Evaluate $ \int_{0}^{1} \ln(x)\ln(1-x)\,dx $ Evaluate the integral, 
$$ \int_{0}^{1} \ln(x)\ln(1-x)\,dx$$ 
I solved this problem, by writing power series and then calculating the series and found the answer to be $ 2 -\zeta(2) $, but I don't think that it is best solution to this problem.   I want to know if it can be solved by any other nice/elegant method. 
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$\ds{\int_{0}^{1}\ln\pars{x}\ln\pars{1 - x}\,\dd x:\ {\large ?}}$.

\begin{align}
&\color{#66f}{\large\int_{0}^{1}\ln\pars{x}\ln\pars{1 - x}\,\dd x}
=\int_{x\ =\ 0}^{x\ =\ 1}\ln\pars{1 - x}\dd\bracks{x\ln\pars{x} - x + 1}
\\[3mm]&=\left.\bracks{x\ln\pars{x} - x + 1}\ln\pars{1 - x}\right\vert_{0}^{1}
-\int_{0}^{1}\bracks{x\ln\pars{x} - x + 1}\,{-1 \over 1 - x}\,\dd x
=\int_{0}^{1}{x\ln\pars{x} \over 1 - x}\,\dd x + 1
\\[3mm]&=-\lim_{\mu\ \to\ 1}\partiald{}{\mu}
\int_{0}^{1}{1 - x^{\mu} \over 1 - x}\,\dd x + 1
=-\lim_{\mu\ \to\ 1}\partiald{\Psi\pars{\mu + 1}}{\mu} + 1
\end{align}
  where $\ds{\Psi\pars{z}}$ is the
  Digamma Function
  $\ds{\bf 6.3.1}$ and we used the identity $\ds{\bf 6.3.22}$.

$$
\color{#66f}{\large\int_{0}^{1}\ln\pars{x}\ln\pars{1 - x}\,\dd x}
=-\Psi'\pars{2} + 1=-\Psi'\pars{1} + 2=-\zeta\pars{2} + 2
$$
Here we used the identities:
$$
\Psi'\pars{z + 1} = \Psi'\pars{z} - {1 \over z^{2}}\,,\qquad
\Psi^{\rm\pars{n}}\pars{1}=\pars{-1}^{n + 1}\,n!\,\zeta\pars{n + 1}\,,\quad n = 1,2,3,\ldots
$$

Since $\ds{\zeta\pars{2} = {\pi^{2} \over 6}}$:
  $$
\color{#66f}{\large\int_{0}^{1}\ln\pars{x}\ln\pars{1 - x}\,\dd x}
=\color{#66f}{\large 2 - {\pi^{2} \over 6}} \approx {\tt 0.3551}
$$

A: You could start from the Beta function
$$ B(p+1,r+1) = \int_0^1 x^p (1-x)^r\; dx = \dfrac{\Gamma(p+1) \Gamma(r+1)}{\Gamma(p+r+2)}$$
take the derivatives with respect to $p$ and $r$, and evaluate at
$p=r=0$.
A: I've found a solution that is interesting, but probably not elegant, and definitely not short.
$I = \displaystyle\int_0^1 \ln(x)\ln(1 - x) dx$
Basic results: 


*

*$\lim\limits_{n \to 0} \dfrac{x^n - 1}{n} = \log x$, or $\lim\limits_{n \to 1}\dfrac{x^{n-1} - 1}{n - 1} = \log x$.

*$\dfrac{d}{dn}\beta(n, n) = 2\beta(n, n)(\psi_0(n) - \psi_0(2n))$ where $\psi_0(n)$ is the digamma function.

*$\dfrac{d^2}{dn^2}\beta(n, n) = 4\beta(n, n)(\psi_0(n) - \psi_0(2n))^2 + 2\beta(n, n)(\psi_1(n) - 2\psi_1(2n))$, where $\psi(1)(n)$ is the polygamma function.

*$\psi_0(1) - \psi_0(2) = -1$ according to the recurrence relation.

*$\psi_1(2) = \psi_1(1) - 1$ according to the recurrence relation.

*$\psi_1(1) = \zeta(2)$.


