Integral of polylogarithms and logs in closed form: $\int_0^1 \frac{du}{u}\text{Li}_2(u)^2(\log u)^2$

Solution 1:

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 $$I=-\frac13\left[\left(S_1+S_2+3S_3+3S_4\right)-2S_1\right]=-\frac{1}{3}\times\frac{\pi^6}{945}=-\frac{\zeta(6)}{3}.$$

Solution 2:

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

$$\int_0^1\frac{\operatorname{Li}_2^2(x)\ln^2x}{x}dx=-20\zeta(7)+12\zeta(2)\zeta(5)$$