Evaluate $ \int_{0}^{1} \ln(x)\ln(1-x)\,dx $

\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}

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Note :

$\displaystyle\ \ \int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\ n=0,1,2,\ldots$

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,

$$ \int_{0}^{1} \ln(x) \ln(1-x) \ dx = \lim_{x \to 1} \left[-x \ln(1-x)+\ln(1-x) \right] + 2 - \text{Li}_{2}(1) = 2 - \zeta(2) .$$

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}