Prove that $\int_0^1 \frac{{\rm{Li}}_2(x)\ln(1-x)\ln^2(x)}{x} \,dx=-\frac{\zeta(6)}{3}$

Question

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.

Answer 1

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

$$ \int_0^1 \frac{{\rm{Li}}_2(x)\ln(1-x)\ln^2(x)}{x} {\rm{d}}x=-\zeta(6)+\frac{25\zeta(6)}{6}-\frac{7\zeta(6)}{2}=-\frac{\zeta(6)}{3}. $$

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$^*$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)].

Answer 2

I will use definitions and methods from P. Freitas' paper here. Sometimes I will refer to the paper as $[\text{F}]$.

$K$ and $J$ functions. First of all we define $K$ and $J$ functions as the following.

$$\begin{align} K(r,p,q):=& \int_0^1\frac{\ln^r(x)\operatorname{Li}_p(x)\operatorname{Li}_q(x)}{x}\,dx, \\ J(m,p,q):=& \int_0^1 x^m\operatorname{Li}_p(x)\operatorname{Li}_q(x)\,dx. \end{align}$$

Note that $K(r,p,q)=K(r,q,p)$ and $J(m,p,q)=J(m,q,p)$. Furthermore $K(0,p,q)=J(-1,p,q)$ and we could reduce it to rational constans and zeta values at positive integers as we will show it in Result $2$. There is a much more general result about $J$ in $[\text{F}]$ Theorem $1$, that we also could simplify it to rational constants and zeta values at positive integers for any $(m,p,q) \in \mathbb{Z}^3$, with $p,q\geq 1$ and $m \geq -2$.

As you have shown your problem is equivalent to find $K(1,2,2).$

$S_{p,q}$ function. Let $S_{p,q}$ be the family of the following linear sums

$$S_{p,q} := \sum_{n=1}^{\infty} \frac{H_n^{(p)}}{n^q},$$

where $H_n^{(p)}:=\sum_{j=1}^n j^{-p}$ finite sums are the generalized harmonic numbers.

We will use some results from Freitas' paper.

Result $1$. For the $K$ function it is true, that

$$K(r,p,q)=-rK(r-1,p,q+1)-K(r,p-1,q+1).$$

For the proof see $[\text{F}]$ Lemma $3.7$ at page $9$, and the proof of Theorems $1$ and $2$ also at page $9$. By applying this repeatedly we see that we are able to reduce the original integral to integrals of the form $K(0,p',q')=J(-1,p',q')$ and integrals $K(i,0,p+q+r-i)$ with $i=1,\dots,r$.

Result $2$. For $p,q \geq 1$ the integral $J(-1,p,q)$ is reducible to zeta values. If $p \geq q$, we have $$J(-1,p,q) = (-1)^{q+1}\left(1+\frac{p+q}{2}\right)\zeta(p+q+1)+2(-1)^q\sum_{j=1}^{\lfloor q/2\rfloor} \zeta(2j)\zeta(p+q-2j+1)+\frac{(-1)^q}{2}\sum_{j=1}^{p-q} \zeta(j+q)\zeta(p-j+1).$$

Since $J(-1,p,q)=J(-1,q,p)$ there is a similar expression for $p<q$. More details at $[\text{F}]$ Theorem $3.3$ at page $6\!-\!7$. You will also find the proof of the statement there.

Result $3$. We also want to simplify $K(r,0,q)$, which expression appers while using Result $1$. For the case when $w=r+q$ weight is even, there is a nice expression at $[\text{F}]$ Theorem $3.1$ at page $5$, with we can simplify $K(r,0,q)$. Sadly according to $[\text{F}]$ there is not a general expression in case when $w=r+q$ is odd. While evaluating your integral we will run into this odd case. But luckily we are not completly lost! As $[\text{F}]$ Lemma $2.1$ at page $4$ says, there is a connection between $S_{r,q}$ and $K(r-1,0,q)$. This is the following. For $q,r \geq 2$ we have

$$S_{r,q}=\zeta(r)\zeta(q)-\frac{(-1)^{r-1}}{(r-1)!} \int_0^1 \frac{\ln^{r-1}(x)\operatorname{Li}_q(x)}{1-x}\,dx=\zeta(r)\zeta(q)-\frac{(-1)^{r-1}}{(r-1)!}K(r-1,0,q),$$

or express it to $K(r-1,0,q)$ we get

$$K(r-1,0,q)=(-1)^{1-r}(r-1)!\left(\zeta(r)\zeta(q)-S_{r,q}\right).$$

You also find the proof at page $4$.

