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I'm not quite sure how to do this one. I've found a simple pole at $i\pi$ and thus the residue $\frac{p(i\pi)}{q'(i\pi)} = \frac{i\pi}{\sinh'{i\pi}} = \frac{i\pi}{\cosh{i\pi}} = \frac{i\pi}{-1} = -i\pi$ as well. When I go to integrate the function, my work so far is $$I' = \int_{-\infty}^{\infty}\frac{x}{\sinh x}dx+\int_{\infty}^{\pi i +\infty}\frac{z}{\sinh z}dz - \int_{-\infty}^{\infty}\frac{x+i\pi}{\sinh{(x+i\pi)}}dx + \int_{-\infty}^{\pi i -\infty}\frac{z}{\sinh z}dz$$ but I am not sure how to evaluate the contour from here. Any help is appreciated!

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    $\begingroup$ there are a lot more poles than that, e.g. $2\pi i$, $-i\pi.$ $\endgroup$ Feb 9 '17 at 4:57
  • $\begingroup$ Are you required to use contour integration for this problem ? $\endgroup$ Feb 9 '17 at 5:18
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    $\begingroup$ Yes I'm required to use contour integration, and I realise there are other poles. However I'm trying to integrate around a rectangle from $-\infty$ to $\infty$ and 0 to $\pi$. I didn't specify that earlier. $\endgroup$
    – John Page
    Feb 9 '17 at 5:41
  • $\begingroup$ @JohnPage Should have picked that up from the question, sorry. $\endgroup$ Feb 9 '17 at 6:09
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If you do the rectangle you specified and round below the pole in a small semicircle with radius $\epsilon$, the contribution from the short sides of the rectangle is zero since $\sinh$ grows exponentially as $x\to\pm\infty .$ Then by Cauchy you have $$ 0 = \int_{-\infty}^\infty \frac{x}{\sinh(x)}dx - \int_{\epsilon}^\infty\frac{x+i\pi}{\sinh(x+i\pi)}dx -\int_{C_\epsilon} \frac{z}{\sinh(z)}dz - \int_{-\infty}^{-\epsilon} \frac{x+i\pi}{\sinh(x+i\pi)}dx $$ where $C_\epsilon$ is counter-clockwise half semicircle under the pole at $i\pi.$

The contribution from the semi-circle is, in the limit $\epsilon \to 0$, $$ \int_\pi^{2\pi} \frac{\epsilon e^{it}+i\pi}{\sinh(\epsilon e^{it}+i\pi)}i\epsilon e^{it}dt \to \int_{\pi}^{2\pi} \pi dt = \pi^2$$ where we expanded $\sinh(x+i\pi) \approx -x$ near $x=i\pi.$

Thus we have $$ \pi^2 = \int_{-\infty}^\infty \frac{x}{\sinh(x)}dx - \lim_{\epsilon\to 0}\left(\int_{-\infty}^{-\epsilon} \frac{x+i\pi}{\sinh(x+i\pi)}dx+\int_{\epsilon}^\infty\frac{x+i\pi}{\sinh(x+i\pi)}dx\right) \\ =\int_{-\infty}^\infty \frac{x}{\sinh(x)}dx -P\int_{-\infty}^\infty\frac{x+i\pi}{\sinh(x+i\pi)}.$$ where $P$ denotes the principal part.

Since $$P\int_{-\infty}^\infty \frac{1}{\sinh(x)}dx = 0$$ (since $\sinh$ is odd), $$P\int_{-\infty}^\infty \frac{x}{\sinh(x)}dx = \int_{-\infty}^\infty \frac{x}{\sinh(x)}dx$$ (since $x/\sinh(x)$ is regular at the origin), and $$\sinh(x+i\pi) = -\sinh(x),$$ we have $$ P\int_{-\infty}^\infty\frac{x+i\pi}{\sinh(x+i\pi)} = -\int_{-\infty}^\infty \frac{x}{\sinh(x)}dx - i\pi P\int_{-\infty}^\infty \frac{1}{\sinh(x)}dx \\ =-\int_{-\infty}^\infty \frac{x}{\sinh(x)}dx.$$

Plugging this in, $$\pi^2 = 2\int_{-\infty}^\infty \frac{x}{\sinh(x)}dx$$ and since the integrand's even, $$ \int_0^\infty \frac{x}{\sinh(x)}dx = \frac{\pi^2}{4}.$$

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