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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

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

Nosrati
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Jia Ming جيا ميڠ
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2 Answers2

9

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}

FDP
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  • (+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. – Mark Viola Oct 14 '21 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 $$\int_{0}^{\infty}\dfrac{\operatorname{csch}x-\frac1x}{x}\ dx=\sum_{n=1}^{\infty}\int_{0}^{\infty}2(-1)^n\dfrac{1}{n^2\pi^2+x^2}\ dx=\sum_{n=1}^{\infty}\dfrac{2(-1)^n}{n\pi}\arctan\dfrac{x}{n\pi}\Big|_{0}^{\infty}=\color{blue}{\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n}}=\color{blue}{\ln\dfrac12}$$

Nosrati
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