Evaluating your integral. First we will use Result $1$ twice. First time for $K(1,2,2)$ and second time $K(1,1,3)$. Thus

$$K(1,2,2)=-K(0,2,3)-K(1,1,3)=-\color{red}{K(0,2,3)}+\color{green}{K(0,1,4)}+\color{blue}{K(1,0,4)}.$$

Evaluating $\color{red}{K(0,2,3)}$ and $\color{green}{K(0,1,4)}$. From the definition of $K$ and $J$ we know that $K(0,2,3)=J(-1,2,3)$ and $K(0,1,4)=J(-1,1,4)$. Now we can use Result $2$. I know that you don't like using CAS, but because Result $2$ contains only elementary operations, such as finite sums, I let this calculation to Maple. I've defined the following function.

Je:=(p,q)->(-1)^(q+1)*(1+(p+q)/2)*Zeta(p+q+1)+2*(-1)^q*sum(Zeta(2*j)*Zeta(p+q-2*j+1),j=1..floor(q/2))+((-1)^q/2)*sum(Zeta(j+q)*Zeta(p-j+1),j=1..p-q);

Je(2,3) and Je(1,4) will do the job. Using Maple or calcaulate it by hand we get

$$\begin{align} J(-1,2,3)=&\frac{1}{2}\zeta^2(3), \\ J(-1,1,4)=&\frac{7}{4}\zeta(6)-\frac{1}{2}\zeta^2(3) = \frac{1}{540}\pi^6 - \frac{1}{2} \zeta^2(3). \end{align}$$

Evaluating $\color{blue}{K(1,0,4)}$. Because $1+4=5$ is odd we could use just the connection to $S_{2,4}$. By Result $3$ we get

$$K(1,0,4)=S_{2,4}-\zeta(2)\zeta(4).$$

This paper by P. Flajolet and B. Salvy $[\text{FS}]$ states that

$$S_{2,4}=\sum_{n=1}^{\infty} \frac{H_n^{(2)}}{n^4} = \zeta^2(3)-\frac{1}{3}\zeta(6)=\zeta^2(3)-\frac{1}{2835}\pi^6.$$

Details about linear sums are also in $[\text{FS}]$. In $[\text{FS}]$ Theorem $3.1$ at page $22$ states that there is an expression to evaluate $S_{p,q}$ when $m=p+q$ weight is odd. This is an analogous situation what we have seen in Result $3$. Fortunately $[\text{FS}]$ page $22\!-\!23$ talks about the even case, which is our case for now, because $2+4=6$ is even. Using this we get

$$K(1,0,4)=\zeta^2(3)-\frac{1}{3}\zeta(6)-\zeta(2)\zeta(4)=\zeta^2(3)-\frac{25}{12}\zeta(6)=\zeta^2(3)-\frac{5}{2268}\pi^6.$$

Putting this all together.

$$K(1,2,2)=-\color{red}{K(0,2,3)}+\color{green}{K(0,1,4)}+\color{blue}{K(1,0,4)}=-\color{red}{\frac{1}{2}\zeta^2(3)}+\color{green}{\frac{7}{4}\zeta(6)-\frac{1}{2}\zeta^2(3)}+\color{blue}{\zeta^2(3)-\frac{25}{12}\zeta(6)}=-\frac{\zeta(6)}{3}=-\frac{\pi^6}{2835},$$

and this completes the proof.

