Problem. Prove that the following dilogarithmic integral has the indicated value: $$\int_{0}^{1}\mathrm{d}x \frac{\ln^2{(x)}\operatorname{Li}_2{(x)}}{1-x}\stackrel{?}{=}-11\zeta{(5)}+6\zeta{(3)}\zeta{(2)}.$$

My attempt:

I began by using the polylogarithmic expansion in terms of generalized harmonic numbers,


Then I switched the order of summation and integration and used the substitution $u=-\ln{x}$ to evaluate the integral:

$$\begin{align} \int_{0}^{1}\mathrm{d}x \frac{\ln^2{(x)}\operatorname{Li}_2{(x)}}{1-x} &=\int_{0}^{1}\mathrm{d}x\ln^2{(x)}\sum_{n=1}^{\infty}H_{n,2}x^n\\ &=\sum_{n=1}^{\infty}H_{n,2}\int_{0}^{1}\mathrm{d}x\,x^n\ln^2{(x)}\\ &=\sum_{n=1}^{\infty}H_{n,2}\int_{0}^{\infty}\mathrm{d}u\,u^2e^{-(n+1)u}\\ &=\sum_{n=1}^{\infty}H_{n,2}\frac{2}{(n+1)^3}\\ &=2\sum_{n=1}^{\infty}\frac{H_{n,2}}{(n+1)^3}. \end{align}$$

So I've reduced the integral to an Euler sum, but unfortunately I've never quite got the knack for evaluating Euler sums. How to proceed from here?


$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}{\ln^2\pars{x}{\rm Li}_2\pars{x} \over 1 - x}\,\dd x\ \stackrel{?}{=}\ -11\zeta\pars{5} + 6\zeta\pars{3}\zeta\pars{2}:\ {\large ?}}$.

$\ds{\large\tt\mbox{The above result is correct !!!}}$.

\begin{align}&\color{#c00000}{\int_{0}^{1}% {\ln^2\pars{x}{\rm Li}_2\pars{x} \over 1 - x}\,\dd x} =\int_{0}^{1}{\ln^2\pars{x} \over 1 - x} \sum_{n = 1}^{\infty}{x^{n} \over n^{2}}\,\dd x \\[3mm]&=\int_{0}^{1}{\ln^2\pars{x} \over 1 - x}\bracks{% \sum_{n = 1}^{\infty}{1 \over n^{2}}- \sum_{n = 1}^{\infty}{1 - x^{n} \over n^{2}}}\,\dd x \\[3mm]&=\zeta\pars{2} \int_{0}^{1}{\ln^2\pars{x} \over 1 - x}\,\dd x -\int_{0}^{1}{\ln^2\pars{x} \over 1 - x} \sum_{n = 1}^{\infty}{1 - x^{n} \over n^{2}}\,\dd x \end{align}

However, \begin{align} \color{#00f}{\int_{0}^{1}{\ln^2\pars{x} \over 1 - x}\,\dd x}&= \int_{0}^{1}\ln\pars{1 - x}\,\bracks{2\ln\pars{x}\,{1 \over x}}\,\dd x =-2\int_{0}^{1}{\rm Li}_{2}'\pars{x}\ln\pars{x}\,\dd x \\[3mm]&=2\int_{0}^{1}{\rm Li}_{2}\pars{x}\,{1 \over x}\,\dd x =2\int_{0}^{1}{\rm Li}_{3}'\pars{x}\,\dd x=2{\rm Li}_{3}\pars{1} =\color{#00f}{2\zeta\pars{3}} \end{align} such that

\begin{align}&\color{#c00000}{\int_{0}^{1}% {\ln^2\pars{x}{\rm Li}_2\pars{x} \over 1 - x}\,\dd x} =2\zeta\pars{2}\zeta\pars{3} -\color{#00f}{\int_{0}^{1}{\ln^2\pars{x} \over 1 - x} \sum_{n = 1}^{\infty}{1 - x^{n} \over n^{2}}\,\dd x}\tag{1} \end{align}

