Evaluate $\int_0^1x\log\left(1+x^2\right)\left[\log\left(\frac{1-x}{1+x}\right)\right]^3\operatorname{d}\!x$

Question

Prove that $$\int_0^1x\log\left(1+x^2\right)\left[\log\left(\frac{1-x}{1+x}\right)\right]^3\operatorname{d}\!x$$ $$I=\frac{105}{8}\zeta(3)-\frac{7\pi^2}{4}\ln2-6\pi{G}+3\pi^2-\frac{3}{8}\pi^3-\frac{7}{64}\pi^4$$ where $G$ is the Catalan's constant. Additional reponse at FDP You are almost at the end. $\mathbf{I}\mathbf{\,continue}$ Using integration by parts, we have: $C=3\displaystyle\int_0^1\dfrac{\log^2x\log(1+x^2)}{x}dx-3\displaystyle\int_0^1\dfrac{\log^2x\log(1+x^2)}{1+x}dx-\displaystyle\int_0^1\dfrac{log^3x}{1+x}dx-\displaystyle\int_0^1\dfrac{log^3x}{1+x^2}dx+\displaystyle\int_0^1\dfrac{x\log^3x}{1+x^2}dx$

$D=\displaystyle\dfrac{3}{2}\int_0^1\dfrac{\log^2x\log(1+x^2)}{x}dx-\displaystyle\dfrac{3}{2}\int_0^1\dfrac{\log^2x\log(1+x^2)}{1+x}dx-\displaystyle\dfrac{3}{2}\int_0^1\dfrac{\log^2x\log(1+x^2)}{(1+x)^2}dx-\displaystyle\dfrac{1}{2}\int_0^1\dfrac{log^3x}{(1+x)^2}dx+\displaystyle\dfrac{1}{2}\int_0^1\dfrac{log^3x}{1+x^2}dx$ $2D-C=\displaystyle\int_0^1\dfrac{(1-x)\log^3x\log(1+x^2)}{(1+x)^3}dx=-3\displaystyle\int_0^1\dfrac{log^2x\log(1+x^2)}{(1+x)^2}dx+\displaystyle\int_0^1\dfrac{log^3x}{1+x}dx-\displaystyle\int_0^1\dfrac{xlog^3x}{1+x^2}dx$. } But,$\displaystyle\int_0^1\dfrac{log^3x}{1+x}dx=-\dfrac{7\pi^4}{120}$; and$ \displaystyle\int_0^1\dfrac{xlog^3x}{1+x^2}dx.=-\dfrac{7\pi^4}{1920}$

Remains to be assessed: $ \displaystyle\int_0^1\dfrac{log^2x\log(1+x^2)}{(1+x)^2}dx$, to be continued . Also, $\displaystyle\int_0^1\dfrac{\log^2x\log(1+x^2)}{(1+x)^2}dx=2\displaystyle\int_0^1\dfrac{\log\,x\log(1+x^2)}{x}dx-2\displaystyle\int_0^1\dfrac{\log\,x\log(1+x^2)}{(1+x)}dx-\displaystyle\int_0^1\dfrac{log^2x}{1+x}dx+\displaystyle\int_0^1\dfrac{log^2x}{1+x^2}dx+\displaystyle\int_0^1\dfrac{x\log^2x}{1+x^2}dx$

These integrals are know

$\displaystyle\int_0^1\dfrac{\log\,x\log(1+x^2)}{x}dx=-\dfrac{3}{16}\zeta(3)$; $\displaystyle\int_0^1\dfrac{\log\,x\log(1+x^2)}{(1+x)}dx=-\dfrac{\pi(G)}{2}-\dfrac{\pi^2}{16}ln2+\dfrac{3}{2}\zeta(3)$

Also, $\displaystyle\int_0^1\dfrac{\log^2x\log(1+x^2)}{(1+x)^2}dx=2\displaystyle\int_0^1\dfrac{\log\,x\log(1+x^2)}{x}dx-2\displaystyle\int_0^1\dfrac{\log\,x\log(1+x^2)}{(1+x)}dx-\displaystyle\int_0^1\dfrac{log^2x}{1+x}dx+\displaystyle\int_0^1\dfrac{log^2x}{1+x^2}dx+\displaystyle\int_0^1\dfrac{x\log^2x}{1+x^2}dx$

