Different ways to Prove $\int_{0}^{1}\frac {{\log(x)} {\log(1-x)}}{x}dx=\zeta(3)$

Question

Question:- Prove that $$\int_0^1 \frac {\log(x) \log(1-x)}{x} \, dx=\zeta(3)$$

That's how I prove it.

Let $a>0$, Consider the following series

$$\sum_{n=1}^\infty \frac {1}{(n+a)^2}=\int_0^1 \int_0^1 \frac {(xy)^a}{1-xy} \, dx \, dy$$

Now, differentiate with respect to $a$ and let $a=0$ to obtain

$$\begin{align} \zeta(3)&=\frac{-1} 2 \int_0^1\int_0^1 \frac {\log(xy)}{1-xy} \, dx\, dy\\\\ &=\frac{-1}{2}\int_0^1 \int_0^1 \frac {\log(x)+\log(y)}{1-xy} \, dx \, dy \end{align}$$

Using symmetry,

$$\zeta(3)=-\int_0^1 \int_0^1 \frac {\log(x)}{1-xy} \, dx$$

$$\zeta(3)=\int_0^1 \frac {\log(x) \log(1-x)}{x} \, dx$$

I want to know what other methods can be used to solve this integral.

Answer 1

Let $I$ be given by the integral

$$I=\int_0^1 \frac{\log(x)\log(1-x)}{x}\,dx$$

Integrating by parts with $u=\log(x)$ and $v=-\text{Li}_2(x)$ yields

$$\begin{align} I&=\int_0^1 \frac{\text{Li}_2(x)}{x}\,dx\\\\ &=\text{Li}_3(1)\\\\ &=\zeta(3) \end{align}$$

as was to be shown!

Answer 2

An alternative approach with integration by parts. $$I= \int_0^1\frac{\ln x\ln(1-x)}{x}dx \underbrace{=}_{IBP}\overbrace{\ln x\ln(1-x)\int_0^1\frac{1}{x}dx}^{0}-\int_0^1\left[\left(\frac{\ln(1-x)}{x}-\frac{\ln x}{1-x}\right)\int\frac{1}{x}dx\right]dx=-\int_0^1\frac{\ln x\ln(1-x)}{x}dx+\int_0^1\frac{\ln^2(x)}{1-x}dx$$ Therefore, $$2I=\int_0^1\frac{\ln^2(x)}{1-x}dx=\sum_{n=0}^{\infty}\color{red}{\int_0^1 x^{n}\ln^2(x)dx} $$ Since

$$\int_0^1 x^p \ln^q dx =(-1)^q \frac{q!}{(p+1)^{q+1}}=(-1)^q\frac{\Gamma(q+1)}{(p+1)^{q+1}}$$ which is proved here

,then for $p=n$ and $q=2$ our integral reduces to $$I= \frac{1}{2}\sum_{n=0}^{\infty}\frac{2!}{(n+1)^{3}}=\zeta(3)$$.

Answer 3

Like @PeterForeman mentioned in the comments, one can use the series for $\log(1-x)$ to compute the integral, given that we are integrating over $[0,1]$:

\begin{equation} I=-\int\limits_{0}^{1}\frac{\log(x)}{x}\sum_{k=1}^{+\infty}\frac{x^{k}}{k}\,\mathrm{d}x \end{equation}

Because both the integral and the sum converge, we can interchange them:

\begin{equation} I=-\sum_{k=1}^{+\infty}\frac{1}{k}\int\limits_{0}^{1}\log(x)x^{k-1}\,\mathrm{d}x \end{equation}

With the substitution $x=e^{-z}$, you will get the following:

\begin{equation} I=\sum_{k=1}^{+\infty}\frac{1}{k}\int\limits_{0}^{+\infty}ze^{-zk}\mathrm{d}z \end{equation}