Also, \begin{align}&\color{#00f}{\int_{0}^{1}{\ln^2\pars{x} \over 1 - x} \sum_{n = 1}^{\infty}{1 - x^{n} \over n^{2}}\,\dd x} =\sum_{n = 1}^{\infty}{1 \over n^{2}} \int_{0}^{1}\ln^2\pars{x}\,{1 - x^{n} \over 1 - x}\,\dd x \\[5mm]&=\sum_{n = 1}^{\infty}{1 \over n^{2}} \int_{0}^{1}\ln^2\pars{x}\sum_{k = 1}^{n}x^{k - 1}\,\dd x \\[3mm]&=\sum_{n = 1}^{\infty}{1 \over n^{2}} \sum_{k = 1}^{n}\ \overbrace{\int_{0}^{1}\ln^2\pars{x}x^{k - 1}\,\dd x} ^{\ds{=\ {2 \over k^{3}}}}\ =\ 2\sum_{n = 1}^{\infty}{H_{n}^{\rm\pars{3}} \over n^{2}}\tag{2} \end{align}

The last sum can be evaluated with the generating function $\ds{\sum_{n = 1}^{\infty}x^{n}H_{n}^{\rm\pars{3}} ={{\rm Li}_{3}\pars{x} \over 1 - x}}$. Namely \begin{align} \sum_{n = 1}^{\infty}{x^{n} \over n}\,H_{n}^{\rm\pars{3}} &=\int_{0}^{x}{{\rm Li}_{3}\pars{t} \over t}\,\dd t +\int_{0}^{x}{{\rm Li}_{3}\pars{t} \over 1 - t}\,\dd t \\[3mm]&={\rm Li}_{4}\pars{x} - \ln\pars{1 - x}{\rm Li}_{3}\pars{x} + \int_{0}^{x}\ln\pars{1 - t}{\rm Li}_{3}'\pars{t}\,\dd t \\[3mm]&={\rm Li}_{4}\pars{x} - \ln\pars{1 - x}{\rm Li}_{3}\pars{x} + \int_{0}^{x}\ln\pars{1 - t}\,{{\rm Li}_{2}\pars{t} \over t}\,\dd t \\[3mm]&={\rm Li}_{4}\pars{x} - \ln\pars{1 - x}{\rm Li}_{3}\pars{x} - \int_{0}^{x}{\rm Li}_{2}\pars{t}{\rm Li}_{2}'\pars{t}\,\dd t \\[3mm]&={\rm Li}_{4}\pars{x} - \ln\pars{1 - x}{\rm Li}_{3}\pars{x} - \half\,{\rm Li}_{2}^{2}\pars{x} \\[5mm]\sum_{n = 1}^{\infty}{H_{n}^{\rm\pars{3}} \over n^{2}} &=\int_{0}^{1}{{\rm Li}_{4}\pars{t} \over t}\,\dd t - \int_{0}^{1}{\ln\pars{1 - t}{\rm Li}_{3}\pars{t} \over t}\,\dd t -\half\int_{0}^{1}{{\rm Li}_{2}^{2}\pars{t} \over t}\,\dd t \\[3mm]&=\zeta\pars{5} + {\rm Li}_{2}\pars{1}{\rm Li}_{3}\pars{1} -\int_{0}^{1}{\rm Li}_{2}\pars{t}\,{{\rm Li}_{2}\pars{t} \over t}\,\dd t -\half\int_{0}^{1}{{\rm Li}_{2}^{2}\pars{t} \over t}\,\dd t \\[3mm]&=\zeta\pars{5} + \zeta\pars{2}\zeta\pars{3} -{3 \over 2}\color{#c00000}{\int_{0}^{1}{{\rm Li}_{2}^{2}\pars{t} \over t}\,\dd t} \\[3mm]&=\zeta\pars{5} + \zeta\pars{2}\zeta\pars{3} -{3 \over 2}\bracks{\color{#c00000}{-3\zeta\pars{5} + 2\zeta\pars{2}\zeta\pars{3}}} \end{align} The $\color{#c00000}{\mbox{red result}}$ has been derived elsewhere such that: $$ \sum_{n = 1}^{\infty}{H_{n}^{\rm\pars{3}} \over n^{2}} ={11 \over 2}\,\zeta\pars{5} - 2\zeta\pars{2}\zeta\pars{3} $$

Expresion $\pars{2}$ becomes: $$ \color{#00f}{\int_{0}^{1}{\ln^2\pars{x} \over 1 - x} \sum_{n = 1}^{\infty}{1 - x^{n} \over n^{2}}\,\dd x} =11\zeta\pars{5} - 4\zeta\pars{2}\zeta\pars{3} $$ which we replace in $\pars{1}$: $$\color{#66f}{\large% \int_{0}^{1}{\ln^2\pars{x}{\rm Li}_2\pars{x} \over 1 - x}\,\dd x\ =-11\zeta\pars{5} + 6\zeta\pars{3}\zeta\pars{2}} \approx {\tt 0.4576} $$

