Fourier transform of $1/\sinh^n(x-\mathrm{i}\epsilon)$ I want to calculate
\begin{equation*}
\int_{-\infty}^{+\infty} \text{d}x \frac{ e^{-\mathrm{i}\omega x} }{ \left( \sinh{( x-\mathrm{i}\epsilon )} \right)^n }
\end{equation*}
with $n\in\text{N}$ and $\epsilon$ as an infinitesimal regularization factor. I tried several tricks and as a last resource I would try to Laurent-expand the $(\sinh(x))^{-n}$ function, but also this seems not possible. How can I calculate this integral?
Edit:
I may have found a way thanks to this article appendix D, but it's strongly non-trivial. I write it here because it can be useful to someone. First there is to do a substitution $x-\mathrm{i}\epsilon\to x$ and after that there is to consider the following path 
inside which there is no singularity. The edges contributions are null for $r\to\pm\infty$ and we obtain
\begin{gather*}
e^{\omega\epsilon}
\int_{-\infty-\mathrm{i}\epsilon}^{+\infty-\mathrm{i}\epsilon} \text{d}x \frac{ e^{-\mathrm{i}\omega x} }{ \sinh^n{x} }
=
e^{\omega\epsilon}
\int_{-\infty - \mathrm{i} \frac{\pi}{2} }^{+\infty - \mathrm{i} \frac{\pi}{2}} \text{d}x \frac{ e^{-\mathrm{i}\omega x} }{ \sinh^n{x} }
\end{gather*}
Regularization parameter was useful to avoid any singularity on the path, but now is not useful anymore and we can put $\epsilon\to 0^+$, following variable substitution $x+\mathrm{i}\pi/2\to x$ leaving us with
\begin{gather*}
\mathrm{i}^n
e^{-\frac{\pi}{2}\omega}
\int_{-\infty}^{+\infty} \text{d}x \frac{ e^{-\mathrm{i}\omega x} }{ \cosh^n{x} }
\end{gather*}
that with variabile change $e^{2x}\to x$ becomes
\begin{gather*}
\mathrm{i}^n 2^{n-1}
e^{-\frac{\pi}{2}\omega}
B \left( \frac{n-\mathrm{i}\omega}{2}, \frac{n+\mathrm{i}\omega}{2} \right)
\end{gather*}
and we can write
\begin{gather*}
B \left( \frac{n-\mathrm{i}\omega}{2}, \frac{n+\mathrm{i}\omega}{2} \right)
=
\left( \Gamma (n) \right)^{-1}
\Gamma \left( \frac{n-\mathrm{i}\omega}{2} \right) 
\Gamma \left( \frac{n+\mathrm{i}\omega}{2} \right)
\end{gather*}
Now, using just basic properties of the gamma function, it is not too difficult to show that
\begin{gather*}
\begin{cases}
\Gamma \left( \frac{n-\mathrm{i}\omega}{2} \right) 
\Gamma \left( \frac{n+\mathrm{i}\omega}{2} \right)
=
\frac{\pi\omega}{\left( e^{\frac{\pi\omega}{2}} - e^{-\frac{\pi\omega}{2}} \right) }
\prod_{k=1}^{\frac{n}{2}-1} \left( \left( \frac{n}{2} - k \right)^2 + \frac{\omega^2}{4} \right)
& n \, \text{even}
\\
\Gamma \left( \frac{n-\mathrm{i}\omega}{2} \right) 
\Gamma \left( \frac{n+\mathrm{i}\omega}{2} \right)
=
\frac{2\pi}{e^{\frac{\pi\omega}{2}} + e^{-\frac{\pi\omega}{2}} }
\prod_{k=1}^{\frac{n-1}{2}} \left( \left( \frac{n}{2} - k \right)^2 + \frac{\omega^2}{4} \right)
& n \, \text{odd}
\end{cases}
\end{gather*}
So that, in the end
\begin{equation*}
\lim_{\epsilon\to 0^+}
\int_{-\infty}^{+\infty} \text{d}x \frac{ e^{-\mathrm{i}\omega x} }{ \left( \sinh{( x-\mathrm{i}\epsilon )} \right)^n }
=
\begin{cases}
\frac{ \mathrm{i}^n 2^{n-1} }{\Gamma(n)}
\frac{\pi\omega}{\left( e^{\pi\omega} - 1 \right) }
\prod_{k=1}^{\frac{n}{2}-1} \left( \left( \frac{n}{2} - k \right)^2 + \frac{\omega^2}{4} \right)
& n \, \text{even}
\\
\frac{ \mathrm{i}^n 2^{n-1} }{\Gamma(n)}
\frac{2\pi}{e^{\pi\omega} + 1 }
\prod_{k=1}^{\frac{n-1}{2}} \left( \left( \frac{n}{2} - k \right)^2 + \frac{\omega^2}{4} \right)
& n \, \text{odd}
\end{cases}
\end{equation*}

Hint $$\sinh(ix)=i\sin (x)$$
and then break down $\sin^n$. Not so difficult.
