In this thread it is mentioned that:
$$\int_0^\infty \left ( \mathrm{arccot} x \right )^3\; \mathrm{d}x = \frac{3 \pi^2 \ln 2}{4} – \frac{21\zeta(3)}{8}$$
where $\zeta$ is the Riemann zeta function.
The steps to the solution I took are:
\begin{align*}
\int_{0}^{\infty} \left ( \mathrm{arccot}x \right )^3 \, \mathrm{d}x &= \int_{0}^{\infty} \left ( x \right ) ' \left ( \mathrm{arccot}x \right )^3 \, \mathrm{d}x \\
&=\left [ x \left ( \mathrm{arccot}x \right )^3 \right ]_0^{\infty} +3 \int_{0}^{\infty} \frac{x\left( \mathrm{arccot }x \right )^2}{ x^2+1} \, \mathrm{d}x \\
&=3 \left [ \frac{\ln \left ( 1+x^2 \right ) \, \left (\mathrm{arccot}(x) \right )^2}{2} \right ]_0^{\infty} + 3 \int_{0}^{\infty} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x^2+1} \, \mathrm{d}x \\
&=3 \left ( \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x^2+1} \, \mathrm{d}x + \int_1^{\infty} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x^2+1} \, \mathrm{d}x \right ) \\
&=3 \left ( \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x^2+1} \, \mathrm{d}x + \int_{0}^{1} \frac{\ln \left ( 1+\frac{1}{x^2} \right ) \arctan x}{1+\frac{1}{x^2}} \cdot \frac{1}{x^2} \, \mathrm{d}x \right ) \\
&=3 \left ( \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x^2+1} \, \mathrm{d}x + \int_{0}^{1} \frac{\left (\ln \left ( 1+x^2 \right ) – 2 \ln x \right ) \arctan x}{1+x^2} \, \mathrm{d}x \right )\\
&=3\left ( \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )\left ( \mathrm{arccot} x + \arctan x \right )}{x^2+1} \, \mathrm{d}x – 2 \int_{0}^{1} \frac{\ln x \arctan x}{1+x^2} \, \mathrm{d}x \right ) \\
&=3\left ( \frac{\pi}{2} \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )}{1+x^2} \, \mathrm{d}x – 2 \int_{0}^{1} \frac{\ln x \arctan x}{1+x^2} \, \mathrm{d}x \right )
\end{align*}
Let us now calculate the the first integral:
\begin{align*}
\int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )}{1+x^2} \, \mathrm{d}x &\overset{x=\tan \theta}{=\! =\! =\! =\! =\!} \int_{0}^{\pi/4} \frac{\ln \left ( 1+\tan^2 \theta \right )}{1+\tan^2 \theta} \cdot \sec^2 \theta \, \mathrm{d}\theta\\
&=\int_{0}^{\pi/4} \ln \left ( 1+ \tan^2 \theta \right ) \, \mathrm{d}\theta \\
&= \int_{0}^{\pi/4} \ln \sec^2 \theta \, \mathrm{d} \theta \\
&=2 \int_{0}^{\pi/4} \ln \sec \theta \, \mathrm{d} \theta \\
&=-2 \int_{0}^{\pi/4} \ln \cos \theta \, \mathrm{d} \theta \\
&=-2 \left ( -\int_{0}^{\pi/4} \left (\sum_{n=1}^{\infty} (-1)^n \frac{\cos 2n \theta}{n} – \ln 2 \right ) \, \mathrm{d}\theta \right )\\
&=2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \int_{0}^{\pi/4} \cos 2n\theta \, \mathrm{d} \theta +2 \int_{0}^{\pi/4} \ln 2 \, \mathrm{d}\theta \\
&= 2 \sum_{n=1}^{\infty} \frac{(-1)^n \sin \frac{n \pi}{2}}{2n^2} +\frac{\pi \ln 2}{2} \\
&= \sum_{n=1}^{\infty} \frac{(-1)^n \sin \frac{n \pi}{2}}{2n^2} + \frac{\pi \ln 2}{2} \\
&=\frac{\pi \ln 2}{2} + \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}\\
&= \frac{\pi \ln 2}{2} – \mathcal{G}
\end{align*}
where $\mathcal{G}$ is the Catalan's constant.
Let's move on to the second integral:
\begin{align*}
\int_{0}^{1} \frac{\ln x \arctan x}{1+x^2} \, \mathrm{d}x &=\int_{0}^{1} \ln x \sum_{n=1}^{\infty} (-1)^{n-1} \left ( \mathcal{H}_{2n} – \frac{\mathcal{H}_n}{2} \right ) x^{2n-1} \, \mathrm{d}x \\
&=\sum_{n=1}^{\infty} (-1)^{n-1} \left ( \mathcal{H}_{2n} – \frac{\mathcal{H}_n}{2} \right ) \int_{0}^{1}x^{2n-1} \ln x \, \mathrm{d}x \\
&=-\frac{1}{4}\sum_{n=1}^{\infty} (-1)^{n-1} \frac{\mathcal{H}_{2n}- \frac{\mathcal{H}_n}{2}}{n^2} \\
&= \frac{1}{4} \sum_{n=1}^{\infty} (-1)^n \frac{\mathcal{H}_{2n} – \frac{\mathcal{H}_n}{2}}{n^2} \\
&=\frac{1}{4} \sum_{n=1}^{\infty} (-1)^n \frac{\mathcal{H}_{2n}}{n^2} -\frac{1}{8} \sum_{n=1}^{\infty} (-1)^{n} \frac{\mathcal{H}_n}{n^2}
\end{align*}
The second sum is an old chestnut evaluating to
$$\sum_{n=1}^{\infty} (-1)^n \frac{\mathcal{H}_n}{n^2} = -\frac{5 \zeta(3)}{8}$$
see for example this link.The first sum should bring the $\mathcal{G}$ in . The question is how?