Answer 3

Here's another approach similar to M.N.C.E.'s post,

We have,

$\displaystyle \begin{align} \int_0^1 \frac{\operatorname{Li_2}(x)\log (1-x) \log^2 x}{x}\,dx &= -\sum\limits_{n=1}^{\infty}\frac{1}{n}\int_0^1x^{n-1}\operatorname{Li_2}(x)\log^2 x\,dx \\ &= -\sum\limits_{n=1}^{\infty}\frac{1}{n}\sum\limits_{m=1}^{\infty} \frac{1}{m^2} \int_0^1 x^{m+n-1}\log^2 x\,dx \\ &= -2\sum\limits_{n,m=1}^{\infty} \frac{1}{nm^2(m+n)^3} \\ &= -\sum\limits_{n,m=1}^{\infty} \frac{1}{nm^2(m+n)^3}-\sum\limits_{n,m=1}^{\infty} \frac{1}{mn^2(m+n)^3} \\ &= -\sum\limits_{n,m=1}^{\infty} \frac{1}{n^2m^2(m+n)^2} = -\frac{1}{3}\zeta(6) \end{align}$

The last double summation is proved here.

Answer 4

By integrating by parts and expanding using $\displaystyle\small{ \sum^\infty_{n=1}H_n^{(p)}x^n=\frac{{\rm Li}_p(x)}{1-x}}$, we obtain \begin{align} \int^1_0\frac{{\rm Li}_2(x)\ln^2{x}\ln(1-x)}{x} &=\frac{1}{3}\int^1_0\frac{\ln^3{x}\ln^2(1-x)}{x}{\rm d}x+\frac{1}{3}\int^1_0\frac{{\rm Li}_2(x)\ln^3{x}}{1-x}{\rm d}x\\ &=\frac{1}{6}\int^1_0\frac{\ln^4{x}\ln(1-x)}{1-x}{\rm d}x+\frac{1}{3}\int^1_0\frac{{\rm Li}_2(x)\ln^3{x}}{1-x}{\rm d}x\\ &=-\frac{1}{6}\sum^\infty_{n=1}H_n\int^1_0x^n\ln^4{x}\ {\rm d}x+\frac{1}{3}\sum^\infty_{n=1}H_n^{(2)}\int^1_0x^n\ln^3{x}\ {\rm d}x\\ &=-4\sum^\infty_{n=1}\frac{H_n}{(n+1)^5}-2\sum^\infty_{n=1}\frac{H_n^{(2)}}{(n+1)^4}\\ &=-4\sum^\infty_{n=1}\frac{H_n}{n^5}-2\sum^\infty_{n=1}\frac{H_n^{(2)}}{n^4}+6\zeta(6)\\ \end{align} Using the following formulae, \begin{align} 2\sum^\infty_{n=1}\frac{H_n}{n^q}&=(q+2)\zeta(q+1)-\sum^{q-2}_{j=1}\zeta(j+1)\zeta(q-j)\tag1\\ \sum^\infty_{n=1}\frac{H_n^{(p)}}{n^q}&=\zeta(p)\zeta(q)+\zeta(p+q)-\sum^\infty_{n=1}\frac{H_n^{(q)}}{n^p}\tag2 \end{align} we can compute the two sums as such: \begin{align} \sum^\infty_{n=1}\frac{H_n}{n^5}=&\frac{7}{4}\zeta(6)-\frac{1}{2}\zeta^2(3)\\ \sum^\infty_{n=1}\frac{H_n^{(2)}}{n^4} =&\zeta(6)+\sum^\infty_{n=1}\frac{H_{n-1}^{(2)}}{n^4}\\\ =&\zeta(6)+\sum^\infty_{n=1}\frac{1}{n^4}\sum^{n-1}_{k=1}\frac{1}{(n-k)^2}\tag3\\ =&\zeta(6)+\sum^\infty_{k=1}\sum^\infty_{n=1}\frac{1}{n^2(n+k)^4}\tag4\\ =&\zeta(6)\color{#E2062C}{-\sum^\infty_{k=1}\frac{4}{k^5}\sum^\infty_{n=1}\left(\frac{1}{n}-\frac{1}{n+k}\right)}+\color{#FF4F00}{\sum^\infty_{k=1}\frac{1}{k^4}\sum^\infty_{n=1}\frac{1}{n^2}}\\ &+\color{#00A000}{\sum^\infty_{k=1}\frac{3}{k^4}\sum^\infty_{n=1}\frac{1}{(n+k)^2}}+\color{#21ABCD}{\sum^\infty_{k=1}\frac{2}{k^3}\sum^\infty_{n=1}\frac{1}{(n+k)^3}}\\ &+\color{#6F00FF}{\sum^\infty_{k=1}\frac{1}{k^2}\sum^\infty_{n=1}\frac{1}{(n+k)^4}}\tag5\\ =&\zeta(6)\color{#E2062C}{-4\sum^\infty_{k=1}\frac{H_k}{k^5}}+\color{#FF4F00}{\zeta(2)\zeta(4)}+\color{#00A000}{3\zeta(2)\zeta(4)-3\sum^\infty_{k=1}\frac{H_k^{(2)}}{k^4}}\\ &+\color{#21ABCD}{2\zeta^2(3)-2\sum^\infty_{k=1}\frac{H_k^{(3)}}{k^3}}+\color{#6F00FF}{\zeta(2)\zeta(4)-\sum^\infty_{k=1}\frac{H_k^{(4)}}{k^2}}\\ =&-8\zeta(6)+3\zeta^2(3)+4\zeta(2)\zeta(4)-2\sum^\infty_{k=1}\frac{H_k^{(2)}}{k^4}\tag6\\ =&-\frac{8}{3}\zeta(6)+\zeta^2(3)+\frac{4}{3}\zeta(2)\zeta(4)\tag7 \end{align} Hence your integral is \begin{align} \int^1_0\frac{{\rm Li}_2(x)\ln^2{x}\ln(1-x)}{x} =\frac{13}{3}\zeta(6)-\frac{8}{3}\zeta(2)\zeta(4) =-\frac{\pi^6}{2835} \end{align}

