integration - Evaluate$intlimits_0^1 [log(x)log(1-x)+operatorname{Li}_2(x)]left[frac{operatorname{Li}_2(x)}{x(1-x)}-frac{zeta(2)}{1-x}right]mathrm dx$

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

This integral haunts me for while and I am still unable to evaluate it which annoys me even more. For the first time I have encountered it within Mathematical Analysis $-$ A collection of Problems by Tolaso J. Kos $($Page $27$, Problem $282$$)$ and I am still stumped by this one.

However, today I am have come 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. I was baffled as I recognized the structure of the final value; moreover reminding 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 similiar 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 of all 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

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

Secondly applying the subsititution $x=1-x$ 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 multplying the two brackets out does not seem like a good idea to me hence 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 concerning the fact that I am not used to e.g. double series and their evaluation. I also tried various ways of Integration by Parts but this seems to be pointless since all variations ended up in producing a divergent term $-$ unless I have missed a sepcial 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 notwithstanding that I would appreciate a solution without involving them. Hence this integral appeared within a collocation of Analysis Probelms I am rather sure that it has been already evaluated somewhere $($maybe even here on MSE where I was not able to find it$)$.

Thanks in advance!

Answer

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 $70$.