Best Answer
\begin{align}J&=\int_0^\infty \frac{\arctan x\ln x}{1+x^2}\,dx\\ K&=\int_0^1 \frac{\arctan x\ln x}{1+x^2}\,dx \end{align}
Define on $[0;+\infty[$ the function, \begin{align}R(x)&=\int_0^x \frac{\ln t}{1+t^2}\,dt\\ &=\int_0^1 \frac{x\ln(tx)}{1+t^2x^2}\,dt\\ \end{align} Observe that, \begin{align}\displaystyle \lim_{x\rightarrow +\infty }R(x)=\int_0^\infty \frac{\ln x}{1+x^2}\,dx=0\end{align}
Perform integration by parts, \begin{align}J&=\Big[R(x)\arctan x\Big]_0^\infty-\int_0^\infty \left(\int_0^1 \frac{x\ln(tx)}{(1+t^2x^2)(1+x^2)}\,dt\right)\,dx\\ &=-\int_0^\infty\left(\int_0^1 \frac{x\ln x}{(1+t^2x^2)(1+x^2)}\,dt\right)\,dx-\int_0^1 \left(\int_0^\infty \frac{x\ln t}{(1+t^2x^2)(1+x^2)}\,dx\right)\,dt\\ &=-\int_0^\infty \ln x\left[\frac{\arctan(tx)}{1+x^2}\right]_{t=0}^{t=1} \,dx-\int_0^1 \ln t\left[\frac{\ln\left(\frac{1+x^2}{1+t^2x^2}\right)}{2(1-t^2)}\right]_{x=0}^{x=\infty}\,dt\\ &=-J+\int_0^1 \frac{\ln^2 t}{1-t^2}\,dt\\ &=-J+\int_0^1 \frac{\ln^2 t}{1-t}\,dt-\int_0^1 \frac{t\ln^2 t}{1-t^2}\,dt\\ \end{align} In the latter integral perform the change of variable variable $\displaystyle y=x^2$, \begin{align}J&=-J+\left(1-\frac{1}{8}\right)\int_0^1 \frac{\ln^2 t}{1-t}\,dt\\ &=-J+\left(1-\frac{1}{8}\right)\times 2\zeta(3)\\ \end{align} Therefore, \begin{align}\boxed{J=\dfrac{7}{8}\zeta(3)}\end{align} But, \begin{align}J&=\int_0^1 \frac{\arctan x\ln x}{1+x^2}\,dx+\int_1^\infty \frac{\arctan x\ln x}{1+x^2}\,dx\\ &=K+\int_1^\infty \frac{\arctan x\ln x}{1+x^2}\,dx\end{align} Perform the change of variable $y=\dfrac{1}{x}$, \begin{align}J&=K-\int_0^1 \frac{\arctan \left(\frac{1}{x}\right)\ln x}{1+x^2}\,dx\\ &=2K+\frac{1}{2}\pi\text{G} \end{align} therefore, \begin{align}\boxed{K=\dfrac{7}{16}\zeta(3)-\frac{1}{4}\pi\text{G}}\end{align}
NB:
For $x>0,\arctan x+\arctan\left(\dfrac{1}{x}\right)=\dfrac{\pi}{2}$
$\displaystyle \int_0^1 \dfrac{\ln x}{1+x^2}\,dx=-\text{G}$
$\text{G}$ the Catalan's constant.
Addendum: \begin{align}L&=\int_0^1 \frac{\ln(1+x^2)}{1+x^2}\,dx\end{align} Perform the change of variable $y=\tan x$, \begin{align}L&=-2\int_0^{\frac{\pi}{4}}\ln(\cos x)\,dx\end{align}
\begin{align}A&=\int_0^{\frac{\pi}{4}}\ln(\cos x)\,dx\\ B&=\int_0^{\frac{\pi}{4}}\ln(\sin x)\,dx\\ A+B&=\int_0^{\frac{\pi}{4}}\ln(\sin x\cos x)\,dx\\ &=\int_0^{\frac{\pi}{4}}\ln\left(\frac{\sin(2x)}{2}\right)\,dx\\ &=\int_0^{\frac{\pi}{4}}\ln\left(\sin(2x)\right)\,dx-\frac{\pi}{4}\ln 2\\ \end{align} Perform the change of variable $\displaystyle y=2x$, \begin{align}A+B&=\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln\left(\sin x\right)\,dx-\frac{\pi}{4}\ln 2\\ &=\frac{1}{2}\times -\frac{\pi}{2}\ln 2-\frac{\pi}{4}\ln 2\\ &=-\frac{\pi}{2}\ln 2\\ B-A&=\int_0^{\frac{\pi}{4}}\ln\left(\tan x\right)\,dx\\ &=-\text{G}\\\end{align} Therefore, \begin{align}A&=\frac{1}{2}\text{G}-\frac{1}{4}\pi\ln 2\\ B&=-\frac{1}{2}\text{G}-\frac{1}{4}\pi\ln 2 \end{align} Therefore, \begin{align}\boxed{L=\frac{1}{2}\pi\ln 2-\text{G}}\end{align}