  • $\begingroup$ Though this answer is detailed and accurate (+1), with all due respect, I feel that it is a little roundabout as neither the sum $\displaystyle \sum_{n \ge 1}\frac{H_n^{(3)}}{n^2}$ nor the integral $\displaystyle \int^1_0\frac{{\rm Li}_2^2(x)}{x}{\rm d}x$ is any easier to evaluate than the original integral itself, and thus referring to previous answers for the result may be considered as cheating slightly. Then again, this is purely my opinion and it certainly does not change the fact that your answer is of great quality. :) $\endgroup$ – SuperAbound Aug 25 '14 at 6:37
  • $\begingroup$ It took me a while to go through this lengthy answer, but I think I'm satisfied with it now. Thanks Felix! $\endgroup$ – David H Aug 28 '14 at 16:12
  • $\begingroup$ @DavidH It took me a while too to find the correct way to arrive to the answer. Thanks. $\endgroup$ – Felix Marin Aug 28 '14 at 16:26

It is easy to see that $$2\sum^\infty_{n=1}\frac{H_n^{(2)}}{(n+1)^3}=2\sum^\infty_{n=1}\frac{H_{n+1}^{(2)}}{(n+1)^3}-2\sum^\infty_{n=1}\frac{1}{(n+1)^5}=2\sum^\infty_{n=1}\frac{H_n^{(2)}}{n^3}-2\zeta(5)$$ Consider $\displaystyle f(z)=\frac{\pi\cot{\pi z} \ \Psi^{(1)}(-z)}{z^3}$. We know that $$\pi\cot{\pi z}=\frac{1}{z-n}-2\zeta(2)(z-n)+O((z-n)^3)$$ and $$\Psi^{(1)}(-z)=\frac{1}{(z-n)^2}+\left(H_n^{(2)}+\zeta(2)\right)+O(z-n)$$ At the positive integers, \begin{align} {\rm Res}(f,n) &=\operatorname*{Res}_{z=n}\left[\frac{1}{z^3(z-n)^3}+\frac{H_n^{(2)}-\zeta(2)}{z^3(z-n)}\right]\\ &=\frac{H_n^{(2)}}{n^3}-\frac{\zeta(2)}{n^3}+\frac{6}{n^5}\\ \end{align} At the negative integers, \begin{align} {\rm Res}(f,-n)&=-\frac{\Psi^{(1)}(n)}{n^3}\\&=\frac{H_{n-1}^{(2)}-\zeta(2)}{n^3}\\&=\frac{H_{n}^{(2)}}{n^3}-\frac{\zeta(2)}{n^3}-\frac{1}{n^5}\tag1 \end{align} At $z=0$, \begin{align} {\rm Res}(f,0)&=[z^2]\left(\frac{1}{z}-2\zeta(2)z\right)\left(\frac{1}{z^2}+\zeta(2)+2\zeta(3)z+3\zeta(4)z^2+4\zeta(5)z^3\right)\\ &=4\zeta(5)-4\zeta(2)\zeta(3) \end{align} Since the sum of the residues $=0$, we conclude that \begin{align} \color\red{\int^1_0\frac{\log^2{x} \ {\rm Li}_2(x)}{1-x}{\rm d}x} &=2\sum^\infty_{n=1}\frac{H_n^{(2)}}{n^3}-2\zeta(5)\\ &=\zeta(2)\zeta(3)-6\zeta(5)+\zeta(2)\zeta(3)+\zeta(5)-4\zeta(5)+4\zeta(2)\zeta(3)-2\zeta(5)\\ &\large{\color\red{=6\zeta(2)\zeta(3)-11\zeta(5)}} \end{align} Explanation
$(1):$ Use the functional equation $\displaystyle \Psi^{(1)}(z+1)=-\frac{1}{z^2}+\Psi^{(1)}(z)$ which is derived by differentiating the functional equation of the digamma function, as well as the fact that $\displaystyle H_n^{(2)}=\frac{1}{n^2}+H_{n-1}^{(2)}$.

