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I am trying to evaluate the following integral: $$\int_0^\infty \frac{\frac{1}{x}-\pi\coth(\pi x)}{x^2+4}dx$$ I'm not sure if a closed-form exists, so far I only know the decimal approximation to be $\approx$-1.83105. I don't think standard complex analysis methods would work, as the integrand is odd, any other insights would be much appreciated.

qncubed3
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    I've made partial progress. There's a seemingly well-known expansion for $(1/x)-\pi\coth(\pi x) = -2x\sum_{n\ge1}\frac{1}{x^2+n^2}$. (See, e.g., Gradeshteyn 1.421.4 or https://math.stackexchange.com/a/2325127/449818.) One may readily evaluate $\int_0^\infty-2x\frac{1}{x^2+a^2}\frac{1}{x^2+n^2} = 2\ln(n/a)/(a^2-n^2)$. One may readily perform the sum $\sum_{n\ge1}-2\ln(a)/(a^2-n^2) = \ln(a)(1-a\pi\cot(\pi a))/a^2$, but I'm at a loss for evaluating $\sum_{n\ge1}2\ln(n)/(a^2-n^2)$. – astronautgravity Dec 27 '21 at 01:11
  • Hmm. It seems that the method I suggested is a dead end; the sum I couldn't evaluate doesn't seem to have a closed form, although it can be put into a rapidly converging sum: https://math.stackexchange.com/q/4123446/449818 – astronautgravity Dec 27 '21 at 03:21
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    @WAH. I agree with you since we end with $$\frac{3 \log (2)-2}{8} -2 \sum _{n=3}^{\infty }\frac{\log (n)}{n^2-4}$$ and the summation has not been solved yet (as far as I know) – Claude Leibovici Dec 27 '21 at 03:41
  • @ClaudeLeibovici: Seems like an opportunity to define a new special function, say $HL'(s,z) \equiv \sum_{n=0}^\infty\log(n)/(n^s+z^s)$. – astronautgravity Dec 27 '21 at 06:24
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    @WAH. It is surprisng that you don't know it : it is very recent but it is known as the Wah's function. Cheers :-) – Claude Leibovici Dec 27 '21 at 06:29

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