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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 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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    @Sam Oh, sorry. It somehow deletes the `*` characters from the last `Iwx` term. If you make it again `I*w*x` it gives the horrible mess, if you are still interested.2011-10-31

1 Answers 1

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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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    @JohnZHANG It is obtained by writing sine as a linear combination of exponents, summing two geometric series and simplifying the result.2014-12-15