Evaluation of a dilogarithmic integral

Question

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

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My attempt:

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

$$\frac{\operatorname{Li}_r{(x)}}{1-x}=\sum_{n=1}^{\infty}H_{n,r}\,x^n;~~r=2.$$

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?

Answer 1

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

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

\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 $\pars{1}$: $$\color{#66f}{\large% \int_{0}^{1}{\ln^2\pars{x}{\rm Li}_2\pars{x} \over 1 - x}\,\dd x\ =-11\zeta\pars{5} + 6\zeta\pars{3}\zeta\pars{2}} \approx {\tt 0.4576} $$

Answer 2

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}

Answer 3

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

$$\operatorname{Li}_2(x)\operatorname{Li}_3(x)=\sum_{n=1}^\infty\left(\frac{6H_n}{n^4}+\frac{3H_n^{(2)}}{n^3}+\frac{H_n^{(3)}}{n^2}-\frac{10}{n^5}\right)x^n$$

set $x=1$ to get

$$\zeta(2)\zeta(3)=6\sum_{n=1}^\infty\frac{H_n}{n^4}+3\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}+\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}-10\zeta(5)\tag{2}$$

Now lets use the well-known identity

$$\sum_{n=1}^\infty\frac{H_n^{(p)}}{n^q}+\sum_{n=1}^\infty\frac{H_n^{(q)}}{n^p}=\zeta(p)\zeta(q)+\zeta(p+q)$$

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

$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\zeta(2)\zeta(3)+\zeta(5)-\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}$$

Plugging this result in $(2)$ yields

$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)$$

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.

Answer 4

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. .