# Problem of the Week #285 - Feb 01, 2019

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#### Euge

##### MHB Global Moderator
Staff member
Here is this week's POTW:

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Evaluate the integral $$\int_0^\infty \frac{\cos(ax)}{\cosh b + \cosh x}\, dx$$ where $a$ is real and $b > 0$.

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#### Euge

##### MHB Global Moderator
Staff member
No one answered this week's problem. You can read my solution below.

Let $r > 0$ such that $\sinh r > 2\cosh b$ and consider the contour integral
$$\oint_{\Gamma(R)} \frac{e^{iaz}}{\cosh b + \cosh z}\, dz$$ where $\Gamma(R)$ is the positively oriented rectangle with vertices at $\pm R, \pm R + 2\pi i$. The integrand, call it $f(z)$ for short, has simple poles at $z = \pm b + \pi i$ lying inside $\Gamma(R)$, so by the residue theorem the integral equals $$2\pi i [\operatorname{Res}_{z = b + \pi i} f(z) + \operatorname{Res}_{z = -b + \pi i} f(z)] = 2\pi i\left[-\frac{e^{iab}e^{-\pi a}}{\sinh b} + \frac{e^{-iab}e^{-\pi a}}{\sinh b} \right] = 4e^{-\pi b}\frac{\pi \sin ab}{\sinh b}$$ The integral of $f(z)$ along the vertical edges of $\Gamma(R)$ is $O(1/\sinh R)$ as $R \to \infty$ since, given $R > r$ and $z$ on a vertical edge, $$\lvert f(z)\rvert \le \frac{e^{-a\operatorname{Im}(z)}}{\sinh R - \cosh b} \le \frac{2}{\sinh R}$$ The sum of the integrals of $f$ along the top and bottom edges is $(1 - e^{-2\pi a})\int_{-R}^R f(x)\, dx$, so in the limit as $R \to \infty$, $$(1 - e^{-2\pi a})\int_{-\infty}^\infty f(x)\, dx = 4e^{-\pi a}\frac{\pi \sin ab}{\sinh b}$$ Taking the real part and exploiting symmetry we deduce

$$\int_0^\infty \frac{\cos ax}{\cosh b + \cosh x}\, dx = \frac{\pi \sin ab}{\sinh \pi a \sinh b}$$

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