Other challenging logarithmic integral $\int_0^1 \frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx$

How can we prove that:

$$\int_0^1\frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx=\frac{\pi^2}{8}\zeta(3)-\frac{27}{16}\zeta(5) $$

Solutions Collecting From Web of "Other challenging logarithmic integral $\int_0^1 \frac{\log^2(x)\log(1-x)\log(1+x)}{x}dx$"

Performing integration by parts by taking $u=\ln(1-x)\ln(1+x)$, then
$$\begin{align}
\int_0^1\frac{\ln^2x\ln(1-x)\ln(1+x)}{x}\ dx&=\frac{1}{3}\bigg[\int_0^1\frac{\ln(1+x)\ln^3x}{1-x}\ dx-\int_0^1\frac{\ln(1-x)\ln^3x}{1+x}\ dx\bigg]\\
&=\frac{1}{3}\bigg[I-J\bigg]\\
\end{align}$$
Evaluation of $I$ :
$$\begin{align}
I&=\int_0^1\frac{\ln^3{x}\big[\ln(1-x^2)-\ln(1-x)\big]}{1-x}\,dx\\
&=\int_0^1\frac{(1+x)\ln^3{x}\ln(1-x^2)}{(1+x)(1-x)}\,dx-\int_0^1\frac{\ln^3{x}\ln(1-x)}{1-x}\,dx\\
&=\int_0^1\frac{\ln^3{x}\ln(1-x^2)}{1-x^2}\,dx+\int_0^1\frac{x\ln^3{x}\ln(1-x^2)}{1-x^2}\,dx-\int_0^1\frac{\ln^3{x}\ln(1-x)}{1-x}\,dx\\
&=\frac{1}{16}\int_0^1\frac{x^{-\frac{1}{2}}\ln^3{x}\ln(1-x)}{1-x}\,dx-\frac{15}{16}\int_0^1\frac{\ln^3{x}\ln(1-x)}{1-x}\,dx\\
&=\frac{1}{16}\int_0^1\frac{x^{-\frac{1}{2}}\ln^3{x}\ln(1-x)}{1-x}\,dx-\frac{15}{16}\int_0^1\frac{\ln^3{x}\ln(1-x)}{1-x}\,dx\\
&=\frac{1}{16}\lim_{\mu \to \frac{1}{2}\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\int_0^1 x^{\mu-1}(1-x)^{\nu-1}\,dx-\frac{15}{16}\lim_{\mu \to 1\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\int_0^1 x^{\mu-1}(1-x)^{\nu-1}\,dx\\
&=\frac{1}{16}\lim_{\mu \to \frac{1}{2}\,,\, \nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\text{B}(\mu,\nu)-\frac{15}{16}\lim_{\mu \to 1\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\text{B}(\mu,\nu)\\
&=12\zeta(5)-\frac{3\pi^2}{8}\zeta(3)-\frac{\pi^4}{8}\ln{2}
\end{align}$$
Evaluation of $J$ :
\begin{align}
J&=\int_0^1\frac{\ln^3{x}\big[\ln(1-x^2)-\ln(1+x)\big]}{1+x}\,dx\\
&=\int_0^1\frac{(1-x)\ln^3{x}\ln(1-x^2)}{(1-x)(1+x)}\,dx-\int_0^1\frac{\ln^3{x}\ln(1+x)}{1+x}\,dx\\
&=\int_0^1\frac{\ln^3{x}\ln(1-x^2)}{1-x^2}\,dx-\int_0^1\frac{x\ln^3{x}\ln(1-x^2)}{1-x^2}\,dx-K\\
&=-\frac{1}{16}\int_0^1\frac{x^{-\frac{1}{2}}\ln^3{x}\ln(1-x)}{1-x}\,dx+\frac{1}{16}\int_0^1\frac{\ln^3{x}\ln(1-x)}{1-x}\,dx-K\\
&=-\frac{1}{16}\lim_{\mu \to \frac{1}{2}\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\int_0^1 x^{\mu-1}(1-x)^{\nu-1}\,dx+\frac{1}{16}\lim_{\mu \to 1\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\int_0^1 x^{\mu-1}(1-x)^{\nu-1}\,dx-K\\
&=-\frac{1}{16}\lim_{\mu \to \frac{1}{2}\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\text{B}(\mu,\nu)+\frac{1}{16}\lim_{\mu \to 1\,,\,\nu \to 0}\frac{\partial^4}{\partial\mu^3\partial\nu}\text{B}(\mu,\nu)-K\\
&=\frac{45}{2}\zeta(5)-\frac{5\pi^2}{4}\zeta(3)-\frac{\pi^4}{8}\ln 2-K
\end{align}
where
\begin{align}
K&=\int_0^1\frac{\ln^3{x}\ln(1+x)}{1+x}\,dx\\
&=\int_0^1\sum_{k=1}^\infty (-1)^{k-1}H_{k}\,x^k\ln^3x\,dx\tag{1}\\
&=\sum_{k=1}^\infty (-1)^{k-1}H_{k}\int_0^1x^k\ln^3x\,dx\\
&=-6\sum_{k=1}^\infty (-1)^{k-1}\frac{H_{k}}{(k+1)^4}\tag{2}\\
&=-6\sum_{k=1}^\infty (-1)^{k-1}\left[\frac{H_{k+1}}{(k+1)^4}-\frac{1}{(k+1)^5}\right]\tag{3}\\
&=6\sum_{k=1}^\infty (-1)^{k-1}\left[\frac{H_{k}}{k^4}-\frac{1}{k^5}\right]\\
&=9\eta(5)-3\zeta(5)-6\eta(2)\zeta(3)\tag{4}\\
&=\frac{87}{16}\zeta(5)-\frac{\pi^2}{2}\zeta(3)\tag{5}
\end{align}

Putting these together, we will get the desired result.


Explanation :

$(1)$ Use generating function $\displaystyle\sum_{k=1}^\infty (-1)^{k-1}H_{k}\,x^k=\frac{\ln(1+x)}{1+x}$

$(2)$ Use formula $\displaystyle\int_0^1 x^k \ln^n x\ dx=\frac{(-1)^n n!}{(k+1)^{n+1}}\quad,\ n\in\mathbb{Z}_{n\ge0}$

$(3)$ Use property $\displaystyle H_{k}=H_{k+1}-\frac{1}{k+1}$

$(4)$ Use formula $\displaystyle \sum_{k=1}^\infty (-1)^{k-1}\frac{H_{k}}{k^{2n}}=\frac{(2n+1)\eta(2n+1)}{2}-\sum_{k=0}^{n-1}\eta(2k)\zeta(2n+1-2k)\,,\,n\in\mathbb{Z}_{n\ge1}$

$(5)$ Use property of Dirichlet eta function $\displaystyle \eta(s)=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^s}=\left(1-2^{1-s}\right)\zeta(s)$