As for how to obtain the laurent series, the series for $\Psi(z)$ was cleverly derived here by Random Variable. In essence, $$\color{blue}{\gamma+\Psi(-z)=\frac{1}{z-n}+H_n+\sum^\infty_{k=1}(-1)^k\left(H_n^{(k+1)}+(-1)^{k+1}\zeta(k+1)\right)(z-n)^k}$$ Differentiating yields $$\color{blue}{\Psi^{(1)}(-z)=\frac{1}{(z-n)^2}+\sum^\infty_{k=1}(-1)^{k+1}k\left(H_n^{(k+1)}+(-1)^{k+1}\zeta(k+1)\right)(z-n)^{k-1}}$$

For $\pi\cot{\pi z}$, \begin{align} \color{blue}{\pi\cot{\pi z}} &=\Psi(1-z)-\Psi(z) \ \ \ \ \ \text{(reflection formula for digamma function)}\\ &=\int^1_0\frac{t^{z-1}-t^{-z}}{1-t}{\rm d}t \ \ \ \ \ \text{(recall that $\Psi(z)=-\gamma+H_{z-1}$)}\\ &=\sum_{k=0}^\infty\int^1_0\left(t^{z+k-1}-t^{-z+k}\right){\rm d}t\\ &=\sum_{k=0}^\infty\left(\frac{1}{z+k}+\frac{1}{z-k-1}\right)\\ &=\frac{1}{z}+\frac{1}{z-1}+\frac{1}{z+1}+\frac{1}{z-2}+\frac{1}{z+2}+\cdots\\ &=\frac{1}{z}+\sum^\infty_{k=1}\left(\frac{1}{z-k}+\frac{1}{z+k}\right)\\ &=\frac{1}{z}+\sum^\infty_{k=1}\frac{2z}{z^2-k^2}\\ &=\frac{1}{z}-2\sum^\infty_{k=1}\sum^\infty_{m=1}\frac{z^{2m-1}}{k^{2m}}\\ &=\color{blue}{\frac{1}{z}-2\sum^\infty_{m=1}\zeta(2m)z^{2m-1}}\\ &=\pi\cot(\pi (z-n)) \ \ \ \ \ \text{(since cotangent has a period of $\pi$)}\\ &=\color{blue}{\frac{1}{z-n}-2\sum^\infty_{m=1}\zeta(2m)(z-n)^{2m-1}}\\ \end{align}

  • 1
    $\begingroup$ I appreciate the hard work you put into this response, SuperAbound. Unfortunately, I do not know enough complex analysis to understand a good bit of it, so I could accept it. But I'm sure it'll be helpful to someone so you have my +1 $\endgroup$ – David H Aug 28 '14 at 16:10

This integral belongs to a wider class of integrals that can always be reduced to single zeta values and to bi-variate zeta values which in turn -- provided the weight of the zeta function in question is not too big-- can also be reduced to single zeta values. My standard way of doing this integrals is as follows. We need to calculate: \begin{equation} {\mathcal I}_{0,2}^{(2)} := \int\limits_0^1 \frac{[\log(1/x)]^2}{2!} \cdot \frac{Li_0(x) Li_2(x)}{x} dx \end{equation} which is just one half of your integral. Now we use the following identity: \begin{equation} \frac{[\log(1/x)]^2}{2!} = \int\limits_{x < \xi_1 < \xi_2 < 1} \prod\limits_{j=1}^2 \frac{d \xi_j}{\xi_j} \end{equation} Inserting this into the above and changing order of integration gives: \begin{eqnarray} {\mathcal I}_{0,2}^{(2)} = \int\limits_{0 < \xi_1 < \xi_2 < 1} \frac{1}{\xi_1} \frac{1}{\xi_2} \underbrace{\int\limits_0^{\xi_1} \frac{Li_0(x) Li_2(x)}{x} d x}_{\left[Li_1(\xi_1) Li_2(\xi_2) - \int\limits_0^{\xi_1} \frac{[Li_1(x)]^2}{x} dx\right]} \cdot d\xi_1 d\xi_2 \end{eqnarray} where we integrated by $x$ using integration by parts. Since two minus zero is even we are left with an irreducible integral that we leave unevaluated for the time being. Now we have: \begin{eqnarray} {\mathcal I}_{0,2}^{(2)} &=& \int\limits_0^1 \frac{[\log(1/\xi_1)]^1}{1!} \left( \frac{1}{2} [Li_2(\xi_1)]^2 \right)^{'} d\xi_1 - \int\limits_0^1 \frac{[\log(1/x)]^2}{2!} \cdot \frac{[Li_1(x)]^2}{x} dx \\ &=& \frac{1}{2} \left( \underbrace{\int\limits_0^1 \frac{[Li_2(\xi)]^2}{\xi} d\xi}_{J_1} - \underbrace{\int\limits_0^1 [\log(1/\xi)]^2 \cdot \frac{[Li_1(\xi)]^2}{\xi} d\xi}_{J_2} \right) \end{eqnarray} Now from Compute an integral containing a product of powers of logarithms. we have that : \begin{eqnarray} J_2&=& -\frac{1}{3} \Psi^{(4)}(1) + 2 \Psi^{(2)}(1) \Psi^{(1)}(1)\\ &=& 8 \zeta(5) - 4 \zeta(3) \zeta(2) \end{eqnarray} where $\Psi^{(j)}(1)$ is the polygamma function at unity and $\Psi^{(j)}(1)=(-1)^{j+1} j! \zeta(j+1)$. On the other hand we have: \begin{eqnarray} J_1 &=& \sum\limits_{m\ge1,n\ge 1} \frac{1}{m^2} \frac{1}{n^2} \frac{1}{(m+n)}\\ & =& \sum\limits_{m\ge 1} \left(\frac{\zeta(2)}{m^3} - \frac{H_m}{m^4}\right) \\ &=& \zeta(2) \zeta(3) - {\bf H}^{(1)}_4(+1)\\ &=& 2 \zeta(2) \zeta(3) - 3 \zeta(5) \end{eqnarray} where in the last line above we used my answer to Calculating alternating Euler sums of odd powers . Therefore: \begin{equation} {\mathcal I}^{(2)}_{0,2}= \frac{1}{2} \left(J_1-J_2\right)= \frac{1}{2} \left(-11 \zeta(5)+6 \zeta(3) \zeta(2) \right) \end{equation} as expected. Note that exactly the same steps can be performed is we replace the power of the logarithm and the orders of the poly-logarithms by any nonnegative integers. The generic result is then given in An integral involving product of poly-logarithms and a power of a logarithm. .