These integrals are know

$\displaystyle\int_0^1\dfrac{\log\,x\log(1+x^2)}{x}dx=-\dfrac{3}{16}\zeta(3)$; $\displaystyle\int_0^1\dfrac{\log\,x\log(1+x^2)}{(1+x)}dx=-\dfrac{\pi}{2}{G}-\dfrac{\pi^2}{16}ln2+\dfrac{3}{2}\zeta(3);\displaystyle\int_0^1\dfrac{log^2x}{1+x}dx=\dfrac{3}{2}\zeta(3); \displaystyle\int_0^1\dfrac{log^2x}{1+x^2}dx=\dfrac{\pi^3}{16}$

Finally,$\displaystyle\int_0^1\dfrac{log^2x\log(1+x^2)}{(1+x)^2}dx=\dfrac{75}{16}\zeta(3)+{\pi}G+\dfrac{\pi^2}{8}ln2+\dfrac{\pi^3}{16}$

And 2D-C=$\dfrac{297}{16}\zeta(3)-3{\pi}G-\dfrac{3}{8}{\pi^2}ln2-\dfrac{3}{16}{\pi^3}-\dfrac{7}{128}{\pi^4}$ We obtain also the result.

Answer 1

I don't have a complete answer.

Let $\displaystyle I=\int_0^1 x\log(1+x^2)\Big[\log\Big(\dfrac{1-x}{1-x}\Big)\Big]^3dx$

First perform the change of variable $y=\dfrac{1-x}{1+x}$.

$\displaystyle I=\int_0^1 \dfrac{2(1-x)}{(1+x)^3}\log\Big(\dfrac{2(1+x^2)}{(1+x)^2}\Big)\log^3xdx$

Notice: $\dfrac{1-x}{(1+x)^3}=\dfrac{2}{(1+x)^2}-\dfrac{1}{(1+x)^3}$

Therefore:

$I=4\log 2\times B-2\log 2\times A+4D-2C-8F+4E$

Where:

$A=\displaystyle \int_0^1\dfrac{\log^3x}{(1+x)^2}dx$

$B=\displaystyle \int_0^1\dfrac{\log^3x}{(1+x)^3}dx$

$C=\displaystyle \int_0^1 \dfrac{\log(1+x^2)\log^3x}{(1+x)^2}dx$

$D=\displaystyle \int_0^1 \dfrac{\log(1+x^2)\log^3x}{(1+x)^3}dx$

$E=\displaystyle \int_0^1 \dfrac{\log(1+x)\log^3x}{(1+x)^2}dx$

$F=\displaystyle \int_0^1 \dfrac{\log(1+x)\log^3x}{(1+x)^3}dx$

One can continue like this, using integration by parts:

$C=\displaystyle\int_0^1 \dfrac{3\log^2x\log(1+x^2)}{x}dx-\int_0^1 \dfrac{3\log^2x\log(1+x^2)}{1+x}dx+\int_0^1 \dfrac{x\log^3 x}{1+x}dx-\int_0^1 \dfrac{x^2\log^3 x}{1+x^2}dx+\int_0^1 \dfrac{x\log^3 x}{1+x^2}dx$

Let's compute $A$:

$\displaystyle A=\int_0^1 \Big[\sum_{n=0}^{+\infty}(n+1)(-1)^nx^n\Big]\log^3 x dx$

$\displaystyle A=-6\sum_{n=0}^{+\infty} \dfrac{(-1)^n}{(n+1)^3}$

$A=-6\times (1-2^{1-3})\zeta(3)=-\dfrac{9}{2}\zeta(3)$

Let's compute $B$:

$\displaystyle B=\dfrac{1}{2}\int_0^1 \Big[(-1)^n(n+2)(n+1)x^n\Big]\log^3 x dx$

$\displaystyle B=-3\sum_{n=0}^{+\infty} \dfrac {(-1)^n}{(n+1)^2}-3\sum_{n=0}^{+\infty} \dfrac {(-1)^n}{(n+1)^3}$

$B=-3\times(1-2^{1-2})\zeta(2)-3\times(1-2^{1-3})\zeta(3)=-\dfrac{3}{2}\zeta(2)-\dfrac{9}{4}\zeta(3)$

$B=-\dfrac{\pi^2}{4}-\dfrac{9}{4}\zeta(3)$

Let's continue.

