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$$mathfrak{I}=intlimits_0^1 left[log(x)log(1-x)+operatorname{Li}_2(x)right]left[frac{operatorname{Li}_2(x)}{x(1-x)}-frac{zeta(2)}{1-x}right]mathrm dx=4zeta(2)zeta(3)-9zeta(5)tag1$$

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

$$mathfrak{J}=intlimits_0^{pi/2}frac{log^2(sin x)log^2(cos x)}{sin xcos x}mathrm dx=frac12zeta(5)-frac14zeta(2)zeta(3)tag2$$

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

$$smallintlimits_0^1 left[log(x)log(1-x)+operatorname{Li}_2(x)right]left[frac{operatorname{Li}_2(x)}{x(1-x)}-frac{zeta(2)}{1-x}right]mathrm dx=intlimits_0^1 left[zeta(2)-operatorname{Li}_2(1-x)right]left[frac{operatorname{Li}_2(x)}{x(1-x)}-frac{zeta(2)}{1-x}right]mathrm dx$$

Secondly applying the subsititution $x=1-x$ after a minor reshape yields to

$$smallbegin{align}
intlimits_0^1 left[zeta(2)-operatorname{Li}_2(1-x)right]left[frac{operatorname{Li}_2(x)}{x(1-x)}-frac{zeta(2)}{1-x}right]mathrm dx&=intlimits_0^1 left[frac{zeta(2)}{1-x}-frac{operatorname{Li}_2(1-x)}{1-x}right]left[frac{operatorname{Li}_2(x)}x-zeta(2)right]mathrm dx\
&=intlimits_0^1 left[zeta(2)-operatorname{Li}_2(1-x)right]left[frac{operatorname{Li}_2(x)}x-zeta(2)right]frac{mathrm dx}{1-x}\
&=intlimits_0^1 left[zeta(2)-operatorname{Li}_2(x)right]left[frac{operatorname{Li}_2(1-x)}{1-x}-zeta(2)right]frac{mathrm dx}x\
&=intlimits_0^1 left[frac{zeta(2)}x-frac{operatorname{Li}_2(x)}xright]left[frac{operatorname{Li}_2(1-x)}{1-x}-zeta(2)right]mathrm dx
end{align}$$

I want to point out the quite interesting one could say "almost"-symmetry of the two integrals

$$begin{align}
mathfrak{I}_1&=intlimits_0^1 left[frac{zeta(2)}{1-x}-frac{operatorname{Li}_2(1-x)}{1-x}right]left[frac{operatorname{Li}_2(x)}x-zeta(2)right]mathrm dx\
mathfrak{I}_2&=intlimits_0^1 left[frac{zeta(2)}x-frac{operatorname{Li}_2(x)}xright]left[frac{operatorname{Li}_2(1-x)}{1-x}-zeta(2)right]mathrm dx
end{align}$$

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!

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
$$int_0^1 frac{ln(1-x) text{Li}_2(x)}{x^2} dx=int_0^1left(1-frac{1}{x}right)left(frac{text{Li}_2(x)}{1-x}+frac{ln^2(1-x)}{x}right) dx$$
$$=-int_0^1 frac{text{Li}_2(x)}{x} dx + int_0^1 frac{ln^2(1-x)}{x} dx - int_0^1 frac{ln^2(1-x)}{x^2} dx=-zeta(3)+int_0^1 frac{ln^2x}{1-x} dx -int_0^1 frac{ln^2(1-x)}{x^2} dx.$$
$$=zeta(3)-int_0^1 frac{ln^2(1-x)}{x^2} dx. (4)$$
Thus, by $(2),(3)$ and $(4),$ we have $K=0$ and hence, by $(1),$
$$I=-J=-int_0^1 frac{text{Li}_2(x)text{Li}_2(1-x)}{x(1-x)} dx=-2int_0^1 frac{text{Li}_2(x)text{Li}_2(1-x)}{x} dx.$$
So integration by parts with $u=text{Li}_2(1-x)$ and $dv=frac{text{Li}_2(x)}{x} dx$ gives
$$I=2int_0^1 frac{text{Li}_3(x)ln x}{1-x} dx=2int_0^1 text{Li}_3(x) ln x sum_{m ge 1}x^{m-1} dx=2sum_{m ge 1} int_0^1 x^{m-1}text{Li}_3(x) ln x dx$$
$$=2sum_{m ge 1} int_0^1x^{m-1}sum_{n ge 1} frac{x^n}{n^3} ln x dx=2sum_{m,n ge 1} frac{1}{n^3}int_0^1x^{n+m-1}ln x dx=-2sum_{m,n ge 1} frac{1}{n^3(n+m)^2}$$
$$=-sum_{m,n ge 1} left(frac{1}{n^3(n+m)^2}+frac{1}{m^3(n+m)^2}right). (5)$$
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
$$I=-sum_{m,n ge 1}left(frac{1}{n^3m^2}-frac{2}{n^2m^3}+frac{3}{m^3n(n+m)}right)=zeta(2)zeta(3)-3sum_{m,n ge 1} frac{1}{m^3n(n+m)}$$
$$=zeta(2)zeta(3)-3sum_{m ge 1} frac{1}{m^4} sum_{n ge 1}left(frac{1}{n}-frac{1}{n+m}right)=zeta(2)zeta(3)-3sum_{m ge 1} frac{H_m}{m^4}, (6)$$
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
$$sum_{m ge 1} frac{H_m}{m^4}=3zeta(5)-zeta(2)zeta(3)$$
and so, by $(6),$

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

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

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