TITLE: Evaluate$\int\limits_0^1 [\log(x)\log(1-x)+\operatorname{Li}_2(x)]\left[\frac{\operatorname{Li}_2(x)}{x(1-x)}-\frac{\zeta(2)}{1-x}\right]\mathrm dx$
QUESTION [14 upvotes]: The following is from Mathematical Analysis $-$ A collection of Problems by Tolaso J. Kos $($Page $27$, Problem $282$$)$

$$\mathfrak{I}=\int\limits_0^1 \left[\log(x)\log(1-x)+\operatorname{Li}_2(x)\right]\left[\frac{\operatorname{Li}_2(x)}{x(1-x)}-\frac{\zeta(2)}{1-x}\right]\mathrm dx=4\zeta(2)\zeta(3)-9\zeta(5)\tag1$$

Today I came across this question asking for the evaluation of the integral

$$\mathfrak{J}=\int\limits_0^{\pi/2}\frac{\log^2(\sin x)\log^2(\cos x)}{\sin x\cos x}\mathrm dx=\frac12\zeta(5)-\frac14\zeta(2)\zeta(3)\tag2$$

Which can done be "rather simple" by invoking the fourth derivative of the Beta Function. The final structure of the result reminded me of the logarithmic integral $(1)$ I was not able to evaluate. It might turn out that this relation is by pure chance but nevertheless it motivated me to look at $(1)$ again. It is hardly probable that $(1)$ can be done in a similar way like $(2)$ in xpaul's answer to the linked question due the involved Dilogarithms$-$but anyway you can prove me wrong.
I have not got that far with $(1)$ but, however, I noticed two, I would say quite interesting, facts about the integral. First, consider the following, well-known functional relation of the Dilogarithm
$$\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\zeta(2)-\log(x)\log(1-x)$$
which can be used in order to get rid of the $\log(x)\log(1-x)$-term within $(1)$ and leading to
$$\small\int\limits_0^1 \left[\log(x)\log(1-x)+\operatorname{Li}_2(x)\right]\left[\frac{\operatorname{Li}_2(x)}{x(1-x)}-\frac{\zeta(2)}{1-x}\right]\mathrm dx=\int\limits_0^1 \left[\zeta(2)-\operatorname{Li}_2(1-x)\right]\left[\frac{\operatorname{Li}_2(x)}{x(1-x)}-\frac{\zeta(2)}{1-x}\right]\mathrm dx$$
Second, applying the substitution $x=1-x$ after a minor reshape yields to
$$\small\begin{align}
\int\limits_0^1 \left[\zeta(2)-\operatorname{Li}_2(1-x)\right]\left[\frac{\operatorname{Li}_2(x)}{x(1-x)}-\frac{\zeta(2)}{1-x}\right]\mathrm dx&=\int\limits_0^1 \left[\frac{\zeta(2)}{1-x}-\frac{\operatorname{Li}_2(1-x)}{1-x}\right]\left[\frac{\operatorname{Li}_2(x)}x-\zeta(2)\right]\mathrm dx\\
&=\int\limits_0^1 \left[\zeta(2)-\operatorname{Li}_2(1-x)\right]\left[\frac{\operatorname{Li}_2(x)}x-\zeta(2)\right]\frac{\mathrm dx}{1-x}\\
&=\int\limits_0^1 \left[\zeta(2)-\operatorname{Li}_2(x)\right]\left[\frac{\operatorname{Li}_2(1-x)}{1-x}-\zeta(2)\right]\frac{\mathrm dx}x\\
&=\int\limits_0^1 \left[\frac{\zeta(2)}x-\frac{\operatorname{Li}_2(x)}x\right]\left[\frac{\operatorname{Li}_2(1-x)}{1-x}-\zeta(2)\right]\mathrm dx
\end{align}$$
I want to point out the quite interesting one could say "almost"-symmetry of the two integrals

$$\begin{align}
\mathfrak{I}_1&=\int\limits_0^1 \left[\frac{\zeta(2)}{1-x}-\frac{\operatorname{Li}_2(1-x)}{1-x}\right]\left[\frac{\operatorname{Li}_2(x)}x-\zeta(2)\right]\mathrm dx\\
\mathfrak{I}_2&=\int\limits_0^1 \left[\frac{\zeta(2)}x-\frac{\operatorname{Li}_2(x)}x\right]\left[\frac{\operatorname{Li}_2(1-x)}{1-x}-\zeta(2)\right]\mathrm dx
\end{align}$$

which might be helpful for the actual evaluation. But from hereon I have no clue how to continue.
Just expanding the brackets out does not seem like a good idea to me. Since one the one hand it is not elegant at all and on the other hand it would lead to to the term $\operatorname{Li}_2(x)\operatorname{Li}_2(1-x)$ for which I have no idea how to deal with (I am not that confident using double series). I also tried various ways of IBP but this seems to be pointless since all variations ended up in producing a divergent term$-$unless I have missed a special choice of $u$ and $\mathrm v$. I have not figured out a suitable substitution nor an appropriate newly introduced parameter (for the application of Feynman's Trick) and the I do not know whether a series expansion would be helpful or not (connected with this issue is the possibility of a double summation with which I cannot really deal).

Thus, I am asking for the closed-form evaluation of $(1)$ hopefully equal to the given value (which works out numerically according to WolframAlpha). Even though I have troubles with double series I would accept an answer invoking these but I would appreciate a solution without involving them. As this integral appeared within a collection of Analysis Problems I am quite sure that it has been already evaluated somewhere (maybe even here on MSE where I was not able to find it!).

Thanks in advance!

REPLY [4 votes]: $$I=\int_0^1\left(\ln x\ln(1-x)+\operatorname{Li}_2(x)\right)\left(\frac{\operatorname{Li}_2(x)}{x}+\frac{\operatorname{Li}_2(x)}{1-x}-\frac{\zeta(2)}{1-x}\right)\ dx$$
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}