Proving that $\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=\frac{\pi^3}{16}$

Another approach,

Perform integration by parts,

\begin{align*} I&=\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)\,dx\\ &=\Big[\ln (x) \ln\left(\frac{1+x^2}{(1-x)^2}\right)\arctan x\Big]_0^1 -\int_0^1 \frac{\ln x}{1+x^2}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx-\int_0^1 \frac{2(1+x)\ln (x)\arctan (x)}{(1-x)(1+x^2)}dx\\ &=-\int_0^1 \frac{\ln x}{1+x^2}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx-2\int_0^1 \frac{(1+x)\ln (x)\arctan (x)}{(1-x)(1+x^2)}dx\\ \end{align*}

For $x\in [0;1]$ define the function $R$ by,

\begin{align*} R(x)=\int_0^x \frac{(1+t)\ln t}{(1-t)(1+t^2)}dt=\int_0^1 \frac{x(1+tx)\ln (tx)}{(1-tx)(1+t^2x^2)}dt\\ \end{align*}

Observe that,

\begin{align*} R(1)=\int_0^1 \frac{t\ln t}{1+t}dt+\int_0^1 \frac{\ln t}{1-t}dt \end{align*} Perform integration by parts,

\begin{align*} I&=-\int_0^1 \frac{\ln x}{1+x^2}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx-2\Big[R(x)\arctan x\Big]_0^1+2\int_0^1\frac{R(x)}{1+x^2}dx\\ &=-\int_0^1 \frac{\ln x}{1+x^2}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx-\frac{\pi}{2}R(1)+2\int_0^1 \int_0^1 \frac{x(1+tx)\ln (tx)}{(1-tx)(1+t^2x^2)(1+x^2)}dtdx\\ &=-\int_0^1 \frac{\ln x}{1+x^2}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx-\frac{\pi}{2}R(1)+\int_0^1 \ln x\left[\frac{1}{1+x^2}\ln\left(\frac{1+t^2x^2}{(1-tx)^2}\right)\right]_{t=0}^{t=1} dx+\\ &\int_0^1 \ln t\left[\frac{1}{1+t^2}\ln\left(\frac{1+x^2}{(1-tx)^2}\right)+\frac{2\arctan (tx)}{1-t^2}-\frac{2t\arctan x}{1+t^2}-\frac{2t\arctan x}{1-t^2}\right]_{x=0}^{x=1} dt\\ &=-\frac{\pi }{2}R(1)+\ln 2\int_0^1 \frac{\ln t}{1+t^2}dt-2\int_0^1 \frac{\ln (1-t)\ln t}{1+t^2}dt+2\int_0^1 \frac{\ln t\arctan t}{1-t^2}dt-\\ &\frac{\pi}{2} \int_0^1 \frac{t\ln t}{1+t^2}dt-\frac{\pi}{2} \int_0^1\frac{t\ln t}{1-t^2} dt\\ \end{align*}

For $x\in [0;1]$ define the function $S$ by,

\begin{align*} S(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*}

Perform integration by parts,

\begin{align*} \int_0^1 \frac{\ln x\arctan x}{1-x^2}dx&=\Big[S(x)\arctan x\Big]_0^1-\int_0^1 \frac{S(x)}{1+x^2}dx\\ &=\frac{\pi}{4}S(1)-\int_0^1 \int_0^1 \frac{x\ln(tx)} {(1-t^2x^2)(1+x^2)} dtdx\\ &=\frac{\pi}{4}S(1)-\frac{1}{2}\int_0^1 \left[ \frac{\ln x}{1+x^2}\ln\left(\frac{1+tx}{1-tx} \right)\right]_{t=0}^{t=1} dx-\\ &\frac{1}{2}\int_0^1 \left[ \frac{\ln t}{1+t^2}\ln\left(\frac{1+x^2}{1-t^2x^2} \right)\right]_{x=0}^{x=1}dt\\ &=\frac{\pi}{4}S(1)-\frac{\ln 2}{2}\int_0^1 \frac{\ln t}{1+t^2}dt+\int_0^1 \frac{\ln(1-x)\ln x}{1+x^2}dx \end{align*}

Therefore,

\begin{align*}I&=\pi\int_0^1\frac{2t\ln t}{t^4-1} dt\end{align*}

Perform the change of variable $y=t^2$,

\begin{align*}I&=\frac{1}{2}\pi \int_0^1 \frac{\ln y}{y^2-1}dy\\ &=\frac{1}{2}\pi\times \frac{3}{4}\zeta(2)\\ &=\frac{\pi^3}{16} \end{align*}

