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So I have the following integral I need to evaluate in the complex plane:

$$\int_{-\infty}^\infty \frac{\sinh(az)}{\sinh(\pi z)}~\mathrm{d}z.$$

This equation will have poles when $z = ni$, so any integer along the imaginary axis. Any tips on how to approach this? Thanks in advance.

Lorago
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Jack
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  • You should compute the residue of the integrand, as done in the page linked by Lorago. Nevertheless, you should be careful with the considered values of $a$, because it may cancel some poles (in the case $a = \pi$ for instance). – Abezhiko Nov 05 '23 at 08:28

1 Answers1

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If you want to evaluate the integral by means of complex integration, the simpler contour can also be used. First, note that $$I_0=\int_{-\infty}^\infty \frac{\sinh(az)}{\sinh(\pi z)}~\mathrm{d}z=\frac12\left(\int_{-\infty}^\infty \frac{e^{az}}{\sinh(\pi z)}~\mathrm{d}z-\int_{-\infty}^\infty \frac{e^{-az}}{\sinh(\pi z)}~\mathrm{d}z\right)=\frac12\left(I(a)-I(-a)\right)$$ where $I(a)$ is evaluated in the principal value sense.

Now let's consider the following closed rectangular contour $C$: $\,\,-R\to R\to R+i\to-R+i\to-R\,\,(R\to\infty)$, with added two small arches around $z=0$ and $z=i$ (clockwise). Denoting the integrals along these arches as $I_{C_1}$ and $I_{C_2}$ $$\oint_C\frac{e^{az}}{\sinh(\pi z)}~\mathrm{d}z=I(a)+I_{C_1}+I(a)e^{ia}+I_{C_2}=0$$ because there are no poles inside the contour. $$I(a)\big(1+e^{ia}\big)=-I_{C_1}-I_{C_1}=\pi i\Big(\underset{z=0}{\operatorname{Res}}\frac{e^{az}}{\sinh(\pi z)}+\underset{z=i}{\operatorname{Res}}\frac{e^{az}}{\sinh(\pi z)}\Big)$$ $$I(a)=i\frac{1-e^{ia}}{1+e^{ia}}=\tan\frac a2$$ $$I_0=\frac12\left(I(a)-I(-a)\right)=\tan\frac a2$$

Svyatoslav
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