# 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.

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.$\qquad\square$