Different ways to Prove $\int_{0}^{1}\frac {{\log(x)} {\log(1-x)}}{x}dx=\zeta(3)$ 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.
 A: 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!
A: 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 $$I= \frac{1}{2}\sum_{n=0}^{\infty}\frac{2!}{(n+1)^{3}}=\zeta(3)$$.
A: 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}
A: 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 $\zeta$
A: 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 :
$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k),\quad q\ge2$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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}
