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


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!

  • $\begingroup$ 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
    Commented Dec 16, 2018 at 11:19
  • 1
    $\begingroup$ 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
    Commented Dec 16, 2018 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
    Commented Dec 16, 2018 at 13:07
  • 1
    $\begingroup$ 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
    Commented Dec 16, 2018 at 14:01
  • $\begingroup$ Have you asked this on AoPS too? Or contact the author of the page? $\endgroup$
    – Zacky
    Commented Jan 30, 2019 at 11:25

2 Answers 2


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 $(6),$


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

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

  • 1
    $\begingroup$ 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
    Commented Jan 31, 2019 at 14:16
  • 1
    $\begingroup$ @mrtaurho I have found it in RMM and linked it. $\endgroup$
    – Zacky
    Commented Feb 2, 2019 at 14:07
  • 1
    $\begingroup$ Wonderful! Thank you for your time and effort. $\endgroup$
    – mrtaurho
    Commented Feb 2, 2019 at 14:11

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

  • $\begingroup$ 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
    Commented Jun 27, 2019 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 $$\sum_{n=1}^\infty \frac{H_n^{(p)}}{n^p}=\frac12\left(\zeta^2(p)+\zeta(2p)\right)$$ $\endgroup$ Commented Jun 27, 2019 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
    Commented Jun 27, 2019 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
    Commented Jun 29, 2019 at 12:34

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