Evaluating $\int_0^\infty\frac{\tan^{-1}av\cot^{-1}av}{1+v^2}\,dv$

Question

The Weierstrass substitution stuck in my head after I used it to prove the rigidity of the braced hendecagon (and tridecagon). Thus I had another look at this question which I eventually answered in a very complicated way, and hit upon a potentially simpler approach. For reference the original question is reproduced below:

Pick three points from the uniform distribution over a unit circle's circumference. What is the probability $P(x)$ of the triangle thus formed containing $(x,0)$ where $0\le x\le1$?

My new approach represents the three points by their Weierstrass parameters – $\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right)$ and similarly for $u$ and $v$ – and then computes the barycentric coordinates of $(x,0)$; they will all be positive iff the triangle contains the point. The raw expressions are a bit messy, but cylindrical algebraic decomposition shows that if $t<u<v$ containment occurs iff $0<v\land t<k/v\land k/v<u<k/t$ where $k=\frac{x-1}{x+1}$, which leads to a single triple integral for the probability: $$P(x)=\frac6{\pi^3}\int_0^\infty\int_{-\infty}^{k/v}\int_{k/v}^{k/t}\frac1{(1+t^2)(1+u^2)(1+v^2)}\,du\,dt\,dv$$ Note that both integration region and integrand are invariant under the involution $(t,u,v)\to(v,-u,t)$, which simplifies the region description and allows solving the first two levels easily: $$P(x)=\frac{12}{\pi^3}\int_0^\infty\int_{-\infty}^{k/v}\int_{k/v}^0\frac1{(1+t^2)(1+u^2)(1+v^2)}\,du\,dt\,dv$$ $$=\frac{12}{\pi^3}\int_0^\infty\frac{\tan^{-1}av\cot^{-1}av}{1+v^2}\,dv\qquad a=\frac{1+x}{1-x}=-\frac1k\tag1$$ I have not been able to solve this last parametric integral; both Mathematica and Rubi fail on it except at $a=1$. Because I found the probability by another method, however, I know what the result must be: $$P(x)=\frac14-\frac3{2\pi^2}\operatorname{Li}_2(x^2)\qquad x=\frac{a-1}{a+1}\tag2$$

How can $(2)$ be obtained from $(1)$?

Answer 1

One common way to solve such integrals is to differentiate under the integral sign w.r.t. a.
Here is another approach using Fourier Series. $$I(a)=\int_0^\infty\frac{\arctan(ax)\operatorname{arccot}(ax)}{1+x^2}dx\overset{\large ax\to \tan(x/2)}=\frac{a}{4}\int_0^\pi\frac{x\left(\pi-x\right)}{1+a^2-(1-a^2)\cos x}dx$$ Refering to this thread we can expand the denominator into Fourier series as: $$\frac{1}{1+a^2-(1-a^2)\cos x}=\frac{1}{2a}+\frac{1}{a}\sum_{n=1}^\infty \left(\frac{1-a}{1+a}\right)^n\cos(nx)$$ $$\Rightarrow I(a)=\frac18\int_0^\pi x(\pi-x)dx+\frac14\sum_{n=1}^\infty \left(\frac{1-a}{1+a}\right)^n\int_0^\pi x(\pi-x)\cos(nx)dx$$ $$=\frac{\pi^3}{48} - \frac{\pi}{4}\sum_{n=1}^\infty \left(\frac{1-a}{1+a}\right)^n\frac{1+(-1)^n}{n^2}=\frac{\pi^3}{48}-\frac{\pi}{8}\operatorname{Li}_2\left(\left(\frac{1-a}{1+a}\right)^2\right)$$ Where $\operatorname{Li_2}(x)$ is the dilogarithm function (from this link the series representation alongside the third functional equation is useful).

