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
I(a,b)=0arccot(x)arccot(ax)arccot(bx) dx, for a>0,b>0
It’s known\require{action}\require{enclose}\texttip{{}^\dagger}{Gradshteyn & Ryzhik, Table of Integrals, Series, and Products, 7th edition, page 599, (4.511)} that
I(a,0)=\frac{\pi^2}4\left[\ln\left(1+\frac1a\right)+\frac{\ln(1+a)}a\right].\tag2
Maple and Mathematica are also able to evaluate
I(1,1)=\frac{3\pi^2}4\ln2-\frac{21}8\zeta(3).\tag3


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

Answer

By the substitution x\mapsto x^{-1} it is not hard to see that
I(\nu^{-1},1)=\int^\infty_0\frac{\arctan^2{x}\arctan(\nu x)}{x^2}{\rm d}x

First, start off by re-expressing the integral
\begin{align}
\int^\pi_0x^2\cos(nx)\cot\left(\frac{x}{2}\right)\ {\rm d}x
&=-{\rm Re}\int_C\frac{z+1}{z-1}z^{n-1}\ln^2{z}\ {\rm d}z\\
&=(-1)^{n}\int^1_0\frac{1-x}{1+x}x^{n-1}(\ln^2{x}-\pi^2)\ {\rm d}x-\int^1_0\frac{1+x}{1-x}x^{n-1}\ln^2{x}\ {\rm d}x\\
\end{align}

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
\begin{align}
I_\nu(\nu^{-1},1)
&=\int^\frac{\pi}{2}_0\frac{x^2}{\tan{x}(\cos^2{x}+\nu^2\sin^2{x})}{\rm d}x\\
&=\frac{1}{4}\int^\pi_0\frac{x^2}{\tan\left(\frac{x}{2}\right)(1+(1-\nu^2)\cos{x}+\nu^2)}{\rm d}x\\
&=\frac{1}{8\nu}\int^\pi_0\left(x^2\cot\left(\frac{x}{2}\right)+2\sum^\infty_{n=1}\left(\frac{\nu-1}{\nu+1}\right)^nx^2\cos(nx)\cot\left(\frac{x}{2}\right)\right){\rm d}x\\
&=\frac{\pi^2}{4\nu}\ln{2}-\frac{7\zeta(3)}{8\nu}-\frac{\xi}{4\nu}\left(\int^1_0\frac{1-x}{(1+x)(1+\xi x)}(\ln^2{x}-\pi^2)+\frac{1+x}{(1-x)(1-\xi x)}\ln^2{x}\ {\rm d}x\right)
\end{align}

Here \xi=\dfrac{\nu-1}{\nu+1}. Utilising the partial fraction decompositions
\frac{1-x}{(1+x)(1+\xi x)}=\frac{\nu+1}{1+x}-\frac{\nu}{1+\xi x}
\frac{1+x}{(1-x)(1-\xi x)}=\frac{\nu+1}{1-x}-\frac{\nu}{1-\xi x}
in tandem with the easily verifiable fact
\int^1_0\frac{\ln^2{x}}{1+\lambda x}{\rm d}x=-\frac{2{\rm Li}_3(-\lambda)}{\lambda}
yields
\begin{align}
I_\nu(\nu^{-1},1)
&=\frac{\pi^2}{4\nu}\ln{2}-\frac{7\zeta(3)}{8\nu}-\frac{\xi}{4\nu}\left(\frac{\pi^2\nu}{\xi}\ln(1+\xi)-\pi^2(\nu+1)\ln{2}+\frac{7\zeta(3)}{2}(\nu+1)-\frac{4\nu}{\xi}\chi_3(\xi)\right)\\
&=\chi_3\left(\frac{\nu-1}{\nu+1}\right)-\frac{7\zeta(3)}{8}+\frac{\pi^2}{4}\ln\left(1+\frac{1}{\nu}\right)
\end{align}

