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?