Integral of polylogarithms and logs in closed form: $\int_0^1 \frac{du}{u}\text{Li}_2(u)^2(\log u)^2$ Is it possible to evaluate this integral in closed form?
$$ \int_0^1 \frac{du}{u}\text{Li}_2(u)^2\log u \stackrel{?}{=} -\frac{\zeta(6)}{3}.$$
I found the possible closed form using an integer relation algorithm.
I found several other possible forms for similar integrals, including
$$ \int_0^1 \frac{du}{u}\text{Li}_2(u)^2(\log u)^2 \stackrel{?}{=} -20\zeta(7)+12\zeta(2)\zeta(5).$$
There doesn't seem to be an equivalent form when the integrand contains $(\log u)^3$, at least not just in terms of $\zeta$.
Does anybody know a trick for evaluating these integrals?
Update. The derivation of the closed form for the second integral follows easily along the ideas O.L. used in the answer for the first integral.
Introduce the functions
$$ I(a,b,c) = \int_0^1 \frac{du}{u}(\log u)^c \text{Li}_a(u)\text{Li}_b(u) $$
and
$$ S(a,b,c) = \sum_{n,m\geq1} \frac{1}{n^am^b(n+m)^c}. $$
Using integration by parts, the expansion of polylogarithms from their power series definition and also that
$$ \int_0^1 (\log u)^s u^{t-1}\,du = \frac{(-1)^s s!}{t^{s+1}},$$
check that
$$ I(2,2,2) = -\frac23 I(1,2,3) = 4S(1,2,4). $$
Now use binomial theorem and the fact that $S(a,b,c)=S(b,a,c)$ to write
$$ 6S(1,2,4) + 2S(3,0,4) = 3S(1,2,4) + 3S(2,1,4)+S(0,3,4)+S(3,0,4) = S(3,3,1). $$
Now, using Mathematica, 
$$ S(3,3,1) = \sum_{n,m\geq1}\frac{1}{n^3m^3(n+m)} = \sum_{m\geq1}\frac{H_m}{m^6} - \frac{\zeta(2)}{m^5} + \frac{\zeta(3)}{m^4}, $$
and
$$ \sum_{m\geq1}\frac{H_m}{m^6} = -\zeta(4)\zeta(3)-\zeta(2)\zeta(5)+4\zeta(7), $$
so
$$ S(3,3,1) = 4\zeta(7)-2\zeta(2)\zeta(5). $$
Also,
$$ S(0,3,4) = \zeta(3)\zeta(4) - \sum_{m\geq1} \frac{H_{n,4}}{m^3} = -17\zeta(7)+10\zeta(2)\zeta(5)+\zeta(3)\zeta(4), $$
from which it follows that
$$ I(2,2,2) = \frac23\left(S(3,3,1)-2S(0,3,4)\right) = -20\zeta(7)+12\zeta(2)\zeta(5). $$
 A: By Cauchy product we have 
$$\operatorname{Li}_2^2(x)=\sum_{n=1}^\infty\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)x^n$$
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
$$\int_0^1\frac{\operatorname{Li}_2^2(x)\ln^2x}{x}dx=8\sum_{n=1}^\infty \frac{H_n}{n^6}+4\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^5}-12\zeta(7)$$
By Euler identity we have $$\sum_{n=1}^\infty \frac{H_n}{n^6}=4\zeta(7)-\zeta(2)\zeta(5)-\zeta(3)\zeta(4)$$ and in my solution here I managed to prove $$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^5}=-10\zeta(7)+5\zeta(2)\zeta(5)+2\zeta(3)\zeta(4)$$
By collecting these results we get
$$\int_0^1\frac{\operatorname{Li}_2^2(x)\ln^2x}{x}dx=-20\zeta(7)+12\zeta(2)\zeta(5)$$
A: I've decided to publish my work so far - I do not promise a solution, but I've made some progress that others may find interesting and/or helpful.
