Calculating the Fourier transform of $\frac{\sinh(kx)}{\sinh(x)}$

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 $$ \begin{eqnarray} \mathcal{F}(\omega, \kappa) &=& \int_{-\infty}^\infty \frac{\sinh(\kappa x)}{\sinh(x)} \mathrm{e}^{i \omega x} \mathrm{d} x = 2 \pi i \sum_{n=1}^\infty \operatorname{Res}_{x = i \pi n} \frac{\sinh(\kappa x)}{\sinh(x)} \mathrm{e}^{i \omega x} \\ &=& \sum_{n=1}^\infty 2 \pi (-1)^{n-1} \mathrm{e}^{-\omega \pi n} \sin(\pi \kappa n) = \frac{2 \pi e^{\pi \omega } \sin (\pi \kappa )}{2 e^{\pi \omega } \cos (\pi \kappa )+e^{2 \pi \omega }+1} \\ &=& \frac{\pi \sin (\pi \kappa )}{\cos (\pi \kappa )+\cosh\left( \pi \omega \right)} \end{eqnarray} $$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}