Residue theorem

Integral involving the Residue theorem

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— 級数bot (@infseriesbot) June 13, 2021

Today we present here the proof of this nice integral. The solution involves Complex Analysis, geometric series, even functions and trigonometry \begin{align*} \int_{0}^{\infty} \frac{\cos \sqrt{x} }{e^{2\pi \sqrt{x}}-1}dx = & \int_{0}^{\infty}\left[ \sum_{n=1}^{\infty} e^{-2\pi n \sqrt{x}} \cos \sqrt{x}\right] dx\\ =& \sum_{n=1}^{\infty} \int_{0}^{\infty}e^{-2\pi n \sqrt{x}} \cos \sqrt{x} dx \\ =& \sum_{n=1}^{\infty} \int_{0}^{\infty}e^{-2\pi n \sqrt{x}} \left(\frac{e^{i\sqrt{x}}+e^{-i\sqrt{x}}}{2}\right) dx \\ =& \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{e^{\sqrt{x}(i-2\pi n)}+e^{\sqrt{x}(-i-2\pi n)}}{2} dx \\ =& \frac{1}{2}\sum_{n=1}^{\infty}\int_{0}^{\infty}e^{\sqrt{x}(i-2\pi n)}dx+\frac{1}{2}\sum_{n=1}^{\infty}\int_{0}^{\infty}e^{\sqrt{x}(-i-2\pi n)}dx\\ =& \frac{1}{2}\sum_{n=1}^{\infty} \tfrac{e^{\sqrt{x}(i-2\pi n)}(-1+(i-2\pi n) \sqrt{x})}{(i-2\pi n)^2} \Big|_0^\infty+\frac{1}{2}\sum_{n=1}^{\infty} \tfrac{e^{\sqrt{x}(-i-2\pi n)}(-1+(-i-2\pi n) \sqrt{x})}{(-i-2\pi n)^2}\Big|_{0}^{\infty}\\ =& \sum_{n=1}^{\infty} \frac{1}{(i-2\pi n)^2}+ \sum_{n=1}^{\infty} \frac{1}{(-i-2\pi n)^2}\\ =& \sum_{n=1}^{\infty} \left[\frac{1}{(i-2\pi n)^2}+\frac{1}{(-i-2\pi n)^2}\right]\\ =& 2+\sum_{n=0}^{\infty} \left[\frac{1}{(i-2\pi n)^2}+\frac{1}{(-i-2\pi n)^2}\right]\\ =& 1+\frac{1}{2}\sum_{n=-\infty}^{\infty} \frac{1}{(i-2\pi n)^2} + \frac{1}{2}\sum_{n=-\infty}^{\infty} \frac{1}{(-i-2\pi n)^2}\\ =& 1- \frac{1}{2}\operatorname{Res}\left(\frac{\pi \cot \pi z}{(i-2\pi z)^2}, \frac{i}{2\pi} \right)- \frac{1}{2}\operatorname{Res}\left(\frac{\pi \cot \pi}{(-i-2\pi z)^2}, -\frac{i}{2\pi} \right)\\ =&1- \frac{1}{2}\lim_{z\to \frac{i}{2\pi}} \left[\frac{d}{dz} \frac{(z-\frac{i}{2\pi})^2(\pi \cot \pi z)}{(i-2\pi z)^2}\right]- \frac{1}{2}\lim_{z\to -\frac{i}{2\pi}} \left[\frac{d}{dz} \frac{(z+\frac{i}{2\pi})^2(\pi \cot \pi z)}{(-i-2\pi z)^2}\right]\\ =&1- \frac{1}{8\pi}\lim_{z\to \frac{i}{2\pi}} \frac{d}{dz}\cot \pi z- \frac{1}{8\pi}\lim_{z\to -\frac{i}{2\pi}} \frac{d}{dz}\cot \pi z\\ =&1+ \frac{1}{8}\csc^2 \left(\frac{i}{2}\right) + \frac{1}{8}\csc^2\left( -\frac{i}{2}\right)\\ =& 1-\frac{1}{8}\left( \frac{2i}{e^{-\frac{1}{2}}-e^{\frac{1}{2}}}\right)^2 -\frac{1}{8}\left( \frac{2i}{e^{\frac{1}{2}}-e^{-\frac{1}{2}}}\right)^2 \\ =& 1- \frac{e}{( e-1)^2} \end{align*}