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,
$$\frac{\operatorname{Li}_r{(x)}}{1-x}=\sum_{n=1}^{\infty}H_{n,r}\,x^n;~~r=2.$$
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
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{\int_{0}^{1}{\ln^2\pars{x}{\rm Li}_2\pars{x} \over 1 - x}\,\dd x\
\stackrel{?}{=}\ -11\zeta\pars{5} + 6\zeta\pars{3}\zeta\pars{2}:\
{\large ?}}$.
$\ds{\large\tt\mbox{The above result is correct !!!}}$.
\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 $\pars{1}$:
$$\color{#66f}{\large%
\int_{0}^{1}{\ln^2\pars{x}{\rm Li}_2\pars{x} \over 1 - x}\,\dd x\
=-11\zeta\pars{5} + 6\zeta\pars{3}\zeta\pars{2}}
\approx {\tt 0.4576}
$$
No comments:
Post a Comment