Evaluating integral $\int_0^{\infty } \left(\coth x-\frac{1}{x}\right) \text{csch} \>x\, dx$

Question

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:

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

Answer 1

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 $$I= (f(0)-f(\infty))\ln\frac12=(0-1)\ln\frac12=\ln2 $$

Answer 2

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

Answer 3

$$\begin{align*} I &= \int_0^\infty \left(\coth (x)-\frac{1}{x}\right) \operatorname{csch}(x) \, dx \\ &= \int_0^\infty \left(\coth(x)\operatorname{csch}(x) - \frac1{x^2} - 2 \sum_{n=1}^\infty \frac {(-1)^n}{x^2+\pi^2 n^2}\right) \, dx \tag1 \\ &= \left[-\operatorname{csch}(x) + \frac1x - 2 \sum_{n=1}^\infty \frac{(-1)^n}{n\pi} \tan^{-1}\left(\frac x{n\pi}\right)\right]_0^\infty \\ &= -2 \sum_{n=1}^\infty \frac{(-1)^n}{n\pi} \cdot \frac\pi2 \tag2 \\ &= \ln(2) \tag3 \end{align*}$$

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• $(1)$ : recall the partial fraction expansion of $\operatorname{csch}(x)$
• $(2)$ : $\dfrac1x-\operatorname{csch}(x)\to0$ as $x\to0^+$ and $x\to\infty$; the sum is uniformly convergent so we may interchange limit with summation (see corollary 7.2)
• $(3)$ : recall the Mercator series for $\ln(1+x)$