Solution:
$\begin{align}
I & = \lim\limits_{n \to 1} \displaystyle\int_0^1 \dfrac{(x^{n - 1} - 1)((1 - x)^{n - 1} - 1)}{(n - 1)^2} dx\\
& = \lim\limits_{n \to 1}\displaystyle\int_0^1 \dfrac{x^{n-1}(1-x)^{n-1} - x^{n - 1} - (1-x)^{n-1} + 1}{(n-1)^2} dx\\
& = \lim\limits_{n \to 1} \dfrac{\beta(n,n) - \frac{1}{n} - \frac{1}{n} + 1}{(n-1)^2}\\
& = \lim\limits_{n \to 1} \dfrac{\beta(n,n)(\psi_0(n)-\psi_0(2n)) + \frac{2}{n^2}}{2(n-1)} \quad [\text{l'Hospital's rule}]\\
& = \lim\limits_{n \to 1} \dfrac{4\beta(n,n)(\psi_0(n)-\psi_0(2n))^2 + 2\beta(n,n)(\psi_1(n)-2\psi_1(2n))- \frac{4}{n^3}}{2}\quad [\text{l'Hospital's rule}]\\
& = 2\beta(1, 1)(\psi_0(1) - \psi_0(2))^2 + \beta(1, 1)(\psi_1(1) - 2\psi_1(2)) - 2\\
& = 2(-1)^2 + 1(\psi_1(1) - 2\psi_1(1) + 2) - 2\\
& = 2 - \psi_1(1)\\
& = 2-\zeta(2)
\end{align}$
A: You could expand
$\ln(1-x)
=-\sum_{n=1}^{\infty} \frac{x^n}{n}
$
and evaluate
$\int_0^1  x^n \ln x\,dx$,
probably by an induction
via integration by parts.
From your description,
you may have already done this.
It sure is easier to 
write this than to do it.
A: \begin{align}
\int_0^1\ln(1-x)\ln x\ dx&=\int_0^1\sum_{n=1}^\infty\frac{x^n}n\ln x\ dx\\
&=\sum_{n=1}^\infty\frac{1}n\int_0^1 x^n\ln x\ dx\\
&=-\sum_{n=1}^\infty\frac{1}n\cdot\frac1{(n+1)^2}\\
&=\sum_{n=1}^\infty\left[\frac{1}n-\frac1{n+1}-\frac1{(n+1)^2}\right]\\
&=1-\left[\sum_{n=1}^\infty\frac1{n^2}-1\right]\\
&=\large\color{blue}{2-\zeta(2)=2-\frac{\pi^2}6}.
\end{align}

Note :
$\displaystyle\ \ \int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\  n=0,1,2,\ldots$
A: Integrating by parts,
$$ \int \ln(x) \ln(1-x) \, dx = x \ln(x) \ln(1-x) - x \ln(1-x)+ \int \frac{x \ln (x)}{1-x} \, dx - \int \frac{x}{1-x} \, dx$$
where
$$  \int \frac{x}{1-x} \, dx = - \int \ dx + \int \frac{1}{1-x} \, dx = -x - \ln(1-x) + C_{1}$$
and $$ \begin{align} \int \frac{x \ln (x)}{1-x} \, dx &= -x \ln (x) - \ln(x) \ln(1-x) + \int dx + \int \frac{\ln (1-x)}{x} \, dx  \\  &= -x \ln (x) - \ln(x) \ln(1-x) + x - \text{Li}_{2}(x) + C_{2}. \end{align}$$
$\text{Li}_{2}(x)$ is the dilogarithm function.
So we have $$ \begin{align} \int \ln(x) \ln(1-x) \, dx &= x \ln(x) \ln(1-x) - x \ln(1-x) - x \ln(x) - \ln(x) \ln(1-x) + 2x  \\ &- \text{Li}_{2}(x) + \ln(1-x) + C . \end{align} $$
Therefore,
$$ \int_{0}^{1} \ln(x) \ln(1-x) \ dx = \lim_{x \to 1} \left[-x \ln(1-x)+\ln(1-x) \right] + 2 - \text{Li}_{2}(1) = 2 - \zeta(2) .$$
A: Using the reflection formula 
$$\log(x)\log(1-x) =\zeta(2)-\mathrm{Li}_2(x)-\mathrm{Li}_2(1-x)  $$
\begin{align}
\int^1_0\log(x)\log(1-x) &=\zeta(2)-\int^1_0\mathrm{Li}_2(x)\,dx-\int^1_0\mathrm{Li}_2(1-x)\,dx\\ &=\zeta(2)-2\int^1_0\mathrm{Li}_2(x)\,dx\\
&=\zeta(2)-2\zeta(2)-2\int^1_0\log(1-x)\,dx\\
&=2-\zeta(2)
\end{align}
A: Noting
$$ \frac{d}{dx}[x(1-\ln(1-x))+\ln(1-x)]=-\ln(1-x) $$
we have
\begin{eqnarray}
\int_0^1\ln x\ln(1-x)dx&=&-\int_0^1\ln xd[x(1-\ln(1-x))+\ln(1-x)]\\
&=&-[x(1-\ln(1-x))+\ln(1-x)]\ln x\bigg|_0^1+\int_0^1\frac{x(1-\ln(1-x))+\ln(1-x)}{x}dx\\
&=&\int_0^1(1-\ln(1-x)+\frac{\ln(1-x)}{x})dx\\
&=&\int_0^1(1-\ln(1-x))dx+\int_0^1\frac{\ln(1-x)}{x}dx\\
&=&2-\zeta(2).
\end{eqnarray}
Here we used the well-known result 
$$ \int_0^1\frac{\ln(1-x)}{x}dx=-\zeta(2). $$
A: Integrate by parts
\begin{align}
 \int_{0}^{1} \ln x\ln(1-x)\,dx
= -2\int_0^1 \ln xdx +\int_0^1\frac{\ln x}{1-x}dx
=2-\zeta(2)
\end{align}
