# 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:

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

• Alternate form $$\int_{0}^{\infty }\!{\frac {x\cosh \left( x \right)-\sinh \left( x \right) }{ x\left( \sinh \left( x \right) \right) ^{2}}}\,{\rm d}x$$ May 3 at 12:01

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 $$I= (f(0)-f(\infty))\ln\frac12=(0-1)\ln\frac12=\ln2$$
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 $$u = \left(\frac{1}{\sinh x} - \frac{1}{x}\right) x, \qquad dv = \frac{dx}{x^2}$$