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.

  • $\begingroup$ Why do we have a restriction on elementary methods when the integral involves a special function? On a sidenote; There are no poles anyway, so you can't use residue method even if you wanted. $\endgroup$ – UserX Oct 12 '14 at 13:07
  • $\begingroup$ @UserX Because I know nothing about residue method since I'm still a junior school student (not a college student) so it's useless for me if the answer using that method. Besides, I love Feynman's style approach $\endgroup$ – Anastasiya-Romanova 秀 Oct 12 '14 at 13:12
  • $\begingroup$ Feynman's method would require a really huge trick for this. Puiseux series and term by term integration looks promising. Second side note; Complex analysis will lend you tools to crack interesting integrals, it's worth looking into. $\endgroup$ – UserX Oct 12 '14 at 13:13
  • $\begingroup$ @UserX I love trick, specially a huge one. It'll enrich my math tools (ô‿ô) $\endgroup$ – Anastasiya-Romanova 秀 Oct 12 '14 at 13:18
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    $\begingroup$ @UserX I said Feynman's style approach, it means the approach of solving integral without using residue/ complex integration method, not only differentiation under integral sign method. $\endgroup$ – Anastasiya-Romanova 秀 Oct 12 '14 at 13:25


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

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

$^*$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)].

  • $\begingroup$ Do you mind explaining how you concluded $\displaystyle \sum_{n,k\ge 1}\frac{2}{k^4(k+n)^2}=-2\zeta^2(3)+\frac{25}{6}\zeta(6)$? Thanks a lot. $\endgroup$ – M.N.C.E. Oct 12 '14 at 22:12
  • $\begingroup$ @M.N.C.E. Please see the edit. Thanks! $\endgroup$ – Olivier Oloa Oct 13 '14 at 6:01
  • $\begingroup$ I tried for 3 hours to understand what @M.N.C.E. asked yet I've not figured it out yet. Even after you gave me a link of paper as reference. +1 for now. I wish there's someone who come with an answer without involving double summation. $\endgroup$ – Anastasiya-Romanova 秀 Oct 13 '14 at 10:29
  • $\begingroup$ Sir, could you elaborate this part, please? $$\!\sum_{n,k\geq 1}^{\infty} \!\frac{1}{k^4n(k+n)}=\!\sum_{k=1}^{\infty}\! \frac{H_k}{k^5}$$How come from the LHS turns out to be the RHS? $\endgroup$ – Anastasiya-Romanova 秀 Oct 15 '14 at 16:34
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    $\begingroup$ @Anastasiya-Romanova Observe that $\frac{1}{n(k+n)}=\frac{1}{k}\left(\frac{1}{n}-\frac{1}{n+k}\right) $ then sum over $n\geq1$ and recall the series representation of the digamma function $$ \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+k} \right)=\gamma+\psi(k+1)= H_k,\quad k\geq1,$$ then sum over $k\geq1$. $\endgroup$ – Olivier Oloa Oct 15 '14 at 20:54

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


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


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


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(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


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


Putting this all together.


and this completes the proof.

  • $\begingroup$ I'm familiar with P. Freitas' paper. This integral came from his paper. +1 for now for your answer $\endgroup$ – Anastasiya-Romanova 秀 Oct 13 '14 at 10:30
  • $\begingroup$ @Anastasiya-Romanova I think this is the most accurate tool to evaluate your integral. As I see using this method is more elegant then evaluate the integral directly with the common methods, such as integrating by parts or by substitutions. These apparatuses are behind the method, hidden in the proofs, what you could read in the paper. $\endgroup$ – user153012 Oct 13 '14 at 12:08
  • $\begingroup$ @Anastasiya-Romanova By understanding and using this "quity easy" process you could evaluate almost every integral in form $K$ and $J$. The only problem is when $K(r,p,q)$ has parameters for that $w=r+p+q$ is odd. But for low odd $w$ weights, such as $w<7$, we could evaluate in common the integeral using the standard results for $S_{p,q}$, because already known to Euler how to evaluate these linear sums for low $p+q$ even cases. $\endgroup$ – user153012 Oct 13 '14 at 12:08
  • $\begingroup$ Okay, I'll take your advice as a consideration ≥Ö‿Ö≤ $\endgroup$ – Anastasiya-Romanova 秀 Oct 13 '14 at 12:12

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}

$(1): \text{See}$ 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}

  • $\begingroup$ Evaluating the sum $\sum_{n=1}^{\infty} \frac{H_n^{(2)}}{n^4}$ is the key in my asnwer too. In general this is the hardest part of the problem, because as a generalization of Anastasiya's question to higher order polylogarithms we will always run into a sum in form $$S_{p,q} := \sum_{n=1}^{\infty} \frac{H_n^{(p)}}{n^q}.$$ If $p+q$ is odd we are always able to put $S_{p,q}$ into a closed-form, which contains a sum of zeta functions at positive integers with rational coefficients, but when $p+q$ is even and $p+q>7$ I'm afraid we stuck in a trap. $\endgroup$ – user153012 Oct 13 '14 at 12:29
  • $\begingroup$ This is a good answer but why your answer in wiki mode? You'll get nothing if someone upvote your answer $\endgroup$ – Anastasiya-Romanova 秀 Oct 13 '14 at 13:42
  • $\begingroup$ Of all the answers, I like your answer but could you please provide me a real analysis approach to evaluate $$\sum^\infty_{n=1}\frac{H_n^{(2)}}{n^4}\,?$$ $\endgroup$ – Anastasiya-Romanova 秀 Oct 15 '14 at 16:28
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    $\begingroup$ @Anastasiya-Romanova Please see the fourth formula under the "Identities: General Manipulations" section here $\endgroup$ – M.N.C.E. Oct 16 '14 at 14:07
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    $\begingroup$ @r9m Thanks. Your proof of that double sum was really clever as well. $\endgroup$ – M.N.C.E. Nov 9 '14 at 6:18

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.

  • 1
    $\begingroup$ (+1) Very good! This proof should be called "ADMIRE AND LEARN"! :-) $\endgroup$ – user 1357113 Nov 9 '14 at 13:57
  • $\begingroup$ I've never understood double sum, so could you elaborate the last three lines? Thanks... $\endgroup$ – Anastasiya-Romanova 秀 Nov 10 '14 at 13:22
  • $\begingroup$ @Anastasiya-Romanova continue the discussion in chat ? :) $\endgroup$ – r9m Nov 10 '14 at 13:25
  • $\begingroup$ @r9m Arrgggghhhhhh, you...!!! In my chatroom, not there $\endgroup$ – Anastasiya-Romanova 秀 Nov 10 '14 at 13:27
  • $\begingroup$ @Anastasiya-Romanova okay .. but where's your chatroom ?! :-) mind posting a link ? $\endgroup$ – r9m Nov 10 '14 at 13:31

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