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I’m trying to evaluate the following integral: $$\int_{0}^{\infty} \frac{4\pi}{16\pi^2 + x^2} \left(\frac{1}{x}+\frac{1}{e^{-x}-1}\right) \, dx$$ I’ve tried using contour integration by using a quarter-circle contour and going around the pole at $z=4\pi i$ with a semi-circular arc, however, I wasn’t able to evaluate the integrals along the imaginary axis.

I wasn’t able to come up with a real or complex method for evaluating this integral, so any help would be appreciated. I’m not sure if a closed-form exists.

KStarGamer
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    When $\alpha \notin 2\pi\mathbb{Z}$, we get $$\int_{0}^{\infty} \frac{\alpha}{x^2+\alpha^2}\left(\frac{1}{x}+\frac{1}{e^{-x}-1}\right),\mathrm{d}x=\sum_{n=1}^{\infty}\frac{2\alpha}{\alpha^2-4\pi^2n^2}\log(2\pi n)+\frac{\log\alpha}{\alpha}-\frac{\log\alpha}{2}\cot\left(\frac{\alpha}{2}\right)-\frac{\pi}{4}.$$ So the desired integral can be realizes as the limit as $\alpha\to4\pi$. – Sangchul Lee Dec 25 '21 at 14:26

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I'm afraid there is no a closed form for this integral. $$I=\int_0^\infty \frac{4\pi}{16\pi^2 + x^2} (\frac{1}{x}+\frac{1}{e^{-x}-1}) dx=4\pi\int_0^\infty \frac{dx}{16\pi^2 + x^2} (\frac{1}{x}+\frac{1}{2}\coth\frac{x}{2}-\frac{1}{2}) dx$$ $$=-\frac{\pi}{4}+\pi\int_0^\infty \frac{dt}{4\pi^2 + t^2}\Big(\frac{1}{t}-\coth t\Big)=-\frac{\pi}{4}-\pi\int_0^\infty \frac{dt}{4\pi^2 + t^2}\frac{d}{dt}\Big(\ln\frac{\sinh t}{t}\Big)$$ Using the representation $\,\,\ln\frac{\sinh t}{t}=\sum_{k=1}^\infty\ln\Big(1+\frac{t^2}{\pi^2k^2}\Big)$ and making the change $x=t^2$ $$I=-\frac{\pi}{4}-\pi\int_0^\infty\frac{dx}{4\pi^2+x}\sum_{k=1}^\infty\frac{1}{\pi^2k^2+x}=-\frac{\pi}{4}-\pi\lim_{N\to\infty}\int_0^N\frac{dx}{4\pi^2+x}\sum_{k=1}^\infty\frac{1}{\pi^2k^2+x}$$ $$=-\frac{\pi}{4}-\pi\lim_{N\to\infty}\sum_{k=1}^\infty\int_0^N\frac{dx}{4\pi^2+x}\frac{1}{\pi^2k^2+x}$$ $$=-\frac{\pi}{4}-\pi\lim_{N\to\infty}\int_0^Ndx\bigg(\frac{1}{3\pi^2}\Big(\frac{1}{\pi^2+x}-\frac{1}{4\pi^2+x}\Big)+\frac{1}{(4\pi^2+x)^2}\bigg)$$ $$-\pi\lim_{N\to\infty}\sum_{k=3}^\infty\int_0^N\frac{dx}{\pi^2(k^2-4)}\Big(\frac{1}{4\pi^2+x}-\frac{1}{\pi^2k^2+x}\Big)$$ Integrating and taking the limit $$=-\frac{\pi}{4}-\frac{2\ln 2}{3\pi}-\frac{1}{4\pi}-\pi \sum_{k=3}^\infty\frac{1}{\pi^2(k^2-4)}\ln\frac{k^2}{4}$$ $$I(n)=-\frac{\pi}{4}+\frac{3\ln 2}{8\pi}-\frac{1}{4\pi}-\frac{2}{\pi}\sum_{k=3}^\infty\frac{\ln k}{k^2-4}\approx-1.36824...$$ It is not clear whether a closed form for the sum $\sum_{k=3}^\infty\frac{\ln k}{k^2-4}$ exists.

Svyatoslav
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