A Challenging Logarithmic Integral $\int_0^1 \frac{\log(x)\log(1-x)\log^2(1+x)}{x}dx$

How can we prove that:

$$\int_0^1 \frac{\log(x)\log(1-x)\log^2(1+x)}{x}dx=\frac{7\pi^2}{48}\zeta(3)-\frac{25}{16}\zeta(5)$$ where $\zeta(z)$ is the Riemann Zeta Function.

The best I could do was to express it in terms of Euler Sums. Let $I$ denote the integral.

$$I=-\frac{\pi^2}{24}\zeta(3)+2\sum_{r=2}^\infty \frac{(-1)^r (H_r)^2}{r^3}-2\sum_{r=2}^\infty \frac{(-1)^r H_r}{r^4}+2 \sum_{r=2}^\infty \frac{(-1)^r H_r^{(2)}H_r}{r^2}-2\sum_{r=2}^\infty \frac{(-1)^r H_r^{(2)}}{r^3}$$

where $\displaystyle H_r^{(n)}=\sum_{n=1}^r \frac{1}{k^n}$. I am unable to simplify these sums further. Does anyone have any idea on how to solve this integral?

Please see here for more details.


Some time ago I was able to solve the simpler integral:

\begin{align*} \int_0^1 \frac{\log(1-x)\log(x)\log(1+x)}{x}dx &=-\frac{3 \pi^4}{160}+\frac{7\log(2)}{4}\zeta(3)-\frac{\pi^2 \log^2(2)}{12} +\frac{\log^4(2)}{12} \\ &\quad+ 2 \text{Li}_4 \left(\frac{1}{2} \right) \end{align*} where $\text{Li}_n(z)$ is the Polylogarithm.

Answer

Let the considered integral be denoted by $I$. Our starting point is to reduce the number of logarithms of different arguments in the integrand. Thus, using the fact
that $6ab^2=(a+b)^3-2a^3+(a-b)^3$ we obtain
\begin{align*}
6I&=\underbrace{\int_0^1\frac{\log x}{x}\log^3(1-x^2)dx}_{x\leftarrow \sqrt{u}}
-2\int_0^1\frac{\log x}{x}\log^3(1-x)dx+
\underbrace{\int_0^1\frac{\log x}{x}\log^3\left(\frac{1-x}{1+x}\right)dx}_{x\leftarrow\frac{1-u}{1+u}}
\\
&= \frac{1}{4}\int_0^1\frac{\log u}{u}\log^3(1-u)du
-2\int_0^1\frac{\log x}{x}\log^3(1-x)dx+
2\int_0^1\frac{\log^3 u}{1-u^2}\log\left(\frac{1-u}{1+u}\right)du \\
&=\frac{-7}{4}\int_0^1\frac{\log x}{x}\log^3(1-x)dx+
2\int_0^1\frac{\log^3 x}{1-x^2}\log\left(\frac{1-x}{1+x}\right)dx \\
\end{align*}
Thus,
$$
I=-\frac{7}{24}J+\frac{1}{3}K \tag{1}
$$
with
$$
J=\int_0^1\frac{\log(1- x)}{1-x}\log^3x\,dx,\quad
K=\int_0^1\frac{1}{1-x^2}\log\left(\frac{1-x}{1+x}\right)\log^3 x\,dx
\tag{2}
$$
Now, for $x\in(-1,1)$, we have
$$
\frac{\log(1-x)}{1-x}=-\left(\sum_{k=0}^\infty x^k\right)\left(\sum_{k=1}^\infty \frac{x^k}{k}\right)
=-\sum_{n=1}^\infty H_nx^n
$$
where $H_n=\sum_{k=1}^n1/k$ is the $n$-th Harmonic number. Similarly,
$$
\frac{1}{1-x^2}\log\left(\frac{1-x}{1+x}\right)
=-\left(\sum_{k=0}^\infty x^{2k}\right)\left(\sum_{k=1}^\infty \frac{-2x^{2k-1}}{2k-1}\right)
=-2\sum_{n=1}^\infty \widetilde{H}_nx^{2n-1}
$$
where
$$\widetilde{H}_n=\sum_{k=1}^n\frac{1}{2k-1}
=H_{2n}-\frac{1}{2}H_n$$
Thus, using the fact that $\int_0^1x^n(-\log x)^3\,dx=-6/(n+1)^4$ we conclude that
\begin{align*}
J&=\sum_{n=1}^\infty H_n\int_0^1x^n(-\log x)^3\,dx=6\sum_{n=1}^\infty
\frac{H_n}{(n+1)^4}\\
&=6\sum_{n=0}^\infty
\frac{1}{(n+1)^4}\left(H_{n+1}-\frac{1}{n+1}\right)=6A-6\zeta(5)\tag{3}
\end{align*}
with
\begin{equation*}
A=\sum_{n=1}^\infty\frac{H_n}{n^4}\tag{4}
\end{equation*}
Similarly,
\begin{align*}
K&=\sum_{n=1}^\infty (2H_{2n}-H_n)\int_0^1x^{2n-1}(-\log x)^3\,dx=6\sum_{n=1}^\infty\frac{2H_{2n}-H_n}{(2n)^4}\\
&=6\sum_{n=1}^\infty\frac{(1+(-1)^{2n})H_{2n}}{(2n)^4}
-\frac{3}{8}\sum_{n=1}^\infty\frac{H_n}{n^4}\\
&=6\sum_{n=1}^\infty\frac{(1+(-1)^{n})H_{n}}{n^4}
-\frac{3}{8}\sum_{n=1}^\infty\frac{H_n}{n^4}\\
&=\frac{45}{8}\sum_{n=1}^\infty\frac{H_{n}}{n^4}
+6\sum_{n=1}^\infty\frac{(-1)^{n}H_{n}}{n^4}=\frac{45}{8}A+6B\tag{5}
\end{align*}
with
\begin{equation*}
B=\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}\tag{6}
\end{equation*}
Combining (1) with (3) and (5) we find that
\begin{equation*}
I=\frac{1}{8}A+2B+\frac{7}{4}\zeta(5)\tag{7}
\end{equation*}
Now, sums $A$ and $B$ are known (see here ) (in a more general setting), and
we have
$$
A=3\zeta(5)-\zeta(2)\zeta(3),\qquad
B=-\frac{59}{32}\zeta(5)+\frac{1}{2}\zeta(2)\zeta(3)
$$
Thus
$$
I=\frac{7}{8}\zeta(2)\zeta(3)-\frac{25}{16}\zeta(5).
$$
which is the announced result.$\qquad\square$

Attribution
Source : Link , Question Author : Shobhit , Answer Author : Omran Kouba

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