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 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?
Calculating the Fourier transform of $\frac{\sinh(kx)}{\sinh(x)}$
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0I thought $(e^{kx} -e^{-kx})/(e^x - e^{-x})$ might help but nothing yet. – 2011-10-30
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0What 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.) – 2011-10-30
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0I got this [horribly messy thing](http://www.wolframalpha.com/input/?i=Integrate+%28Exp%5Bkx%5D+-+Exp%5B-kx%5D%29%2F%28Exp%5Bx%5D+-+Exp%5B-x%5D%29Exp%5B-I*w*x%5Ddx)... – 2011-10-30
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0Out of curiosity: is this homework or something you bumped into in an application? – 2011-10-30
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1@J.M. I now have added some motivation above. – 2011-10-30
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0@HansLundmark: Thanks for your suggestion. I will try that! Sounds like it might work. =) – 2011-10-30
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0@percusse: Wolframalpha unfortunately doesn't show me anything under your link. – 2011-10-30
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0@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
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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0Very 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! – 2011-10-30
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0I just managed to derive the expression I wanted. =) Thanks to your help. Great! – 2011-10-30
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1I 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. – 2013-01-16
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0May I know how to get the closed form of the inifite series? – 2014-12-15
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0@JohnZHANG It is obtained by writing sine as a linear combination of exponents, summing two geometric series and simplifying the result. – 2014-12-15