ @J.M. I now have added some motivation above. $\endgroup$
    – Sam
    Oct 30, 2011 at 16:26

2 Answers 2

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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 $$ \begin{eqnarray} \mathcal{F}(\omega, \kappa) &=& \int_{-\infty}^\infty \frac{\sinh(\kappa x)}{\sinh(x)} \mathrm{e}^{i \omega x} \mathrm{d} x = 2 \pi i \sum_{n=1}^\infty \operatorname{Res}_{x = i \pi n} \frac{\sinh(\kappa x)}{\sinh(x)} \mathrm{e}^{i \omega x} \\ &=& \sum_{n=1}^\infty 2 \pi (-1)^{n-1} \mathrm{e}^{-\omega \pi n} \sin(\pi \kappa n) = \frac{2 \pi e^{\pi \omega } \sin (\pi \kappa )}{2 e^{\pi \omega } \cos (\pi \kappa )+e^{2 \pi \omega }+1} \\ &=& \frac{\pi \sin (\pi \kappa )}{\cos (\pi \kappa )+\cosh\left( \pi \omega \right)} \end{eqnarray} $$

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  • $\begingroup$ 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! $\endgroup$
    – Sam
    Oct 30, 2011 at 17:17
  • $\begingroup$ I just managed to derive the expression I wanted. =) Thanks to your help. Great! $\endgroup$
    – Sam
    Oct 30, 2011 at 17:51
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    $\begingroup$ 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. $\endgroup$
    – user55315
    Jan 16, 2013 at 23:09
  • $\begingroup$ May I know how to get the closed form of the inifite series? $\endgroup$
    – John
    Dec 15, 2014 at 16:18
  • $\begingroup$ @JohnZHANG It is obtained by writing sine as a linear combination of exponents, summing two geometric series and simplifying the result. $\endgroup$
    – Sasha
    Dec 15, 2014 at 17:30
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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