Now, with the substitution $zk=s$, you will be able to express the integral in terms of the gamma function:

\begin{equation} I=\sum_{k=1}^{+\infty}\frac{1}{k^{3}}\underbrace{\int\limits_{0}^{+\infty}se^{-s}\mathrm{d}s}_{\Gamma(2)} \end{equation}

\begin{equation} I=\sum_{k=1}^{+\infty}\frac{1}{k^{3}} \end{equation}

\begin{equation} \int\limits_{0}^{1}\frac{\log(x)\log(1-x)}{x}\,\mathrm{d}x=\zeta(3) \end{equation}

Answer 4

Using the Beta Integral and Telescoping Series $$ \begin{align} \int_0^1\frac{\log(x)\log(1-x)}x\,\mathrm{d}x &=\sum_{j=1}^\infty\sum_{k=1}^\infty\frac1j\frac1k\int_0^1x^{j-1}(1-x)^k\,\mathrm{d }x\tag1\\ &=\sum_{j=1}^\infty\sum_{k=1}^\infty\color{#C00}{\frac1j}\frac{\color{#C00}{(j-1)!}\,\color{#090}{(k-1)!}}{\color{#090}{(j+k)!}}\tag2\\ &=\sum_{j=1}^\infty\sum_{k=1}^\infty\color{#C00}{\frac{(j-1)!}j}\color{#090}{\frac1j\left(\frac{(k-1)!}{(j+k-1)!}-\frac{k!}{(j+k)!}\right)}\tag3\\ &=\sum_{j=1}^\infty\frac{(j-1)!}{j^2}\frac1{j!}\tag4\\ &=\sum_{j=1}^\infty\frac1{j^3}\tag5\\[6pt] &=\zeta(3)\tag6 \end{align} $$ Explanation:
$(1)$: apply the series for $\log(1-x)$ and $\log(x)=\log(1-(1-x))$
$(2)$: apply the Beta Integral
$(3)$: $\frac{(k-1)!}{(j+k)!}=\frac1j\left(\frac{(k-1)!}{(j+k-1)!}-\frac{k!}{(j+k)!}\right)$
$(4)$: Telescoping Series
$(5)$: simplify
$(6)$: apply definition of $\zeta$

Answer 5

First let $1-x\to x$ then add the integral to both sides

$$I=\int_0^1\frac{\ln x\ln(1-x)}{x}dx=\int_0^1\frac{\ln(1-x)\ln x}{1-x}dx$$

$$\Longrightarrow 2I=\int_0^1\frac{\ln x\ln(1-x)}{x(1-x)}dx=-\sum_{n=1}^\infty H_n\int_0^1 x^{n-1}\ln xdx=\sum_{n=1}^\infty\frac{H_n}{n^2}=2\zeta(3)$$

The last result follows from Euler sum :

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k),\quad q\ge2$$

Answer 6

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{1}{\ln\pars{x}\ln\pars{1 - x} \over x}\,\dd x} \\[5mm] \stackrel{x\ \mapsto\ 1 - x}{=}\,\,\,&\ \left. {\partial^{2} \over \partial\mu\,\partial\nu}\int_{0}^{1}x^{\mu - 1} \bracks{\pars{1 - x}^{\nu} - 1}\,\dd x \,\right\vert_{{\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}} \\[5mm] = &\ {\partial^{2} \over \partial\mu\,\partial\nu} \bracks{{\Gamma\pars{\mu}\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu + 1}} - {1 \over \mu}} _{{\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}} \\[5mm] = &\ {\partial^{2} \over \partial\mu\,\partial\nu}\braces{{1 \over \mu} \bracks{{\Gamma\pars{\mu + 1}\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu + 1}} - 1}} _{{\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}} \\[5mm] = &\ {1 \over 2}\,{\partial^{3} \over \partial\mu^{2}\,\partial\nu} \bracks{{\Gamma\pars{\mu + 1}\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu + 1}}} _{{\large\mu\ =\ 0^{+}} \atop {\large\nu\ =\ 0}} \\[5mm] = &\ {1 \over 2}\,\partiald[2]{}{\mu} \bracks{-\gamma - \Psi\pars{\mu + 1}}_{\ \mu\ =\ 0^{+}} = -\,{1 \over 2}\,\Psi\, ''\pars{1} = \bbx{\zeta\pars{3}} \\ & \end{align}