Evaluate$\int\limits_0^1 [\log(x)\log(1-x)+\operatorname{Li}_2(x)]\left[\frac{\operatorname{Li}_2(x)}{x(1-x)}-\frac{\zeta(2)}{1-x}\right]\mathrm dx$

Cross-posting this integral on AoPS brings Y. Sharifi's solution here after a day. Quite amazing one!

I will copy here his entire solution:

Let $I$ be your integral. Using the identity $\ln x \ln(1-x)+\text{Li}_2(x)=\zeta(2)-\text{Li}_2(1-x),$ we have $$I=-\int_0^1 \frac{\text{Li}_2(x)\text{Li}_2(1-x)}{x(1-x)} \ dx + \zeta(2)\int_0^1 \left(\frac{\text{Li}_2(x)}{x(1-x)}-\frac{\zeta(2)}{1-x}+\frac{\text{Li}_2(1-x)}{1-x}\right)dx.$$ Let $$J=\int_0^1 \frac{\text{Li}_2(x)\text{Li}_2(1-x)}{x(1-x)} \ dx, \ \ \ \ \ K:=\int_0^1 \left(\frac{\text{Li}_2(x)}{x(1-x)}-\frac{\zeta(2)}{1-x}+\frac{\text{Li}_2(1-x)}{1-x}\right)dx.$$ So $$I=\zeta(2)K - J. \ \ \ \ \ \ \ \ \ (1)$$ We first show that $K=0.$ Start with using integration by parts in $K,$ with $u=\frac{\text{Li}_2(x)}{x}-\zeta(2)+\text{Li}_2(1-x)$ and $dv=\frac{dx}{1-x}.$ Then $$K=\int_0^1 \ln(1-x)\left(\frac{\ln x}{1-x}-\frac{\ln(1-x)}{x^2}-\frac{\text{Li}_2(x)}{x^2}\right)dx. \ \ \ \ \ \ \ \ \ \ (2)$$ Using the Maclaurin series of $\ln(1-x),$ we quickly find the first integral in $K$
$$\int_0^1 \frac{\ln x \ln(1-x)}{1-x} \ dx = \int_0^1 \frac{\ln x \ln(1-x)}{x} \ dx=\zeta(3). \ \ \ \ \ \ \ \ \ \ (3)$$ Next, we ignore the second integral in $K$ for now and we look at the third one, i.e. $\int_0^1 \frac{\ln(1-x) \text{Li}_2(x)}{x^2} \ dx.$ In this integral, we use integration by parts with $u=\ln(1-x)\text{Li}_2(x)$ and $dv=\frac{dx}{x^2};$ notice that we need to choose $v=1-\frac{1}{x}.$ So $$\int_0^1 \frac{\ln(1-x) \text{Li}_2(x)}{x^2} \ dx=\int_0^1\left(1-\frac{1}{x}\right)\left(\frac{\text{Li}_2(x)}{1-x}+\frac{\ln^2(1-x)}{x}\right) dx$$ $$=-\int_0^1 \frac{\text{Li}_2(x)}{x} \ dx + \int_0^1 \frac{\ln^2(1-x)}{x} \ dx - \int_0^1 \frac{\ln^2(1-x)}{x^2} \ dx=-\zeta(3)+\int_0^1 \frac{\ln^2x}{1-x} \ dx -\int_0^1 \frac{\ln^2(1-x)}{x^2} \ dx.