(it's a bad idea to try to compute $E$ by the way. happily, we're lucky, we don't have to do this)

$\displaystyle E=3\int_0^1 \dfrac{\log^2(x)\log(1+x)}{x}dx-3\int_0^1 \dfrac{\log^2(x)\log(1+x)}{1+x}dx+A$

Let $E_1:=\displaystyle \int_0^1 \dfrac{\log^2(x)\log(1+x)}{x}dx$

Let $\displaystyle E_2:=\int_0^1 \dfrac{\log^2(x)\log(1+x)}{1+x}dx$

(believe me, you don't want to compute $E_2$)

Using integration by parts and partial fraction decomposition: $\displaystyle -8F=12E_2-12E_1+24\int_0^1 \dfrac{\log x\log(1+x)}{x}dx-24\int_0^1 \dfrac{\log x\log(1+x)}{1+x}dx+12\times\int_0^1\dfrac{\log^2x}{(1+x)^2}dx-4B$

Let $\displaystyle F_1=\int_0^1 \dfrac{\log x\log(1+x)}{x}dx$

$\displaystyle F_1=\int_0^1 \Big[\sum_{n=0}^{+\infty}(-1)^n\dfrac{x^{n+1}}{n+1}\Big]\dfrac{\log x}{x}dx$

$F_1=\displaystyle -\sum_{n=0}^{+\infty}\dfrac{(-1)^n}{(n+1)^3}=-(1-2^{1-3})\zeta(3)=-\dfrac{3}{4}\zeta(3)$

Let $\displaystyle F_2=\int_0^1 \dfrac{\log x\log(1+x)}{1+x}dx=-\dfrac{1}{8}\zeta(3)$

(based upon the computation of $K$ in: Other challenging logarithmic integral $\int_0^1 \frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx$ )

Another way to compute $F_2$ knowing that:

$Li_2\Big(\dfrac{1}{2}\Big)=\dfrac{1}{12}\Big(\pi^2-6(\log 2)^2\Big)$

$Li_3\Big(\dfrac{1}{2}\Big)=\dfrac{1}{24}\Big(4(\log 2)^3-2\pi^2\log 2+21\zeta(3)\Big)$

Performing the change of variable $y=\log(1+x)$:

$\displaystyle F_2=\int_0^{\log 2} x\log(e^x-1)dx=\int_0^{\log 2} x^2dx+\int_0^{\log 2} x\log(1-e^{-x})dx$

$\displaystyle F_2=\dfrac{(\log 2)^3}{3}-\sum_{n=1}^{+\infty}\Big(\int_0^{\log 2}\dfrac{xe^{-nx}}{n}dx\Big)$

$\displaystyle F_2=\dfrac{(\log 2)^3}{3}-\sum_{n=1}^{+\infty}\dfrac{1}{n^3}+\log 2\sum_{n=1}^{+\infty}\dfrac{2^{-n}}{n^2}+\sum_{n=1}^{+\infty}\dfrac{2^{-n}}{n^3}$

$\displaystyle F_2=\dfrac{(\log 2)^3}{3}-\zeta(3)+\log 2 Li_2\Big(\dfrac{1}{2}\Big)+Li_3\Big(\dfrac{1}{2}\Big)=-\dfrac{1}{8}\zeta(3)$

Let $\displaystyle F_3=\int_0^1\dfrac{\log^2x}{(1+x)^2}dx$

$\displaystyle F_3=\int_0^1 \Big[\sum_{n=0}^{+\infty}(n+1)(-1)^nx^n\Big]\log^2 x dx$

$\displaystyle F_3=2\sum_{n=0}^{+\infty} \dfrac{(-1)^n}{(n+1)^2}$

$F_3=2\times (1-2^{1-2})\zeta(2)=\zeta(2)=\dfrac{\pi^2}{6}$

Therefore:

$-8F+4E=24F_1-24F_2+12F_3-4B+4A=-24\zeta(3)+3\pi^2$

The harder part remains to do !