Compute $ \int_{0}^{1}\frac{\ln(x) \ln^2 (1-x)}{x} dx $
I'm looking for some nice proofs at this problem. One idea would be to use Taylor expansion and then integrating term by term. What else can we do? Thanks.
Compute $ \int_{0}^{1}\frac{\ln(x) \ln^2 (1-x)}{x} dx $
I'm looking for some nice proofs at this problem. One idea would be to use Taylor expansion and then integrating term by term. What else can we do? Thanks.
This answer is from my old calculation.
First, assume we are well aware of the following famous result.
$\zeta(2) =\frac{\pi^{2}}{6}, \quad \zeta(4) =\frac{\pi^{4}}{90}$
Next, by a simple calculation we obtain
$ H_{n} := \sum_{k=1}^{n} \frac{1}{k} =\int_{0}^{1}\frac{1-t^{n}}{1-t}\, dt. $
and
$ \frac{\log (1-x)}{1-x}\ =\ -\sum_{n=1}^{\infty}H_{n}x^{n}. $
Finally, define the polylogarithm as
$ \mathrm{Li}_{s}(x) :=\sum_{n=1}^{\infty} \frac{x^n}{n^s}, $
so that it satisfies the recurrence relation
$ \mathrm{Li}_{1}(x) =-\log (1-x) , \quad \mathrm{Li}_{s+1}(x) =\int_{0}^{x}\frac{\mathrm{Li}_{s}(t)}{t}\, dt $
and the identity
$ \mathrm{Li}_{s}(1) =\zeta(s). $
The the all-in-one straight calculation goes as follows:
\begin{align*} \int_{0}^{1}\frac{\log x\log^{2}(1-x)}{x}\, dx & = \int_{0}^{1}\frac{\log (1-x)\log^{2}x}{1-x}\, dx = -\sum_{n=1}^{\infty}H_{n}\int_{0}^{1}x^{n}\log^{2}x\, dx\\ & = -2\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)^{3}}\\ & = 2\sum_{n=1}^{\infty}\left[\frac{1}{(n+1)^{4}}-\frac{H_{n+1}}{(n+1)^{3}}\right] = 2\sum_{n=0}^{\infty}\left[\frac{1}{(n+1)^{4}}-\frac{H_{n+1}}{(n+1)^{3}}\right]\\ & = 2\zeta(4)-2\sum_{n=1}^{\infty}\frac{H_{n}}{n^{3}}\\ & = 2\zeta(4)-2\sum_{n=1}^{\infty}\frac{1}{n^{3}}\int_{0}^{1}\frac{1-t^{n}}{1-t}\, dt = 2\zeta(4)-2\int_{0}^{1}\frac{\zeta(3)-\mathrm{Li}_{3}(t)}{1-t}\, dt\\ & = 2\zeta(4)+\left[2 (\zeta(3)-\mathrm{Li}_{3}(t))\log (1-t)\right]_{0}^{1}+2\int_{0}^{1}\frac{\mathrm{Li}_{2}(t)\log (1-t)}{t}\, dt\\ & = 2\zeta(4)-2\int_{0}^{1}\mathrm{Li}_{2}(t)\frac{d\mathrm{Li}_{2}(t)}{dt}\, dt\\ & = 2\zeta(4)-\left[\mathrm{Li}_{2}^{2}(t)\right]_{0}^{1} = 2\zeta(4)-\zeta(2)^{2} = \frac{\pi^{4}}{45}-\frac{\pi^{4}}{36} = -\frac{\pi^{4}}{180}\\ & = -\frac{1}{2}\zeta(4). \end{align*}
The Taylor expansion approach gives you $-2 \sum_{k=1}^\infty H_k/(k+1)^3$ where $H_k = \sum_{n=1}^k 1/n$. Wolfram Alpha says this is $-\pi^4/180$, but I don't know how it gets that.
