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