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Evaluating $$\int_0^\infty \left( \frac{x}{e^x-e^{-x}}-\frac{1}{2} \right) \frac{dx}{x^2}$$

I tried to calculate it by Mathematica, but it failed to give me an answer.

Then I got interested in this problem because it actually can be well evaluated.

My attempt

Put $$ g(x)=\frac{1}{x}\left(\frac{1}{e^x-1}-\frac{1}{x}+\frac{1}{2}e^{-x}\right) $$

Considering that $$ \left( \frac{x}{e^x-e^{-x}}-\frac{1}{2} \right) \frac{1}{x^2} = -\frac{1}{2x}\left(e^{-x}-e^{-2x}\right) + g(x)-2g(2x) $$

and $$ \int_0^\infty g(x) dx = 2 \int_0^\infty g(2x) dx $$

thus via Frullani's integral we have $$ \int_0^\infty \left( \frac{x}{e^x-e^{-x}}-\frac{1}{2} \right) \frac{dx}{x^2} = -\frac{1}{2} \int_0^\infty \frac{1}{x}\left(e^{-x}-e^{-2x}\right) =-\frac{1}{2} \log 2 $$

But I am looking forward to other approaches, beacuse this method doesn't seem quite natural. And I would highly appreciate it if you could share any thoughts on how to solve this problem. Thanks in advance!

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This integral is $$ \frac{1}{4}\int_{-\infty}^\infty \frac{x\,\mathrm{csch}(x)-1}{x^2}dx $$ This function is analytic on $\mathbb R$, but $\mathrm{csch}$ still has an infinite number of poles in $\mathbb C$ at $i\pi \mathbb Z$. The residue at each of these poles is $(-1)^n/(in\pi)$, so doing the standard contour around the upper half-plane gives $$ \frac{1}{4}\int_{-\infty}^\infty \frac{x\,\mathrm{csch}(x)-1}{x^2}dx = \frac{2\pi i}{4}\sum_{n=1}^\infty\frac{(-1)^n}{in\pi} = \frac{1}{2}\sum_{n=1}^\infty \frac{(-1)^n}{n} = -\frac{1}{2}\ln(2). $$ (This is admittedly trickier than I made it out to be as the error estimates on the contour need to be done more carefully than the usual case, but it's not too hard to work out.)

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  • $\begingroup$ It seems to me that you need to be a bit careful in showing that the integral over the semicircle of large radius is $0$ in the limit. What if you accidentally pick a large radius that passes near (or, worse, through) one of the poles? $\endgroup$ – Barry Cipra May 12 at 6:14
  • $\begingroup$ @BarryCipra You treat it as the limit of a sequence of contours of radius $(n+1/2)\pi$ to avoid going through a pole. The bigger issue that I glossed over is you need to make separate error estimates near the imaginary axis because the integrand only decays as $1/r$ there. $\endgroup$ – eyeballfrog May 12 at 6:18
  • $\begingroup$ Yes, that's what I was getting at. It's been a long time (like almost 50 years) since I've done a contour integral in earnest, but I recall there are occasionally subtle points that need attending to. Experts, of course, know what can be safely glossed over, as here. $\endgroup$ – Barry Cipra May 12 at 6:58
  • $\begingroup$ @BarryCipra. Then, you are a young man ! For me 61 years. Cheers. $\endgroup$ – Claude Leibovici May 12 at 7:38
  • $\begingroup$ @ClaudeLeibovici, or perhaps we are both young men, and you were a child prodigy. To your health! $\endgroup$ – Barry Cipra May 12 at 10:10

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