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?


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 enter image description here 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*}

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    $\begingroup$ This doesn't work for $n=1$. $\endgroup$
    – Mark Viola
    Nov 30, 2022 at 17:12
  • $\begingroup$ @MarkViola I think that in that case just get rid of the infinite product. Also in the $n=2$ case there could be a problem, because $k$ in the product goes from $1$ to $0$, but if you solve explicitely you find $-2\pi\omega/(e^{\pi\omega}-1)$, that is exactly the term in front of the infinite product; I think that with also $n=1$ is the same $\endgroup$
    – Rob Tan
    Nov 30, 2022 at 17:52


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