$$ $$=\zeta(3)-\int_0^1 \frac{\ln^2(1-x)}{x^2} \ dx. \ \ \ \ \ \ \ \ \ (4)$$ Thus, by $(2),(3)$ and $(4),$ we have $K=0$ and hence, by $(1),$ $$I=-J=-\int_0^1 \frac{\text{Li}_2(x)\text{Li}_2(1-x)}{x(1-x)} \ dx=-2\int_0^1 \frac{\text{Li}_2(x)\text{Li}_2(1-x)}{x} \ dx.$$ So integration by parts with $u=\text{Li}_2(1-x)$ and $dv=\frac{\text{Li}_2(x)}{x} \ dx$ gives $$I=2\int_0^1 \frac{\text{Li}_3(x)\ln x}{1-x} \ dx=2\int_0^1 \text{Li}_3(x) \ln x \sum_{m \ge 1}x^{m-1} dx=2\sum_{m \ge 1} \int_0^1 x^{m-1}\text{Li}_3(x) \ln x \ dx$$ $$=2\sum_{m \ge 1} \int_0^1x^{m-1}\sum_{n \ge 1} \frac{x^n}{n^3} \ln x \ dx=2\sum_{m,n \ge 1} \frac{1}{n^3}\int_0^1x^{n+m-1}\ln x \ dx=-2\sum_{m,n \ge 1} \frac{1}{n^3(n+m)^2}$$ $$=-\sum_{m,n \ge 1} \left(\frac{1}{n^3(n+m)^2}+\frac{1}{m^3(n+m)^2}\right). \ \ \ \ \ \ \ \ \ (5)$$ So $(5)$ and the following identity $$\frac{1}{n^3(n+m)^2}+\frac{1}{m^3(n+m)^2}=\frac{1}{n^3m^2}-\frac{2}{n^2m^3}+\frac{3}{m^3n(n+m)}$$ together give $$I=-\sum_{m,n \ge 1}\left(\frac{1}{n^3m^2}-\frac{2}{n^2m^3}+\frac{3}{m^3n(n+m)}\right)=\zeta(2)\zeta(3)-3\sum_{m,n \ge 1} \frac{1}{m^3n(n+m)}$$ $$=\zeta(2)\zeta(3)-3\sum_{m \ge 1} \frac{1}{m^4} \sum_{n \ge 1}\left(\frac{1}{n}-\frac{1}{n+m}\right)=\zeta(2)\zeta(3)-3\sum_{m \ge 1} \frac{H_m}{m^4}, \ \ \ \ \ \ \ \ \ (6)$$ where, as usual, $H_m:=\sum_{j=1}^m \frac{1}{j}$ is the $m$-th harmonic number. Now we use Euler's formula $$\sum_{m \ge 1} \frac{H_m}{m^k}=\left(1+\frac{k}{2}\right)\zeta(k+1)-\frac{1}{2}\sum_{i=1}^{k-2}\zeta(i+1)\zeta(k-i), \ \ \ \ k \ge 2,$$ with $k=4$ to get $$\sum_{m \ge 1} \frac{H_m}{m^4}=3\zeta(5)-\zeta(2)\zeta(3)$$ and so, by $(6),$

