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

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!

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  • $\begingroup$ Could the downvoter please elaborate on why he downvoted? Is my question missing some details or effort; can I add something which would make the downvote redundant? $\endgroup$ – mrtaurho Dec 16 '18 at 11:19
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    $\begingroup$ I can't speak for the downvoter, but perhaps they noticed that your proposed value for $\mathfrak{I}$ in equation $(1)$ is way off numerically. The correct value should be $\mathfrak{I}=4\,\zeta{\left(2\right)}\,\zeta{\left(3\right)}-\color{red}{9}\,\zeta{\left(5\right)}$. $\endgroup$ – David H Dec 16 '18 at 12:55
  • $\begingroup$ @David H Oh gosh. I have forgotten to write down the $9$. I will add this in the post hence it is like this within the source I have given. $\endgroup$ – mrtaurho Dec 16 '18 at 13:07
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    $\begingroup$ Antiderivative of \begin{align}\frac{1}{1-x}\left[\frac{\operatorname{Li}_2(x)}x-\zeta(2)\right]\end{align} is computable. This is the part corresponding to $\zeta(2)\zeta(3)$ $\endgroup$ – FDP Dec 16 '18 at 14:01
  • $\begingroup$ Have you asked this on AoPS too? Or contact the author of the page? $\endgroup$ – Zacky Jan 30 at 11:25
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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 $70$.

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    $\begingroup$ I am surprised over and over again by his ability of evaluating different and complex integrals. It always seems so easy seeing his approach but I could have never come up with this by myself! However, thank you for cross posting the question on AoPS, I already gave him a thumps up there, and for sharing ysharifi's solution here. Anyway I am feeling a little bit uneasy about the extensive usage of double sums in the end even though there are quite simple in this case. $\endgroup$ – mrtaurho Jan 31 at 14:16
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    $\begingroup$ @mrtaurho I have found it in RMM and linked it. $\endgroup$ – Zacky Feb 2 at 14:07
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    $\begingroup$ Wonderful! Thank you for your time and effort. $\endgroup$ – mrtaurho Feb 2 at 14:11
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$$I=\int_0^1\left(\ln x\ln(1-x)+\operatorname{Li}_2(x)\right)\left(\frac{\operatorname{Li}_2(x)}{x}+\frac{\operatorname{Li}_2(x)}{1-x}-\frac{\zeta(2)}{1-x}\right)\ dx$$ using the Dilogarithmic identity for the first and second parenthesis, 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}

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  • $\begingroup$ Nice one! (+1) Looks less messy than the solution provided by ysharifi and almost like it was designed for you (judging from what I have seen so far from you, here on MSE). As you constantly refer to sums of the type $\frac{H^{(p)}_n}{n^s}$ as well-known I guess you are referring to Euler's Forumla? Is there another way to obtain those values (without integrating the generating function repeately, maybe by using SBP, as we recently talked about it)? Furthermore, how did you learned to deal that well with Harmonic Numbers? Any good references? $\endgroup$ – mrtaurho Jun 27 at 20:07
  • $\begingroup$ Thank you mrtaurho. That sum is well known and i evaluated it using series manipulation and I'll provide the proof soon. I learned harmonic series on my won and with help of Cornel he is the mater of the harmonic series. I would recommend his book , almost impossible integrals , sums and series. using Abel's summation, we can prove $$\sum_{n=1}^\infty \frac{H_n^{(p)}}{n^q}+\sum_{n=1}^\infty \frac{H_n^{(q)}}{n^p}=\zeta(p)\zeta(q)+\zeta(p+q)$$ and we have a special case when $q=p, $ we have $$\sum_{n=1}^\infty \frac{H_n^{(p)}}{n^p}=\frac12\left(\zeta^2(p)+\zeta(2p)\right)$$ $\endgroup$ – Ali Shather Jun 27 at 21:02
  • $\begingroup$ I have got the feeling that I really need this book, as I encountered it already quoted quite often by our fellow user Zacky. Your given identity looks interesting (aswell as helpful though); I will give it a try as soon as I have time for it. $\endgroup$ – mrtaurho Jun 27 at 21:07
  • $\begingroup$ Interesting: I was aware of the solution in the RMM (as Zacky did some great research in finding the proposal of this problem) but was completely oblivious to the fact, that it was actually you, who solved it there! $\endgroup$ – mrtaurho Jun 29 at 12:34

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