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Explanation:
$(1): \text{See}$ here.
$(3): \text{Expand $H_{n-1}^{(2)}$}$.
$(4): \text{Reverse the order of summation}$.
$(5): \text{Partial Fractions}$.
$(6): \text{Simplify the sums with $(1)$ and $(2)$}$.
$(7): \text{Move the unknown sum over to the other side and divide by $3$}$.

Proof of $(2)$: \begin{align} \sum^\infty_{n=1}\frac{H_n^{(p)}}{n^q} &=\sum^\infty_{n=1}\frac{1}{n^q}\sum^n_{k=1}\frac{1}{n^p}\\ &=\sum^\infty_{n=1}\frac{1}{n^q}\left(\sum^\infty_{k=1}\frac{1}{k^p}-\sum^\infty_{k=n}\frac{1}{k^p}+\frac{1}{n^p}\right)\\ &=\sum^\infty_{n=1}\frac{1}{n^q}\sum^\infty_{k=1}\frac{1}{k^p}+\sum^\infty_{n=1}\frac{1}{n^{p+q}}-\sum^\infty_{k=1}\frac{1}{k^p}\sum^k_{n=1}\frac{1}{n^q}\\ &=\zeta(p)\zeta(q)+\zeta(p+q)-\sum^\infty_{n=1}\frac{H_n^{(q)}}{n^p} \end{align}

Answer 5

here is a solution without using the value of $\ \displaystyle \sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}$ or $\ \displaystyle\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}$: \begin{align} I&=\int_0^1 \frac{\operatorname{Li}_2(x)\ln(1-x)\ln^2x}{x}\ dx\\ &=\sum_{n=1}^\infty\frac1{n^2}\int_0^1x^{n-1}\ln(1-x)\ln^2x\ dx\\ &=\sum_{n=1}^\infty\frac1{n^2}\frac{\partial^2}{\partial n^2}\int_0^1x^{n-1}\ln(1-x)\ dx\\ &=\sum_{n=1}^\infty\frac1{n^2}\frac{\partial^2}{\partial n^2}\left(-\frac{H_n}{n}\right)\\ &=\sum_{n=1}^\infty\frac1{n^2}\left(\frac{2\zeta(2)}{n^2}+\frac{2\zeta(3)}{n}-\frac{2H_n}{n^3}-\frac{2H_n^{(2)}}{n^2}-\frac{2H_n^{(3)}}{n}\right)\\ &=\frac72\zeta(6)+2\zeta^2(3)-2\sum_{n=1}^\infty\frac{H_n}{n^5}-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}-2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^3}\tag{1} \end{align}