$$I=\zeta(2)\zeta(3)-3(3\zeta(5)-\zeta(2)\zeta(3))=4\zeta(2)\zeta(3)-9\zeta(5).$$

Edit. This integral was proposed two years ago in RMM and it appeared as problem UP $089$.

See in this link, at the page $70$.

$$I=\int_0^1\left(\ln x\ln(1-x)+\operatorname{Li}_2(x)\right)\left(\frac{\operatorname{Li}_2(x)}{x}+\frac{\operatorname{Li}_2(x)}{1-x}-\frac{\zeta(2)}{1-x}\right)\ dx$$ using the reflection dilogarithmic formula $\operatorname{Li}_2(x)-\operatorname{Li}_2(1-x)=\zeta(2)-\ln x\ln(1-x)$ for the first parentheses and the second and third term of the second parentheses we get \begin{align} I&=\int_0^1\left(\zeta(2)-\operatorname{Li}_2(1-x)\right)\left(\frac{\operatorname{Li}_2(x)}{x}-\frac{\ln x\ln(1-x)}{1-x}-\frac{\operatorname{Li}_2(1-x)}{1-x}\right)\ dx\\ &=\zeta(2)\int_0^1\left(\frac{\operatorname{Li}_2(x)}{x}-\frac{\ln x\ln(1-x)}{1-x}-\frac{\operatorname{Li}_2(1-x)}{1-x}\right)\ dx\\ &\quad-\int_0^1\operatorname{Li}_2(1-x)\left(\frac{\operatorname{Li}_2(x)}{x}-\frac{\ln x\ln(1-x)}{1-x}-\frac{\operatorname{Li}_2(1-x)}{1-x}\right)\ dx\\ &=I_1-I_2\\ I_1&=\zeta(2)\left(\zeta(3)-\int_0^1\frac{\ln x\ln(1-x)}{1-x}\ dx-\zeta(3)\right)\\ &=-\zeta(2)\int_0^1\frac{\ln x\ln(1-x)}{x}\ dx=\zeta(2)\sum_{n=1}^\infty\frac1n\int_0^1x^{n-1}\ln x\ dx=\boxed{-\zeta(2)\zeta(3)}\\ I_2&=\int_0^1\frac{\operatorname{Li}_2(1-x)\operatorname{Li}_2(x)}{x}\ dx-\int_0^1\frac{\operatorname{Li}_2(1-x)\ln x\ln(1-x)}{1-x}\ dx-\int_0^1\frac{\operatorname{Li}_2^2(1-x)}{1-x}\ dx\\ &=\int_0^1\frac{\operatorname{Li}_2(1-x)\operatorname{Li}_2(x)}{x}\ dx-\int_0^1\frac{\operatorname{Li}_2(x)\ln x\ln(1-x)}{x}\ dx-\int_0^1\frac{\operatorname{Li}_2^2(x)}{x}\ dx\\ &=\int_0^1\frac{\operatorname{Li}_2(x)}{x}\left(\operatorname{Li}_2(1-x)-\ln x\ln(1-x)-\operatorname{Li}_2(x)\right)\ dx, \quad \text{apply IBP}\\ &=-\zeta(2)\zeta(3)-\int_0^1\operatorname{Li}_3(x)\left(\frac{\ln x}{1-x}+\frac{\ln x}{1-x}-\frac{\ln(1-x)}{x}+\frac{\ln(1-x)}{x}\right)\ dx\\ &=-\zeta(2)\zeta(3)-2\int_0^1\frac{\operatorname{Li}_3(x)\ln x}{1-x}\ dx\\ &=-\zeta(2)\zeta(3)-2\sum_{n=1}^\infty H_n^{(3)}\int_0^1x^n\ln x\ dx\\ &=-\zeta(2)\zeta(3)+2\sum_{n=1}^\infty \frac{H_n^{(3)}}{(n+1)^2}\\ &=-\zeta(2)\zeta(3)+2\left(\sum_{n=1}^\infty \frac{H_n^{(3)}}{n^2}-\zeta(5)\right)\\ &=-\zeta(2)\zeta(3)+2\left(\frac{11}2\zeta(5)-2\zeta(2)\zeta(3)-\zeta(5)\right)\\ &=\boxed{9\zeta(5)-5\zeta(2)\zeta(3)} \end{align} Plugging the boxed results, we get $\ \displaystyle I=4\zeta(2)\zeta(3)-9\zeta(5)$

Note: I solved this problem in a different approach and can be found here page $68-69$

here is a detailed proof to compute $\displaystyle\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}$ and lets start with a different sum: \begin{align} S&=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=\sum_{n=1}^\infty\frac1{n^3}\left(\zeta(2)-\sum_{k=1}^\infty\frac1{(n+k)^2}\right)=\zeta(2)\zeta(3)-\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{n^3(n+k)^2}\\ &=\zeta(2)\zeta(3)-\sum_{k=1}^\infty\sum_{n=1}^\infty\left(\frac{3}{k^4}\left(\frac1{n}-\frac1{n+k}\right)-\frac2{k^3n^2}-\frac1{k^3(n+k)^2}+\frac1{k^2n^3}\right)\\ &=\zeta(2)\zeta(3)-\sum_{k=1}^\infty\left(\frac{3H_k}{k^4}-\frac{2\zeta(2)}{k^3}-\frac{\zeta(2)-H_k^{(2)}}{k^3}+\frac{\zeta(3)}{k^2}\right)\\ &=\zeta(2)\zeta(3)-3\sum_{k=1}^\infty\frac{H_k}{k^4}+2\zeta(2)\zeta(3)+\zeta(2)\zeta(3)-S-\zeta(2)\zeta(3)\\ 2S&=3\zeta(2)\zeta(3)-3\sum_{k=1}^\infty\frac{H_k}{k^4}\\ &=3\zeta(2)\zeta(3)-3\left(3\zeta(5)-\zeta(2)\zeta(3)\right)\\ &=6\zeta(2)\zeta(3)-9\zeta(5)\\ S&=3\zeta(2)\zeta(3)-\frac92\zeta(5) \end{align} 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)\ $, we get \begin{align} \sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}&=\zeta(2)\zeta(3)+\zeta(5)-S\\ &=\frac{11}2\zeta(5)-2\zeta(2)\zeta(3) \end{align}