Prove $2\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx+\int_0^1 \frac{\ln^2(1+x)}{1+x^2}dx=\frac{3}{16}\pi\ln^22-2G\ln2+\frac{7}{64}\pi^3$

Question

How prove that $$2\int_0^1 \dfrac{(\ln(1-x))^2}{1+x^2}dx+\int_0^1 \dfrac{(\ln(1+x))^2}{1+x^2}dx=\dfrac{3}{16}\pi\left(\ln2\right)^2-2G\ln2+\dfrac{7}{64}\pi^3$$ Where G is the Catalan's Constant. $$ \int_0^1 \dfrac{(\ln(1+x))^2}{1+x^2}dx=\Big[\arctan x\left(\ln\left(1+x\right)\right)^2\Big]_0^1-2\int_0^1 \dfrac{\arctan x\ln(1+x)}{1+x}dx=\dfrac{1}{4}\pi\left(\ln2\right)^2-2J$$ see users FDP about Evaluating $$\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$$

Answer 1

Let,

$\displaystyle A=\int_0^1 \dfrac{\left(\ln(1-x)\right)^2}{1+x^2}dx$

$\displaystyle B=\int_0^1 \dfrac{\left(\ln(1+x)\right)^2}{1+x^2}dx$

In integral defining $A$ perform the change of variable $y=\dfrac{1-x}{1+x}$,

$\begin{align} A&=\int_0^1 \dfrac{\left(\ln\left(\tfrac{2x}{1+x}\right)\right)^2}{1+x^2}dx\\ &=B-2\int_0^1 \dfrac{\ln x\ln(1+x)}{1+x^2}dx-\dfrac{1}{4}\pi(\ln 2)^2+\dfrac{\pi^3}{16}-2G\ln 2+\dfrac{1}{4}\pi(\ln 2)^2\\ &=B+\dfrac{\pi^3}{16}-2G\ln 2-2\int_0^1 \dfrac{\ln x\ln(1+x)}{1+x^2}dx \end{align}$

From Evaluating $\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$

using (6),

$\boxed{\displaystyle \int_0^1 \dfrac{\ln x\ln(1+x)}{1+x^2}dx=\dfrac{\pi^3}{256}-\dfrac{1}{2}G\ln 2+\dfrac{9}{64}\pi(\ln 2)^2-\dfrac{3}{2}J}$

using (4),

$\boxed{B=\dfrac{1}{4}\pi(\ln 2)^2-2J}$

Therefore,

$\boxed{A=J+\dfrac{7}{128}\pi^3-\dfrac{1}{32}\pi(\ln 2)^2-G\ln 2}$

Therefore,

$\begin{align}2A+B&=\left(2J+\dfrac{7}{64}\pi^3-\dfrac{1}{16}\pi(\ln 2)^2-2G\ln 2\right)+\dfrac{1}{4}\pi(\ln 2)^2-2J\\ &=\boxed{\dfrac{7}{64}\pi^3+\dfrac{3}{16}\pi(\ln 2)^2-2G\ln 2} \end{align}$