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:
$$I_n=\int_0^{\infty } \left(\coth^n (x)-\frac{1}{x^n}\right) x^{n-1}\text{csch}(x) \, dx\tag{3}$$
Best Answer
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 $$I= (f(0)-f(\infty))\ln\frac12=(0-1)\ln\frac12=\ln2 $$