Answer 2

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Just for fun, we can in fact solve a more general problem without too much more effort. Define the function $\mathcal{I}:\mathbb{R}_{>0}^{2}\rightarrow\mathbb{R}$ via the parametric integral

$$\mathcal{I}{\left(a,b\right)}:=\int_{0}^{\infty}\mathrm{d}x\,\frac{\arctan{\left(ax\right)}\operatorname{arccot}{\left(bx\right)}}{1+x^{2}},$$

where here the arctangent and arccotangent functions are defined respectively by

$$\arctan{\left(z\right)}=\int_{0}^{z}\mathrm{d}x\,\frac{1}{1+x^{2}};~~~\small{z\in\mathbb{R}},$$

$$\operatorname{arccot}{\left(z\right)}=\int_{z}^{\infty}\mathrm{d}x\,\frac{1}{1+x^{2}};~~~\small{z\in\mathbb{R}}.$$

Then, given fixed but arbitrary $\left(a,b\right)\in\mathbb{R}_{>0}^{2}$, we find

$$\begin{align} \mathcal{I}{\left(a,b\right)} &=\int_{0}^{\infty}\mathrm{d}x\,\frac{\arctan{\left(ax\right)}\operatorname{arccot}{\left(bx\right)}}{1+x^{2}}\\ &=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{a}\mathrm{d}t\int_{b}^{\infty}\mathrm{d}u\,\frac{x^{2}}{\left(1+x^{2}\right)\left(1+x^{2}t^{2}\right)\left(1+x^{2}u^{2}\right)}\\ &=\int_{0}^{a}\mathrm{d}t\int_{b}^{\infty}\mathrm{d}u\int_{0}^{\infty}\mathrm{d}x\,\frac{x^{2}}{\left(1+x^{2}\right)\left(1+x^{2}t^{2}\right)\left(1+x^{2}u^{2}\right)}\\ &=\int_{0}^{a}\mathrm{d}t\int_{b}^{\infty}\mathrm{d}u\int_{0}^{\infty}\mathrm{d}y\,\frac{y^{2}}{\left(y^{2}+1\right)\left(y^{2}+t^{2}\right)\left(y^{2}+u^{2}\right)};~~~\small{\left[x=\frac{1}{y}\right]}\\ &=\int_{0}^{a}\mathrm{d}t\int_{b}^{\infty}\mathrm{d}u\,\frac{1}{\left(1-t^{2}\right)\left(1-u^{2}\right)\left(t^{2}-u^{2}\right)}\\ &~~~~~\times\int_{0}^{\infty}\mathrm{d}y\,\frac{\left(1-t^{2}\right)\left(1-u^{2}\right)\left(t^{2}-u^{2}\right)y^{2}}{\left(y^{2}+1\right)\left(y^{2}+t^{2}\right)\left(y^{2}+u^{2}\right)}\\ &=\int_{0}^{a}\mathrm{d}t\int_{b}^{\infty}\mathrm{d}u\,\frac{1}{\left(1-t^{2}\right)\left(1-u^{2}\right)\left(t^{2}-u^{2}\right)}\\ &~~~~~\times\int_{0}^{\infty}\mathrm{d}y\,\left[\frac{\left(1-u^{2}\right)t^{2}}{\left(y^{2}+t^{2}\right)}-\frac{\left(1-t^{2}\right)u^{2}}{\left(y^{2}+u^{2}\right)}-\frac{\left(t^{2}-u^{2}\right)}{\left(y^{2}+1\right)}\right]\\ &=\int_{0}^{a}\mathrm{d}t\int_{b}^{\infty}\mathrm{d}u\,\frac{1}{\left(1-t^{2}\right)\left(1-u^{2}\right)\left(t^{2}-u^{2}\right)}\\ &~~~~~\times\left[\int_{0}^{\infty}\mathrm{d}y\,\frac{\left(1-u^{2}\right)t^{2}}{\left(y^{2}+t^{2}\right)}-\int_{0}^{\infty}\mathrm{d}y\,\frac{\left(1-t^{2}\right)u^{2}}{\left(y^{2}+u^{2}\right)}-\int_{0}^{\infty}\mathrm{d}y\,\frac{\left(t^{2}-u^{2}\right)}{\left(y^{2}+1\right)}\right]\\ &=\int_{0}^{a}\mathrm{d}t\int_{b}^{\infty}\mathrm{d}u\,\frac{1}{\left(1-t^{2}\right)\left(1-u^{2}\right)\left(t^{2}-u^{2}\right)}\\ &~~~~~\times\left[\left(1-u^{2}\right)t\frac{\pi}{2}-\left(1-t^{2}\right)u\frac{\pi}{2}-\left(t^{2}-u^{2}\right)\frac{\pi}{2}\right]\\ &=\frac{\pi}{2}\int_{0}^{a}\mathrm{d}t\int_{b}^{\infty}\mathrm{d}u\,\frac{\left(1-t\right)\left(1-u\right)\left(t-u\right)}{\left(1-t^{2}\right)\left(1-u^{2}\right)\left(t^{2}-u^{2}\right)}\\ &=\frac{\pi}{2}\int_{0}^{a}\mathrm{d}t\int_{b}^{\infty}\mathrm{d}u\,\frac{1}{\left(1+t\right)\left(1+u\right)\left(t+u\right)}\\ &=\frac{\pi}{2}\int_{0}^{a}\mathrm{d}t\,\frac{1}{\left(1-t^{2}\right)}\int_{b}^{\infty}\mathrm{d}u\,\frac{\left(1-t\right)}{\left(1+u\right)\left(t+u\right)}\\ &=\frac{\pi}{2}\int_{0}^{a}\mathrm{d}t\,\frac{1}{1-t^{2}}\int_{b}^{\infty}\mathrm{d}u\,\left[\frac{1}{\left(t+u\right)}-\frac{1}{\left(1+u\right)}\right]\\ &=\frac{\pi}{2}\int_{0}^{a}\mathrm{d}t\,\frac{1}{1-t^{2}}\int_{b}^{\infty}\mathrm{d}u\,\frac{d}{du}\left[\ln{\left(t+u\right)}-\ln{\left(1+u\right)}\right]\\ &=\frac{\pi}{2}\int_{0}^{a}\mathrm{d}t\,\frac{1}{1-t^{2}}\int_{b}^{\infty}\mathrm{d}u\,\frac{d}{du}\ln{\left(\frac{t+u}{1+u}\right)}\\ &=\frac{\pi}{2}\int_{0}^{a}\mathrm{d}t\,\frac{1}{1-t^{2}}\left[-\ln{\left(\frac{t+b}{1+b}\right)}\right]\\ &=-\frac{\pi}{4}\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}u\,\frac{1}{u}\ln{\left(\frac{\left(\frac{1-u}{1+u}\right)+b}{1+b}\right)};~~~\small{\left[t=\frac{1-u}{1+u}\right]}\\ &=\frac{\pi}{4}\int_{p}^{1}\mathrm{d}u\,\frac{1}{u}\ln{\left(\frac{1+u}{1-qu}\right)};~~~\small{\left[p:=\frac{1-a}{1+a}\land q:=\frac{1-b}{1+b}\right]}\\ &=\frac{\pi}{4}\int_{p}^{1}\mathrm{d}u\,\left[-\frac{\ln{\left(1-qu\right)}}{u}+\frac{\ln{\left(1+u\right)}}{u}\right]\\ &=\frac{\pi}{4}\int_{p}^{1}\mathrm{d}u\,\frac{d}{du}\left[\operatorname{Li}_{2}{\left(qu\right)}-\operatorname{Li}_{2}{\left(-u\right)}\right]\\ &=\frac{\pi}{4}\left[-\operatorname{Li}_{2}{\left(-1\right)}+\operatorname{Li}_{2}{\left(-p\right)}+\operatorname{Li}_{2}{\left(q\right)}-\operatorname{Li}_{2}{\left(pq\right)}\right]\\ &=\frac{\pi}{4}\left[\frac{\pi^{2}}{12}+\operatorname{Li}_{2}{\left(-\frac{1-a}{1+a}\right)}+\operatorname{Li}_{2}{\left(\frac{1-b}{1+b}\right)}-\operatorname{Li}_{2}{\left(\frac{1-a}{1+a}\cdot\frac{1-b}{1+b}\right)}\right].\blacksquare\\ \end{align}$$

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