  • $\begingroup$ Oh hey, it's you! I may as well take this opportunity to confess that I have been silently stalking every single one of your polylogarithm posts recently with great interest. Polylogs have been an on-again-off-again hobby of mine for a while now. If I may ask, is there any reason in particular why you seem to be so preoccupied with them recently? Cheers! $\endgroup$ – David H Jun 20 '17 at 15:46
  • $\begingroup$ @David H: One reason would be the following. Consider computing an integral: \begin{equation} \int\limits_0^1 \frac{P(\xi)}{Q(\xi)} \cdot \prod\limits_{j=1}^d [\log(1+\xi/a_j )]^{b_j} d\xi\end{equation} where $P$ and $Q$ are polynomials. It is not hard to see that all those integrals are reduced to Euler sums which in turn are reduced to multiple zeta values(MZVs). It is therefore essential to be able to compute MZVs for any given weight or depth. Unfortunately I have not seen any generic results on that.I want to create an algorithm for doing that. $\endgroup$ – Przemo Jun 20 '17 at 16:39
  • $\begingroup$ That's essentially the exact same problem that got me started down the polylogarithm rabbit hole, probably three years ago. I can't even tell you how many weekends I spent scouring the internet for just one source referencing the problem. I thought surely even if no general results exist, someone would have bothered writing paper describing the roadblocks. But at the end of the day, I'm mostly a self-taught hobbyist when it comes to formal mathematics, so there are likely plenty of better ways to search for results that I didn't think of. Do you have any decent papers to suggest? $\endgroup$ – David H Jun 20 '17 at 17:21
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    $\begingroup$ @David H:To be honest I am also not a formal mathematician but I simply need a solid background in discrete mathematics. When I look back at my math courses in the university I only realize how little I know when confronted with real life problems I need to solve now. So I also have to learn all that stuff myself many times with great pain. As for references I found this usna.edu/Users/math/meh/biblio.html . There is also an interface for calculating zeta values wayback.cecm.sfu.ca/cgi-bin/EZFace/zetaform.cgi . $\endgroup$ – Przemo Jun 21 '17 at 9:39

From $$\sum_{n=1}^\infty H_n^{(2)}x^n=\frac{\operatorname{Li}_2(x)}{1-x}$$

it follows that

$$I=\int_0^1\frac{\ln^2(x)\operatorname{Li}_2(x)}{1-x}dx=\sum_{n=1}^\infty H_n^{(2)}\int_0^1 x^n \ln^2(x)dx\\=2\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)^3}=2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}-2\zeta(5)\tag1$$

By Cauchy product we have


set $x=1$ to get


Now lets use the well-known identity


set $p=2$ and $q=3$


Plugging this result in $(2)$ yields


and fianlly, plugging this result in $(1)$, the closed form of $I$ follows. Note that the value $\sum_{n=1}^\infty \frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$ was used in our calculations which can be found using Euler identity.


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