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$$ \int_0^\infty \frac{\text{csch}(x)-\frac1x}{x} {\rm d}x. $$

This integral was from a recent contest like two weeks ago and I still can't crack it. Well, to be exact it was in the form of

$$ \int_0^\infty \frac{2}{x^2} \left( \frac{x}{e^x - e^{-x}} - \frac12 \right) {\rm d}x. $$

The hint was to turn it into Frullani integral, but nothing i've tried worked out, by-parts leaves you with something that doesn't converge and I can't find a way to turn the numerator into $f(ax)-f(bx)$. I noted that it can also be written in the form

$$\int_0^\infty \frac{\text{csch}(\frac1x) - x}{x} {\rm d}x.$$

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  • $\begingroup$ It's a local contest facebook.com/utarNMC $\endgroup$ Jun 26, 2019 at 6:39
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    $\begingroup$ Do you interest alternative way? $\endgroup$
    – Nosrati
    Jun 26, 2019 at 6:42
  • $\begingroup$ yeah of course, anything that works $\endgroup$ Jun 26, 2019 at 6:43
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    $\begingroup$ The result below lead to think that $a=2,b=1$ $\endgroup$
    – FDP
    Jun 26, 2019 at 8:12
  • $\begingroup$ Can't we contour it? $\endgroup$ Jun 26, 2019 at 11:31

2 Answers 2

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Define the function $F$ for $x>0$ by: \begin{align}F(x)=\text{cotanh}\left(\frac{x}{2}\right)-\frac{2}{x}\end{align} Observe that, \begin{align}\lim_{x\rightarrow 0} F(x)&=0\\ \lim_{x\rightarrow \infty} F(x)&=1\\ F(x)-F(2x)&=\frac{1}{\sinh x}-\frac{1}{x} \end{align} On can use Frullani's theorem: \begin{align}\int_0^\infty \frac{\text{csch}(x)-\frac1x}{x} {\rm d}x&=\int_0^\infty \frac{F(x)-F(2x)}{x}\,dx\\ &=\left(F(0)-F(\infty)\right)\ln\left(\frac{2}{1}\right)\\ &=\boxed{-\ln 2} \end{align}

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  • $\begingroup$ (+1) The only answer of the posted that actually uses Frullani. You might be interested in THIS, which extends the commonly seen version of Frullani, and THIS, which converts the original integral to a Frullani integral. $\endgroup$
    – Mark Viola
    Oct 14, 2021 at 18:25
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One way is using the expansion of $\operatorname{csch}x$, that is $$\operatorname{csch}x=\dfrac{1}{x}+\sum_{n=1}^{\infty}\dfrac{2(-1)^nx}{n^2\pi^2+x^2}$$ then $$\int_{0}^{\infty}\dfrac{\operatorname{csch}x-\frac1x}{x}\ dx=\sum_{n=1}^{\infty}\int_{0}^{\infty}2(-1)^n\dfrac{1}{n^2\pi^2+x^2}\ dx=\sum_{n=1}^{\infty}\dfrac{2(-1)^n}{n\pi}\arctan\dfrac{x}{n\pi}\Big|_{0}^{\infty}=\color{blue}{\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n}}=\color{blue}{\ln\dfrac12}$$

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  • $\begingroup$ This doesn't use a Frullani integral, which was the thrust of the question. $\endgroup$
    – Mark Viola
    Oct 14, 2021 at 18:25

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