Calculating the Fourier transform of $\frac{\sinh(kx)}{\sinh(x)}$ 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?
 A: 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
$$ \begin{eqnarray}
  \mathcal{F}(\omega, \kappa) &=& \int_{-\infty}^\infty \frac{\sinh(\kappa x)}{\sinh(x)} \mathrm{e}^{i \omega x} \mathrm{d} x = 2 \pi i \sum_{n=1}^\infty \operatorname{Res}_{x = i \pi n} \frac{\sinh(\kappa x)}{\sinh(x)} \mathrm{e}^{i \omega x}  \\
  &=& \sum_{n=1}^\infty 2 \pi (-1)^{n-1} \mathrm{e}^{-\omega \pi n} \sin(\pi \kappa n) =  
    \frac{2 \pi  e^{\pi  \omega } \sin (\pi  \kappa )}{2 e^{\pi  \omega } \cos (\pi 
   \kappa )+e^{2 \pi  \omega }+1}
 \\ &=& 
    \frac{\pi   \sin (\pi  \kappa )}{\cos (\pi 
   \kappa )+\cosh\left( \pi  \omega \right)} \end{eqnarray}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
