Evaluating integral $\int_0^{\infty } \left(\coth x-\frac{1}{x}\right) \text{csch} \>x\, dx$ I am having trouble evaluating the following integral:
$$\int_0^{\infty } \left(\coth (x)-\frac{1}{x}\right) \text{csch}(x) \, dx\tag{1}$$
Numerically the integral appears to evaluate to $\log 2$.
The Weierstrass substitution doesn't help me other than allow a series approximation to be calculated using Mathematica. The Weierstrass substitution results in
$$\frac{1}{2} \int_0^1 \left(\frac{1}{t^2}-\frac{1}{t \tanh ^{-1}(t)}+1\right) \, dt\tag{2}$$
Any ideas?
Incidentally there seem to be a sequence of such integrals with closed forms:
$$I_n=\int_0^{\infty } \left(\coth^n (x)-\frac{1}{x^n}\right) x^{n-1}\text{csch}(x) \, dx\tag{3}$$
 A: Rewrite the integral as
$$I=\int_0^{\infty } \left(\coth x-\frac{1}{x}\right) \text{csch}x\, dx
= \int_0^{\infty } \frac{f(x)-f(\frac x2)}x dx
$$
where
$f(x) = \coth x - x \>\text{csch}^2x$.
Then, apply the Frullani's theorem to obtain
$$I= (f(0)-f(\infty))\ln\frac12=(0-1)\ln\frac12=\ln2
$$
A: Merely a comment.  Find it in the literature.
Gradshteyn, I. S.; Ryzhik, I. M.; Zwillinger, Daniel (ed.); Moll, Victor (ed.), Table of integrals, series, and products. Translated from the Russian. Translation edited and with a preface by Victor Moll and Daniel Zwillinger, Amsterdam: Elsevier/Academic Press (ISBN 978-0-12-384933-5/hbk; 978-0-12-384934-2/ebook). xlv, 1133 p. (2015). ZBL1300.65001.
3.529.1 is
$$
\int_0^\infty \left(\frac{1}{\sinh x} - \frac{1}{x}\right)\frac{dx}{x} = -\ln 2
\tag{$*$}$$
We get the desired integral
$$\int_{0}^{\infty }\!{\frac {x\cosh \left( x \right)-\sinh \left( x
 \right) }{ x\left( \sinh \left( x \right)  \right) ^{2}}}\,{\rm d}x = \ln 2$$
when we integrate $(*)$ by parts using
$$
u = \left(\frac{1}{\sinh x} - \frac{1}{x}\right) x,
\qquad
dv = \frac{dx}{x^2}
$$
