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

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

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

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

Solution 1:

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}

Solution 2:

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 $$\int_{0}^{\infty}\dfrac{\operatorname{csch}x-\frac1x}{x}\ dx=\sum_{n=1}^{\infty}\int_{0}^{\infty}2(-1)^n\dfrac{1}{n^2\pi^2+x^2}\ dx=\sum_{n=1}^{\infty}\dfrac{2(-1)^n}{n\pi}\arctan\dfrac{x}{n\pi}\Big|_{0}^{\infty}=\color{blue}{\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n}}=\color{blue}{\ln\dfrac12}$$