$$\text{Let } I_{n,k}=\int_{0}^{1}\frac{\text{Li}_{k}(u)}{u}\log(u)^{n}du$$
Integrating by parts gives $$I_{n,k}=\left[\text{Li}_{k+1}(u)\log(u)^{n}\right]_{u=0}^{u=1}-\int_{0}^{1}\frac{\text{Li}_{k+1}(u)}{u}n\log(u)^{n-1}du$$
$$\text{Hence, }I_{n,k}=-nI_{n-1,k+1} \implies I_{n,k}=(-1)^{r}\frac{n!}{(n-r)!}I_{n-r,k+r}$$
Taking $r=n$ gives $I_{n,k}=(-1)^{n}n!I_{0,n+k}$.
$$\text{But obviously } I_{0,n+k}=\int_{0}^{1}\frac{\text{Li}_{n+k}(u)}{u}du=\text{Li}_{n+k+1}(1)-\text{Li}_{n+k+1}(0)=\zeta(n+k+1)$$
$$\text{Now consider }J_{n,k,l}=\int_{0}^{1}\frac{\text{Li}_{k}(u)}{u}\text{Li}_{l}(u)\log(u)^{n}du$$
Integrating by parts again,
$$J_{n,k,l}=\left[\text{Li}_{k+1}(u)\text{Li}_{l}(u)\log(u)^{n}\right]_{0}^{1}-\int_{0}^{1}\frac{\text{Li}_{l-1}(u)}{u}\text{Li}_{k+1}(u)\log(u)^{n}-\int_{0}^{1}\frac{n\log(u)^{n-1}}{u}\text{Li}_{k+1}(u)\text{Li}_{l}(u) du$$
So $J_{n,k,l}=-J_{n,k+1,l-1}-nJ_{n-1,k+1,l}$; continuing in the spirit of the first part suggests that we ought to try to increase the first and second indices, while decreasing the third. If we can succeed in this, we have found a closed form.
A: Now it is a proof.

Let us integrate once by parts to replace the (first) integral by
$$I=\int_0^1\frac{\ln u\,\mathrm{Li}_2(u)^2du}{u}=\int_0^1\frac{\ln^2 u\ln(1-u)}{u}\mathrm{Li}_2(u)\,du.$$
Next replace $\mathrm{Li}_2(u)=\sum_{m=1}^{\infty}u^m/m^2$ and $\ln(1-u)=-\sum_{n=1}^{\infty}u^{n}/n$ by the corresponding Taylor series. Exchanging the order of summation and integration, evaluate the integrals with respect to $u$. This can be done using that
$$\int_0^1 u^{s-1}\ln^2u\,du=\frac{2}{s^3}.$$
So $I$ can be written as a double series
$$I=-2\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^2n(m+n)^3}.$$
Now let us introduce the following sums:
\begin{align}
&S_1=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^3(m+n)^3},\\
&S_2=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{n^3(m+n)^3},\\
&S_3=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^2n(m+n)^3},\\
&S_4=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{mn^2(m+n)^3}.
\end{align}
It is obvious that $S_1=S_2$ and $S_3=S_4$. What is more funny (but still obvious to prove) is that
$$S_1+S_2+3S_3+3S_4=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^3n^3}=\zeta(3)^2.\tag{1}$$
Therefore, if we manage to compute $S_1=S_2$, we will be able to compute $I$. But
$$S_1=-\sum_{m=1}^{\infty}\frac{\psi''(1+m)}{2m^3}=\frac12\left(\zeta(3)^2-\frac{\pi^6}{945}\right).\tag{2}$$
Here the first equality follows from the recursion relation $\psi''(z+1)-\psi''(z)={2}/{z^3}$ and telescoping argument, whereas the second was obtained using Mathematica.
Now combining (1), (2) and the fact that $I=-(S_3+S_4)$, we find
$$I=-\frac13\left[\left(S_1+S_2+3S_3+3S_4\right)-2S_1\right]=-\frac{1}{3}\times\frac{\pi^6}{945}=-\frac{\zeta(6)}{3}.$$
A: You might find Two New Zeta Constants: Fractal String, Continued Fraction, and Hypergeometric Aspects of the Riemann Zeta Function interesting . It relates hypergeometric functions, polylogarithms, and the zeta function
