I have spent my holiday on Sunday to crack several integral & series problems and I am having trouble to prove the following integral
\begin{equation}
\int_0^1 \frac{{\rm{Li}}_2(x)\ln(1-x)\ln^2(x)}{x} \,dx=-\frac{\zeta(6)}{3}
\end{equation}
Using integration by parts, $u=\ln^2(x)$ and $dv=\displaystyle\frac{{\rm{Li}}_2(x)\ln(1-x)}{x} \,dx$, I manage to obtain that the integral is equivalent to
\begin{equation}
\int_0^1 \frac{{\rm{Li}}_2^2(x)\ln(x)}{x} \,dx
\end{equation}
where ${\rm{Li}}_2^2(x)={\rm{Li}}_2(x)^2$, square of dilogarithm of $x$.
Could anyone here please help me to prove the above integral preferably with elementary ways (high school methods/ not residue method)? Any help would be greatly appreciated. Thank you.
Best Answer
Hint.
Observe that $$ \frac{d}{dx}{\rm{Li}}_{p+1}(x)=\frac{{\rm{Li}}_{p}(x)}{x} \tag1$$ and $$ \int_0^1x^\alpha\:{\rm{Li}}_p(x) \:{\rm{d}}x = \sum_{n=1}^{\infty} \frac{1}{n^p}\int_0^1x^{\alpha+n}\:{\rm{d}}x=\sum_{n=1}^{\infty} \frac{1}{n^p(n+\alpha+1)}, \quad \alpha>-2.\tag2 $$ Differentiating $(2)$ twice w. r. t. $\alpha$ gives $$ \int_0^1x^\alpha\:{\rm{Li}}_p(x) \ln x \:{\rm{d}}x =-\sum_{n=1}^{\infty} \frac{1}{n^p(n+\alpha+1)^2}\tag3 $$ $$ \int_0^1x^\alpha\:{\rm{Li}}_p(x) \ln^2 x \:{\rm{d}}x =2\sum_{n=1}^{\infty} \frac{1}{n^p(n+\alpha+1)^3}.\tag4 $$ Now set $\displaystyle I:=\int_0^1 \frac{{\rm{Li}}_2(x)\ln(1-x)\ln^2(x)}{x} {\rm{d}}x .$
We may write $$ \begin{align} I &=\int_0^1 {\rm{Li}'}_3(x)\ln(1-x)\ln^2x \:{\rm{d}}x\\ &=\left.{\rm{Li}}_3(x)\ln(1-x)\ln^2x \right|_0^1-\int_0^1 {\rm{Li}}_3(x)\left(\ln(1-x)\ln^2x \:{\rm{d}}x\right)'\:{\rm{d}}x\\ &=\int_0^1 {\rm{Li}}_3(x)\frac{\ln^2 x}{1-x} {\rm{d}}x-2\int_0^1 \frac{{\rm{Li}}_3(x)}{x}\ln(1-x)\ln x \:{\rm{d}}x\\ &=\int_0^1 {\rm{Li}}_3(x)\frac{\ln^2 x}{1-x} {\rm{d}}x-2\left(\left.{\rm{Li}}_4(x)\ln(1-x)\ln x \right|_0^1-\int_0^1 {\rm{Li}}_4(x)\left(\ln(1-x)\ln x \:{\rm{d}}x\right)'\:{\rm{d}}x\right)\\ &=\int_0^1 {\rm{Li}}_3(x)\frac{\ln^2 x}{1-x} {\rm{d}}x-2\int_0^1 {\rm{Li}}_4(x)\frac{\ln x}{1-x} {\rm{d}}x+2\int_0^1 {\rm{Li}}_4(x)\frac{\ln (1-x)}{x} {\rm{d}}x\\ &=\int_0^1 {\rm{Li}}_3(x)\frac{\ln^2 x}{1-x} {\rm{d}}x-2\int_0^1 {\rm{Li}}_4(x)\frac{\ln x}{1-x} {\rm{d}}x+2\int_0^1 \frac{{\rm{Li}}_5(x)-\zeta(5)}{1-x} {\rm{d}}x. \end{align} $$ Then, using $(4)$, $$ \int_0^1\!\! {\rm{Li}}_3(x)\frac{\ln^2 x}{1-x} {\rm{d}}x =\! \sum_{n=0}^{\infty}\!\int_0^1 x^n{\rm{Li}}_3(x)\ln^2 x {\rm{d}}x=2\sum_{n,k\geq 1}^{\infty} \!\frac{1}{k^3(k+n)^3}=\zeta^2(3)-\zeta(6) $$ similarly, using $(3)$, $$ -2\!\int_0^1\!\! \!\! {\rm{Li}}_4(x)\frac{\ln x}{1-x} {\rm{d}}x =-2\! \sum_{n=0}^{\infty}\!\int_0^1\!\! x^n{\rm{Li}}_4(x)\ln x {\rm{d}}x=2\!\sum_{n,k\geq 1}^{\infty} \!\frac{1}{k^4(k+n)^2}=-2\zeta^2(3)+\!\frac{25\zeta(6)}{6}^{*} $$ and using $(2)$, $$ 2\!\!\int_0^1 \frac{{\rm{Li}}_5(x)-\zeta(5)}{1-x} {\rm{d}}x =\! 2\!\sum_{n=0}^{\infty}\!\int_0^1 \!\!x^n\left({\rm{Li}}_5(x)-\zeta(5)\right) {\rm{d}}x =\!-2\!\sum_{n,k\geq 1}^{\infty} \!\frac{1}{k^4n(k+n)}=-2\!\sum_{k=1}^{\infty}\! \frac{H_k}{k^5} $$ From Euler's standard formula we get $$ -2\sum_{k=1}^{\infty}\! \frac{H_k}{k^5}=\zeta^2(3)-\frac{7\zeta(6)}{2} $$ Putting all this together, we end up with
$^*$We readily have $$ \begin{align} 2\sum_{n,k\geq 1}^{\infty} \!\frac{1}{k^4(k+n)^2}&=2\sum_{k\geq 1}^{\infty} \!\frac{1}{k^4}\sum_{n=1}^{\infty}\frac{1}{(n+k)^2} \\&=2\sum_{k\geq 1}^{\infty} \!\frac{1}{k^4}\left(\zeta(2)-\sum_{n=1}^{k}\frac{1}{n^2}\right) \\&=2\sum_{k\geq 1}^{\infty} \!\frac{1}{k^4}\left(\zeta(2)-\frac{1}{k^2}-\sum_{n=1}^{k-1}\frac{1}{n^2}\right) \\&=2\zeta(2)\zeta(4)-2\zeta(6)-2\zeta(4,2) \\&=\frac32\zeta(6)-2\zeta(4,2) \end{align} $$ where $\zeta(4,2)$ denotes a Multi Zeta Values (MZVs) namely $$ \zeta(4,2):=\sum_{k\geq 1}^{\infty}\frac{1}{k^4}\sum_{n=1}^{k-1}\frac{1}{n^2}. $$ Obtaining a reduction formula for $\zeta(4,2)$ is a tough part in this evaluation. We have $$ \color{purple}{\zeta(4,2)=\zeta^2(3)-\frac43\zeta(6)}, $$ and a proof, using MZVs algebra, may be found here [p. 8 (3.10)].