lets do more simplifications with these sums: \begin{align} S&=\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^3}=\sum_{n=1}^\infty\frac1{n^3}\sum_{k=1}^\infty\left(\zeta(3)-\frac1{(k+n)^3}\right)\\ &=\zeta^2(3)-\sum_{k=1}^\infty\sum_{n=1}^\infty\left(\frac{6}{k^5}\left(\frac1n-\frac1{n+k}\right)-\frac{3}{k^4n^2}-\frac{3}{k^4(k+n)^2}+\frac1{k^3n^3}-\frac1{k^3(k+n)^3}\right)\\ &=\zeta^2(3)-\sum_{k=1}^\infty\left(\frac{6H_k}{k^5}-\frac{3\zeta(2)}{k^4}-\frac{3}{k^4}\left(\zeta(2)-H_k^{(2)}\right)+\frac{\zeta(3)}{k^3}-\frac1{k^3}\left(\zeta(3)-H_k^{(3)}\right)\right)\\ &=\zeta^2(3)-6\sum_{k=1}^\infty\frac{H_k}{k^5}+3\zeta(2)\zeta(4)+3\zeta(2)\zeta(4)-3\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}-\zeta^2(3)+\zeta^2(3)-S\\ &=\frac12\zeta^2(3)+\frac{21}4\zeta(6)-3\sum_{k=1}^\infty\frac{H_k}{k^5}-\frac32\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4} \end{align} or $$\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}=\frac13\zeta^2(3)+\frac{21}6\zeta(6)-2\sum_{k=1}^\infty\frac{H_k}{k^5}-\frac23\sum_{k=1}^\infty\frac{H_k^{(3)}}{k^3}\tag{2}$$ Plugging (2) in (1), we have $$\boxed{I=\frac43\zeta^2(3)-\frac72\zeta(6)+2\sum_{k=1}^\infty\frac{H_k}{k^5}-\frac23\sum_{k=1}^\infty\frac{H_k^{(3)}}{k^3}}$$

using $\displaystyle\sum_{n=1}^\infty \frac{H_n^{(p)}}{n^q}+\sum_{n=1}^\infty \frac{H_n^{(q)}}{n^p}=\zeta(p)\zeta(q)+\zeta(p+q)$ (can be proved using Abel's summation or as proved by M.N.C.E) and the special case is when $q=p$, we get $\displaystyle\sum_{n=1}^\infty \frac{H_n^{(p)}}{n^p}=\frac12\left(\zeta^2(p)+\zeta(2p)\right)$ which gives $\displaystyle\sum_{k=1}^\infty \frac{H_k^{(3)}}{k^3}=\frac12\left(\zeta^2(3)+\zeta(6)\right)$, plugging this sum along with $\displaystyle\sum_{k=1}^\infty \frac{H_k}{k^5}=\frac74\zeta(6)-\frac12\zeta^2(3)\ $ in the boxed result, we get $\ \displaystyle I=-\frac13\zeta(6)$

As a bonus, plug $\sum_{k=1}^\infty \frac{H_k^{(3)}}{k^3}=\frac12\left(\zeta^2(3)+\zeta(6)\right)$ in (2) we get $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}=\zeta^2(3)-\frac13\zeta(6)$

Answer 6

Another approach:

By Cauchy product we have

$$\operatorname{Li}_2(x)\ln(1-x)=-\sum_{n=1}^\infty\left(\frac{2H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac3{n^3}\right)x^n$$

Multiply both sides by $\frac{\ln^2x}{x}$ then integrate from $x=0$ to $1$ and use the fact that $\int_0^1 x^{n-1}\ln^2xdx=\frac{2}{n^3}$

we get

$$\int_0^1\frac{\operatorname{Li}_2(x)\ln(1-x)\ln^2x}{x}dx=-4\sum_{n=1}^\infty\frac{H_n}{n^5}-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}+6\zeta(6)$$

By Euler identity we have $\sum_{n=1}^\infty\frac{H_n}{n^5}=\frac74\zeta(6)-\frac12\zeta^2(3)$ and in my previous solution above we proved $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}=\zeta^2(3)-\frac13\zeta(6)$. By collecting these results, the answer follows.

