# Integrals of the form ∫∞0arccot(x)⋅arccot(ax)⋅arccot(bx) dx{\large\int}_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx

I’m interested in integrals of the form

It’s known$\require{action}\require{enclose}\texttip{{}^\dagger}{Gradshteyn & Ryzhik, Table of Integrals, Series, and Products, 7th edition, page 599, (4.511)}$ that

Maple and Mathematica are also able to evaluate

Is it possible to find a general closed form for $I(a,1)$? Or, at least, for $I(2,1)$ or $I(3,1)$?

By the substitution $x\mapsto x^{-1}$ it is not hard to see that

First, start off by re-expressing the integral

where $C$ is the arc joining $z=1$ to $z=-1$. (The first equality follows from $z=e^{ix}$, whereas the second follows from the fact that the integral along $[-1,\epsilon]\cup\epsilon\exp(i[\pi,0])\cup[\epsilon,1]\cup C$ is $0$ since $z=1$ is a removable singularity and the indent around $z=0$ vanishes.)

Next, note that

Here $\xi=\dfrac{\nu-1}{\nu+1}$. Utilising the partial fraction decompositions

in tandem with the easily verifiable fact

yields

where $\displaystyle\chi_s(z)=\sum_{n\ \text{odd}}\frac{z^n}{n^s}=\frac{1}{2}\left({\rm Li}_s(z)-{\rm Li}_s(-z)\right)$ is the Legendre-chi function.

Integrating back,

Repeatedly integrating by parts, it is not hard to derive that

Therefore, we get, for $I(\nu^{-1},1)$,

If no mistakes were made, this formula should hold for $0<\nu\le1$. Further simplifications to the formula may be possible through some polylogarithm identities.