# About the integral $\int\arctan\left(\frac{1}{\sinh^2 x}\right)dx$, some idea or feedback

While I was playing with Wolfram Alpha calculator I wondered if it is known a closed-form for $$\int_0^\infty\arctan\left(\frac{1}{\sinh^2 x}\right)dx.\tag{1}$$

Wolfram Alpha provide me the coressponding indefinite integral using this code

int arctan(1/sinh^2(x))dx

but it seems like as science fiction that I can to understand what did this CAS since the integral is very difficult.

Question. Can you provide me an idea to get such indefinite integral $$\int\arctan\left(\frac{1}{\sinh^2 x}\right)dx?$$ Of course, if it is a known integral and closed-form $(1)$ answer as a reference request, refering the literature and I try to find and read such exercise from the literature. Thanks you in advance.

• I don't require all details, just some idea about how to attack it.
– user243301
Mar 28, 2018 at 19:43

By substituting $x=\text{arcsinh}(u)$ we have $$\int_{0}^{+\infty}\arctan\left(\frac{1}{\sinh t}\right)\,dt = \int_{0}^{+\infty}\frac{\arctan\frac{1}{u}}{\sqrt{1+u^2}}\,du \stackrel{u\mapsto 1/v}{=}\int_{0}^{+\infty}\frac{\arctan v}{v\sqrt{1+v^2}}\,dv$$ and by substituting $v=\tan\theta$ we are left with $$\int_{0}^{\pi/2}\frac{\theta\,d\theta}{\sin\theta}$$ which is a notorious integral, equal to twice the Catalan constant $G$: $$\int_{0}^{+\infty}\arctan\left(\frac{1}{\sinh t}\right)\,dt = 2\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2}\approx 1.831931188354438.$$ The indefinite integral is similarly related to the dilogarithm function $\text{Li}_2$.
If we have an extra square there is little to change: $$\int_{0}^{+\infty}\arctan\left(\frac{1}{\sinh^2 t}\right)\,dt = \int_{0}^{+\infty}\frac{\arctan u}{2u\sqrt{1+u}}\,du=\int_{0}^{1}\int_{0}^{+\infty}\frac{1}{2(1+a^2 u^2)\sqrt{1+u}}\,du\,da$$ we "simply" have a linear combination of squared logarithms evaluated at ugly points.