Answer 7

\begin{align*} J&=\int_0^1 \frac{{\rm{Li}}_2(x)\ln(1-x)\ln^2(x)}{x} \,dx\\=&\overset{\text{IBP}}=\frac{1}{3}\Big[\ln^3 x{\rm{Li}}_2(x)\ln(1-x)\Big]_0^1+\frac{1}{3}\int_0^1 \frac{{\rm{Li}}_2(x)\ln^3 x}{1-x}\,dx+\\&\frac{1}{3}\int_0^1 \frac{\ln^3 x\ln^2(1-x)}{x}\,dx\\ &=\frac{1}{3}\int_0^1 \frac{{\rm{Li}}_2(x)\ln^3 x}{1-x}\,dx+\frac{1}{3}\int_0^1 \frac{\ln^3 x\ln^2(1-x)}{x}\,dx\\ &\overset{\text{IBP}}=\frac{1}{3}\int_0^1 \frac{{\rm{Li}}_2(x)\ln^3 x}{1-x}\,dx+\frac{1}{12}\Big[\ln^4 x\ln^2(1-x)\Big]_0^1+\\&\frac{1}{6}\int_0^1 \frac{\ln^4 x\ln(1-x)}{1-x}\,dx\\ &=\frac{1}{3}\int_0^1 \frac{{\rm{Li}}_2(x)\ln^3 x}{1-x}\,dx+\frac{1}{6}\int_0^1 \frac{\ln^4 x\ln(1-x)}{1-x}\,dx\\ \end{align*} On the other hand, \begin{align*} A&=\int_0^1 \frac{\ln x}{1-x}dx,B=\int_0^1 \frac{\ln^2 x}{1-x}dx,C=\int_0^1 \frac{\ln^3 x}{1-x}dx\\ D&=\int_0^1 \frac{\ln^4 x}{1-x}dx,F=\int_0^1 \frac{\ln^5 x}{1-x}dx\\ \frac{B^2}{2}&=\int_0^1 \frac{\ln^2 x}{1-x}\left(\int_0^x \frac{\ln^2 t}{1-t}dt\right)dx\\ &=\int_0^1 \int_0^1\left(\frac{x\ln^2 x\ln^2 (tx)}{(1-t)(1-tx)}\right)dt dx\\ &=\int_0^1 \int_0^1\left(\frac{\ln^2 x\ln^2 (tx)}{(1-t)(1-x)}-\frac{\ln^2 x\ln^2 (tx)}{(1-t)(1-tx)}\right)dt dx\\ &=\int_0^1 \int_0^1\left(\frac{\ln^4 x+\ln^2 t\ln^2 x+2\ln^3 x\ln t}{(1-t)(1-x)}+\frac{2\ln t\ln^3(tx)-\ln^4(tx)-\ln^2 t\ln^2(tx)}{(1-t)(1-tx)}\right)dt dx\\ &=B^2+2AC+\\&\int_0^1 \int_0^1\left(\frac{\ln^4 x}{(1-t)(1-x)}+\frac{2\ln t\ln^3(tx)-\ln^4(tx)-\ln^2 t\ln^2(tx)}{(1-t)(1-tx)}\right)dt dx\\ &=B^2+2AC+\\&\int_0^1 \left(\frac{D}{1-t}+\frac{2\ln t}{t(1-t)}\int_0^t \frac{\ln^3 u}{1-u}du-\frac{1}{t(1-t)}\int_0^t \frac{\ln^4 u}{1-u}du-\frac{\ln^2 t}{t(1-t)}\int_0^t \frac{\ln^2 u}{1-u}du\right)dt\\ &0\leq \alpha<1\\ K(\alpha)&=B^2+2AC+\\&\int_0^\alpha \left(\frac{D}{1-t}+\frac{2\ln t}{t(1-t)}\int_0^t \frac{\ln^3 u}{1-u}du-\frac{1}{t(1-t)}\int_0^t \frac{\ln^4 u}{1-u}du-\frac{\ln^2 t}{t(1-t)}\int_0^t \frac{\ln^2 u}{1-u}du\right)dt\\ &=B^2+2AC+-D\ln(1-\alpha)+\\&2\int_0^\alpha \frac{\ln t}{t}\left(\int_0^t \frac{\ln^3 u}{1-u}dt\right)dt+2\int_0^\alpha \frac{\ln t}{1-t}\left(\int_0^t \frac{\ln^3 u}{1-u}dt\right)dt-\\&\int_0^\alpha \frac{1}{t(1-t)}\left(\int_0^t \frac{\ln^4 u}{1-u}du\right)dt-\int_0^\alpha\frac{\ln^2 t}{t}\left(\int_0^t \frac{\ln^2 u}{1-u}du\right)dt-\int_0^\alpha\frac{\ln^2 t}{1-t}\left(\int_0^t \frac{\ln^2 u}{1-u}du\right)dt\\ &\overset{\text{IBP}}=B^2+2AC-D\ln(1-\alpha)\\&+\ln^2 \alpha \int_0^\alpha \frac{\ln^3 u}{1-u}du-\int_0^\alpha \frac{\ln^5 t}{1-t}dt+2 \left(\int_0^\alpha\frac{\ln t}{1-t}dt\right)\left(\int_0^\alpha\frac{\ln^3 u}{1-u}du\right)-\\&2\int_0^\alpha \frac{\ln^3 t}{1-t}\left(\int_0^t\frac{\ln u}{1-u}du\right)dt+\ln(1-\alpha)\int_0^\alpha \frac{\ln^4 u}{1-u}du-\ln(\alpha)\int_0^\alpha \frac{\ln^4 u}{1-u}du+\\&\int_0^a \frac{\ln\left(\frac{t}{1-t}\right)\ln^4 t}{1-t}dt-\frac{1}{3}\ln^3(\alpha)\int_0^\alpha \frac{\ln^2 u}{1-u}du+\frac{1}{3}\int_0^\alpha \frac{\ln^5 t}{1-t}dt-\int_0^\alpha\frac{\ln^2 t}{1-t}\left(\int_0^t \frac{\ln^2 u}{1-u}du\right)dt\\ \frac{1}{2}B^2&=\lim_{\alpha\rightarrow 1} K(\alpha)\\ &=B^2+2AC-F+2AC-2\int_0^1 \frac{\ln^3 t}{1-t}\left(\int_0^t \frac{\ln u}{1-u}\right)dt+F-\\&\int_0^1 \frac{\ln(1-t)\ln^4 t}{1-t}\,dt+\frac{ 1}{3}F-\frac{1}{2}B^2\\\end{align*} Therefore, \begin{align*} 0&=4AC+\frac{1}{3}F-2\int_0^1 \frac{\ln^3 t}{1-t}\left(\int_0^t \frac{\ln u}{1-u}\right)dt-\int_0^1 \frac{\ln(1-t)\ln^4 t}{1-t}\,dt \end{align*} Moreover,