Put \begin{equation*} I=\int_{0}^1\dfrac{\arctan x}{x}\ln\left(\dfrac{1+x^2}{(1-x)^2}\right)\, \mathrm{d}x. \end{equation*} Via the substitution $ x=\dfrac{z}{z+1}$ we get \begin{equation*} I = \int_{0}^{\infty}\dfrac{\arctan \frac{z}{z+1}\ln(2z^2+2z+1)}{z^2+z}\, \mathrm{d}z. \end{equation*} Put \begin{equation*} \log z=\ln|z|+i\arg z, \quad -\pi<\arg z <\pi. \end{equation*} Then \begin{equation*} \arctan \frac{z}{z+1}\ln(2z^2+2z+1) = \text{Im}\left(\log^2(1+z+iz)\right). \end{equation*} Consequently \begin{equation*} I = \text{Im}\left(\int_{0}^{\infty}\dfrac{\log^2(1+z+iz)}{z^2+z}\right)\mathrm{d}z. \end{equation*} However, $ \log(z) $ is an analytic function in $ \text{Re} z>0 $. According to Cauchys integral theorem we get the same value if we integrate along the curve with the parametrization $ z=(1-i)s, s>0 $. \begin{gather*} I = \text{Im}\left(\int_{0}^{\infty}\dfrac{\ln^2(2s+1)}{s(s+1-is)}\, \mathrm{d}s\right) = \int_{0}^{\infty}\dfrac{\ln^2(2s+1)}{2s^2+2s+1}\, \mathrm{d}s = \\[2ex] \int_{0}^{\infty}\dfrac{2\ln^2(2s+1)}{(2s+1)^2+1}\, \mathrm{d}s = [t=2s+1] = \\[2ex] \int_{1}^{\infty}\dfrac{\ln^2(t)}{t^2+1}\, \mathrm{d}t =[u= 1/t] = \int_{0}^{1}\dfrac{\ln^2(u)}{u^2+1}\, \mathrm{d}u. \end{gather*} Thus \begin{equation*} 2I = \int_{0}^{\infty}\dfrac{\ln^2(u)}{u^2+1}\, \mathrm{d}u \end{equation*} In order to evaluate this integral we integrate $ \dfrac{\log^3(z)}{z^2+1} $ along a keyhole contour and use residue calculus. In this case $ \log z =\ln |z|+i\arg z, \quad 0<\arg z < 2\pi $. We get \begin{equation*} I = \dfrac{\pi^3}{16}. \end{equation*}

Starting with breaking the integral

$\displaystyle I=\int_0^1\frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)\ dx=\int_0^1\frac{\arctan x}{x}\ln(1+x^2)dx-2\int_0^1\frac{\arctan x}{x}\ln(1-x)dx$

then using the identity$\ \displaystyle\arctan x\ln(1+x^2)=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}} {2n+1}x^{2n+1}$ for the first integral and series-expanding $\displaystyle\arctan x$ of the second integral, we get \begin{align*} I&=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}}{2n+1}\int_0^1x^{2n}\ dx-2\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^1x^{2n}\ln(1-x)\ dx\\ &=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}}{(2n+1)^2}-2\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left(-\frac{H_{2n+1}}{2n+1}\right)\\ &=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}}{(2n+1)^2}-2\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left(-\frac{H_{2n}}{2n+1}-\frac{1}{(2n+1)^2}\right)\\ &=2\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^3}=2\beta(3)=\frac{\pi^3}{16} \end{align*}

where $\beta(3)=\frac{\pi^3}{32}$ is the Dirichlet beta function.

Note that we used the classical result $\int_0^1 x^{n-1}\ln(1-x)dx=-\frac{H_n}{n}$ which can be proved as follows:

$$\int_0^1 x^{n-1}\ln(1-x)dx=-\sum_{k=1}^\infty\frac1k\int_0^1 x^{n+k-1}dx=-\sum_{k=1}^\infty\frac{1}{k(n+k)}\\=-\frac1n\sum_{k=1}^\infty\left(\frac1k-\frac1{n+k}\right)=-\frac1n\sum_{k=1}^n\frac1k=-\frac{H_n}{n}$$