@Chri's sister: see here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=353720&p=1921474&hilit=Borwein#p1921474
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\Huge\left. a\right)}$ \begin{align} &\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x = \left.\partiald[2]{}{\mu}\int_{0}^{1}\ln\pars{x} \,{\pars{1 - x}^{\mu} -1 \over x}\,\dd x\,\right\vert_{\large\ \mu\ =\ 0^{+}} \\[5mm] & = \left.\partiald[2]{}{\mu}\sum_{n = 1}^{\infty}{\mu \choose n}\pars{-1}^{n}\ \int_{0}^{1}\ln\pars{x}x^{n - 1}\,\dd x\,\right\vert_{\large\ \mu\ =\ 0^{+}} = \left.-\,\partiald[2]{}{\mu}\sum_{n = 1}^{\infty}{\mu \choose n} {\pars{-1}^{n} \over n^{2}}\,\right\vert_{\large\ \mu\ =\ 0^{+}} \\[5mm] & = \left.-\,\partiald[2]{}{\mu}\sum_{n = 1}^{\infty}{1 \over n!}\, {\Gamma\pars{\mu + 1} \over \Gamma\pars{1 + \mu - n}}\, {\pars{-1}^{n} \over n^{2}}\,\right\vert_{\large\ \mu\ =\ 0^{+}}\label{1}\tag{1} \end{align}
Note that
\begin{align} {\Gamma\pars{\mu + 1} \over \Gamma\pars{1 + \mu - n}} & = {\Gamma\pars{\mu + 1} \over \pi/\braces{\Gamma\pars{n - \mu}\sin\pars{\pi\bracks{n - \mu}}}} = {\pars{-1}^{n + 1} \over \pi}\,\sin\pars{\pi\mu} \Gamma\pars{\mu + 1}\Gamma\pars{n - \mu} \\[5mm] & = \pars{-1}^{n + 1}\,\mu \braces{\Gamma\pars{1}\Gamma\pars{n} + \bracks{\Gamma'\pars{1}\Gamma\pars{n} - \Gamma\pars{1}\Gamma'\pars{n}}\mu} + \,\mrm{O}\pars{\mu^{3}} \\[5mm] & = \pars{-1}^{n + 1}\,\Gamma\pars{n}\mu + \pars{-1}^{n + 1}\bracks{-\gamma\pars{n - 1}! -\pars{n - 1}!\,\Psi\pars{n}}\color{#f00}{\mu^{2}} + \,\mrm{O}\pars{\mu^{3}} \\[5mm] & = \pars{-1}^{n + 1}\,\Gamma\pars{n}\mu - \pars{-1}^{n + 1}\pars{n - 1}!\,H_{n - 1}\,\color{#f00}{\mu^{2}} + \,\mrm{O}\pars{\mu^{3}} \end{align}
such that $\ds{\left.\partiald[2]{}{\mu}{\Gamma\pars{\mu + 1} \over \Gamma\pars{1 + \mu - n}}\right\vert_{\ \mu\ =\ 0^{+}} = -2\pars{-1}^{n + 1}\pars{n - 1}!\,H_{n - 1}}$
Expression \eqref{1} becomes \begin{align} &\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x = -\sum_{n = 1}^{\infty}{1 \over n!}\, \bracks{-2\pars{-1}^{n + 1}\pars{n - 1}!\,H_{n - 1}}\, {\pars{-1}^{n} \over n^{2}} = -2\sum_{n = 1}^{\infty}{H_{n - 1} \over n^{3}} \\[5mm] & = -2\sum_{n = 1}^{\infty}{H_{n} \over n^{3}} + 2\sum_{n = 1}^{\infty}{1 \over n^{4}} = -2\pars{\pi^{4} \over 72} + 2\zeta\pars{4} = -\,{5 \over 2}\,\zeta\pars{4} + 2\zeta\pars{4} = \bbx{-\,{1 \over 2}\,\zeta\pars{4}} \end{align}
Note that $\ds{\sum_{n = 1}^{\infty}{H_{n} \over n^{3}} = {\pi^{4} \over 72} = {5 \over 4}\,\zeta\pars{4}}$ is a well known identity. See expression $\pars{19}$ in this page.
Note that
\begin{align} {1 \over 3}\int_{0}^{1}{\ln^{3}\pars{1 - x} \over x}\,\dd x = {1 \over 3}\int_{0}^{1}{\ln^{3}\pars{x} \over 1 - x}\,\dd x = \int_{0}^{1}\ln\pars{1 - x}{\ln^{2}\pars{x} \over x}\,\dd x = -\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x \end{align} such that \eqref{2} is reduced to \begin{align} \int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x & = \underbrace{{1 \over 3}\int_{0}^{1}\bracks{% \ln^{3}\pars{x \over 1 - x} - \ln^{3}\pars{x}}\,{\dd x \over x}} _{\ds{\mc{I}_{1}}} \\[5mm] & -\ \underbrace{2\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x} _{\ds{\mc{I}_{2}}}\ =\ \mc{I}_{1} - \,\mc{I}_{2}\label{3}\tag{3} \end{align}
Hereafter, I'll evaluate $\ds{\,\mc{I}_{1}}$ and $\ds{\,\mc{I}_{2}}$.
In this answer I will make use of a Maclaurin series expansion for the term $\ln^2 (1 - x)$, which I show here to be $\ln^2 (1 - x) = 2 \sum_{n = 2}^\infty \frac{H_{n - 1} x^n}{n},$ and the well-known Euler sum of $\sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{1}{2} \zeta^2 (2),$ several proofs for which can be found here.
From the above Maclaurin series expansion for $\ln^2 (1 - x)$ the integral can be written as \begin{align*} \int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \sum_{n = 2}^\infty \int_0^1 x^{n - 1} \ln x \, dx. \end{align*} The integral that appears to the right can be readily found by parts. The result is $\int_0^1 x^{n - 1} \ln x \, dx = -\frac{1}{n^2}.$ Thus $\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = -2 \sum_{n = 2}^\infty \frac{H_{n - 1}}{n^3}.$
From properties of harmonic numbers, since $H_n = H_{n - 1} + \frac{1}{n},$ the integral becomes $\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \sum_{n = 2}^\infty \frac{1}{n^4} - 2 \sum_{n = 2}^\infty \frac{H_n}{n^3} = 2 \sum_{n = 1}^\infty \frac{1}{n^4} - 2 \sum_{n = 1}^\infty \frac{H_n}{n^3}.$ As $\sum_{n = 1}^\infty \frac{1}{n^4} = \zeta (4) \quad \text{and} \quad \sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{1}{2} \zeta^2 (2),$ we have $\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \zeta (4) - \zeta^2 (2) = - \frac{\pi^4}{180} = -\frac{1}{2} \zeta (4),$ as expected.