\begin{align*} \int_0^1 \frac{\ln^3 t}{1-t}\left(\int_0^t \frac{\ln u}{1-u}\right)dt&\overset{\text{IBP}}=\int_0^1 \frac{\ln^3 t}{1-t}\left(-\Big[\ln(1-t)\ln t\Big]_0^t+\int_0^t\frac{\ln(1-u)}{u}du\right)dt\\ &=-\int_0^1 \frac{\ln(1-t)\ln^4 t}{1-t}\,dt-\int_0^1 \frac{{\rm{Li}}_2(t)\ln^3 t}{1-t}dt \end{align*} Therefore, \begin{align*}0&=4AC+\frac{1}{3}F+\int_0^1 \frac{\ln(1-t)\ln^4 t}{1-t}\,dt+2\int_0^1 \frac{{\rm{Li}}_2(t)\ln^3 t}{1-t}dt\\ -\frac{2}{3}AC-\frac{1}{18}F&=\frac{1}{6}\int_0^1 \frac{\ln(1-t)\ln^4 t}{1-t}\,dt+\frac{1}{3}\int_0^1 \frac{{\rm{Li}}_2(t)\ln^3 t}{1-t}dt\\ J&=-\frac{2}{3}AC-\frac{1}{18}F\\ &=\boxed{-\frac{1}{2835}\pi^6} \end{align*} i assume that, \begin{align}A&=\int_0^1 \frac{\ln x}{1-x}dx=-\frac{\pi^2}{6}\\ C&=\int_0^1 \frac{\ln^3 x}{1-x}dx=-\frac{\pi^4}{15}\\ F&=\int_0^1 \frac{\ln^5 x}{1-x}dx=-\frac{8\pi^6}{63}\\ \end{align}

NB: If $f$ is derivable on $[0,1]$, $\displaystyle \int_0^1 f(x)f^\prime(x)dx=\frac{1}{2}f(1)^2-\frac{1}{2}f(0)^2$