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,
\begin{align}
I_\nu(\nu^{-1},1)
&=\underbrace{\frac{\pi^2}{4}\ln\left(\frac{(1+\nu)^{1+\nu}}{\nu^\nu}\right)-\frac{7\zeta(3)}{8}\nu}_{\text{Let this be C}}+2\int^\frac{1-\nu}{1+\nu}_1\frac{\chi_3(v)}{(1+v)^2}{\rm d}v\\
&=C-\left.\frac{2\chi_3(v)}{1+v}\right|^\frac{1-\nu}{1+\nu}_1+2\int^\frac{1-\nu}{1+\nu}_1\frac{\chi_2(v)}{v(1+v)}{\rm d}v\\
&=C+(1-\nu)\chi_3\left(\frac{1-\nu}{1+\nu}\right)-\frac{7\zeta(3)}{8}-\left.\color{white}{\frac{1}{1}}2\chi_2(v)\ln(1+v)\right|^\frac{1-\nu}{1+\nu}_1+\int^\frac{1-\nu}{1+\nu}_1\frac{\ln(1+v)\ln\left(\frac{1+v}{1-v}\right)}{v}{\rm d}v\\
&=C+(1-\nu)\chi_3\left(\frac{1-\nu}{1+\nu}\right)-\frac{7\zeta(3)}{8}+2\chi_2\left(\frac{1-\nu}{1+\nu}\right)\ln\left(\frac{1+\nu}{2}\right)+\frac{\pi^2}{4}\ln{2}\\
&\ \ \ \ +\frac{1}{2}\int^\frac{1-\nu}{1+\nu}_1\frac{\ln^2(1+v)-\ln^2(1-v)+\ln^2\left(\frac{1-v}{1+v}\right)}{v}{\rm d}v
\end{align}

Repeatedly integrating by parts, it is not hard to derive that
\begin{align}
\frac{1}{2}\int^\frac{1-\nu}{1+\nu}_1\frac{\ln^2(1+v)}{v}{\rm d}v
=&\ -\frac{1}{6}\ln^3\left(\frac{1+\nu}{2}\right)+\frac{7\zeta(3)}{8}-{\rm Li}_3\left(\frac{1+\nu}{2}\right)+{\rm Li}_2\left(\frac{1+\nu}{2}\right)\ln\left(\frac{1+\nu}{2}\right)\\
&\ +\frac{1}{2}\ln\left(\frac{1-\nu}{2}\right)\ln^2\left(\frac{1+\nu}{2}\right)\\
-\frac{1}{2}\int^\frac{1-\nu}{1+\nu}_1\frac{\ln^2(1-v)}{v}{\rm d}v
=&\ {\rm Li}_3\left(\frac{2\nu}{1+\nu}\right)-{\rm Li}_2\left(\frac{2\nu}{1+\nu}\right)\ln\left(\frac{2\nu}{1+\nu}\right)-\frac{1}{2}\ln\left(\frac{1-\nu}{1+\nu}\right)\ln^2\left(\frac{2\nu}{1+\nu}\right)\\
\frac{1}{2}\int^\frac{1-\nu}{1+\nu}_1\frac{\ln^2\left(\frac{1-v}{1+v}\right)}{v}{\rm d}v
=&\ -2\chi_3(\nu)+2\chi_2(\nu)\ln{\nu}+\frac{1}{2}\ln^2\nu\ln\left(\frac{1-\nu}{1+\nu}\right)
\end{align}

Therefore, we get, for I(\nu^{-1},1),
\begin{align}
I(\nu^{-1},1)
=&\color{brown}{\ (1-\nu)\chi_3\left(\frac{1-\nu}{1+\nu}\right)-2\chi_3(\nu)+{\rm Li}_3\left(\frac{2\nu}{1+\nu}\right)-{\rm Li}_3\left(\frac{1+\nu}{2}\right)-\frac{7\zeta(3)}{8}\nu}\\
&\ \color{brown}{+2\chi_2(\nu)\ln{\nu}+2\chi_2\left(\frac{1-\nu}{1+\nu}\right)\ln\left(\frac{1+\nu}{2}\right)-{\rm Li}_2\left(\frac{2\nu}{1+\nu}\right)\ln\left(\frac{2\nu}{1+\nu}\right)}\\
&\ \color{brown}{+{\rm Li}_2\left(\frac{1+\nu}{2}\right)\ln\left(\frac{1+\nu}{2}\right)+\frac{\pi^2}{4}\ln\left(\frac{(1+\nu)^{1+\nu}}{\nu^\nu}\right)+\frac{\pi^2}{4}\ln{2}}\\
&\ \color{brown}{+\frac{1}{2}\ln^2\nu\ln\left(\frac{1-\nu}{1+\nu}\right)+\frac{1}{2}\ln\left(\frac{1-\nu}{2}\right)\ln^2\left(\frac{1+\nu}{2}\right)-\frac{1}{6}\ln^3\left(\frac{1+\nu}{2}\right)}\\
&\ \color{brown}{-\frac{1}{2}\ln\left(\frac{1-\nu}{1+\nu}\right)\ln^2\left(\frac{2\nu}{1+\nu}\right)}
\end{align}

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.

Attribution
Source : Link , Question Author : Vladimir Reshetnikov , Answer Author : M.N.C.E.

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