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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:

$$I_n=\int_0^{\infty } \left(\coth^n (x)-\frac{1}{x^n}\right) x^{n-1}\text{csch}(x) \, dx\tag{3}$$

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    $\begingroup$ Alternate form $$\int_{0}^{\infty }\!{\frac {x\cosh \left( x \right)-\sinh \left( x \right) }{ x\left( \sinh \left( x \right) \right) ^{2}}}\,{\rm d}x$$ $\endgroup$
    – GEdgar
    May 3 at 12:01
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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 $$I= (f(0)-f(\infty))\ln\frac12=(0-1)\ln\frac12=\ln2 $$

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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 $$ u = \left(\frac{1}{\sinh x} - \frac{1}{x}\right) x, \qquad dv = \frac{dx}{x^2} $$

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