Problem. Prove that the following dilogarithmic integral has the indicated value:
∫10dxln2(x)Li2(x)1−x?=−11ζ(5)+6ζ(3)ζ(2).
My attempt:
I began by using the polylogarithmic expansion in terms of generalized harmonic numbers,
Lir(x)1−x=∞∑n=1Hn,rxn; r=2.
Then I switched the order of summation and integration and used the substitution u=−lnx to evaluate the integral:
∫10dxln2(x)Li2(x)1−x=∫10dxln2(x)∞∑n=1Hn,2xn=∞∑n=1Hn,2∫10dxxnln2(x)=∞∑n=1Hn,2∫∞0duu2e−(n+1)u=∞∑n=1Hn,22(n+1)3=2∞∑n=1Hn,2(n+1)3.
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