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?

  • $\begingroup$ I thought $(e^{kx} -e^{-kx})/(e^x - e^{-x})$ might help but nothing yet. $\endgroup$ – user13838 Oct 30 '11 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.) $\endgroup$ – Hans Lundmark Oct 30 '11 at 15:10
  • $\begingroup$ I got this horribly messy thing... $\endgroup$ – user13838 Oct 30 '11 at 15:25
  • $\begingroup$ Out of curiosity: is this homework or something you bumped into in an application? $\endgroup$ – J. M. is a poor mathematician Oct 30 '11 at 15:58
  • 1
    $\begingroup$ @J.M. I now have added some motivation above. $\endgroup$ – Sam Oct 30 '11 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 $$ \begin{eqnarray} \mathcal{F}(\omega, \kappa) &=& \int_{-\infty}^\infty \frac{\sinh(\kappa x)}{\sinh(x)} \mathrm{e}^{i \omega x} \mathrm{d} x = 2 \pi i \sum_{n=1}^\infty \operatorname{Res}_{x = i \pi n} \frac{\sinh(\kappa x)}{\sinh(x)} \mathrm{e}^{i \omega x} \\ &=& \sum_{n=1}^\infty 2 \pi (-1)^{n-1} \mathrm{e}^{-\omega \pi n} \sin(\pi \kappa n) = \frac{2 \pi e^{\pi \omega } \sin (\pi \kappa )}{2 e^{\pi \omega } \cos (\pi \kappa )+e^{2 \pi \omega }+1} \\ &=& \frac{\pi \sin (\pi \kappa )}{\cos (\pi \kappa )+\cosh\left( \pi \omega \right)} \end{eqnarray} $$

  • $\begingroup$ 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 '11 at 17:17
  • $\begingroup$ I just managed to derive the expression I wanted. =) Thanks to your help. Great! $\endgroup$ – Sam Oct 30 '11 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 '13 at 23:09
  • $\begingroup$ May I know how to get the closed form of the inifite series? $\endgroup$ – John Dec 15 '14 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 '14 at 17:30

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