Can you find this integral?

$\int (\log x) (\log (1-x)) \, dx=-\text{Li}_2(x)-x (\log (x)-2)+(x-1) \log (1-x) (\log (x)-1)+\text{constant}$ .

$\underset{x\to 0}{\text{lim}}\left(-\text{Li}_2(x)-x (\log (x)-2)+(x-1) \log (1-x) (\log (x)-1)\right) \Rightarrow 0$ .

$\underset{x\to 1}{\text{lim}}\left(-\text{Li}_2(x)-x (\log (x)-2)+(x-1) \log (1-x) (\log (x)-1)\right) \Rightarrow 2-\frac{\pi ^2}{6}$ .

Both limits of the indefinite integral are necessary.

The integral in the question is undefined. It should be $\displaystyle I=\int_0^1\ln x\ln\left(1-x\right)dx$

Now, to solve the integral, simply expand it using $\displaystyle \ln\left(1-x\right)=-\sum_{n=1}^{\infty}\frac{x^n}{n}$ and then, switching the summation and integral sign, we have

$\displaystyle \int_0^1\ln x\ln\left(1-x\right)dx=-\sum_{n=1}^{\infty}\left(\int_0^1\frac{x^n\ln x}{n}dx\right)$

Substituting $\displaystyle \ln x=-t$ we have

$\displaystyle I=\sum_{n=1}^{\infty}\left(\int_0^{\infty}\frac{te^{-\left(n+1\right)t}}{n}dt\right)$

$\displaystyle =\sum_{n=1}^{\infty}\frac{1}{\left(n+1\right)^2n}$

$\displaystyle =\sum_{n=1}^{\infty}\frac{1}{n\left(n+1\right)}-\sum_{n=2}^{\infty}\frac{1}{n^2}$

$\displaystyle =1-\left(\frac{\pi^2}{6}-1\right)$

Thus making the answer $\displaystyle \boxed{2-\frac{\pi^2}{6}}$

First, we need to turn the integrand function into a series and integrate everything term by term. We have

$\displaystyle\ln{x} \ln{(1-x)} = \ln{x} (-x-\frac{x^2}{2}-\frac{x^3}{3} - ...) = \sum^{\infty}_{n=1} \frac{-\ln{x}x^n}{n}$ . $(1)$

Now it is possible to integrate every term of the sequence by parts:

$\displaystyle\int\limits_0^1 \frac{-\ln{x}x^n}{n}dx = - \frac{1}{n(n+1)}\int\limits_0^1 \ln{x}d(x^n+1) =$

$\displaystyle= - \frac{1}{n(n+1)}(\ln1\cdot 1^{n+1} - \lim_{x\to 0+} \ln{x} \cdot x^{n+1}) +$

$\displaystyle + \frac{1}{n(n+1)} \int_0^1 x^{n+1}d(\ln{x}) = \frac{1}{n(n+1)} \int_0^1 x^ndx = \frac{1}{n(n+1)^2}$ . $(2)$

Can we integrate the series? In order to make sure in the validity of our actions $(1)$ , we should show that for all $\varepsilon > 0$ there is such $m_{0}\in{N}$ , that

$\displaystyle\Bigg|\int\limits_0^1 \ln{x} \Big(\ln{(1-x)} + \sum^{m}_{n=1} \frac{x^n}{n}\Big) dx\Bigg|< \varepsilon,$ for all $m \geq m_0$ . $(3)$

On given $\varepsilon > 0$ , let's choose $\delta \in (0, 1/2)$ , so that:

$\displaystyle\Big|\Big(\int\limits_0^{\delta} + \int\limits_{1-\delta}^{0}\Big) \ln{x} \ln{(1-x)}dx\Big|< \varepsilon/2.$ .

Taking into account this double inequality

$\displaystyle\ln{(1-x)} < \ln{(1-x)} + \sum^{m}_{n=1} \frac{x^n}{n} < 0,$

we will get

$\displaystyle\Big|\Big(\int\limits_0^{\delta} + \int\limits_{1-\delta}^{0}\Big) \ln{x} \Big(\ln{(1-x)} + \sum^{m}_{n=1} \frac{x^n}{n}\Big) dx\Big|< \varepsilon/2.$ $(4)$

Since in the interval $[\delta, 1-\delta]$ the function $\displaystyle\ln{x} \Big(\ln{(1-x)} + \sum^{m}_{n=1} \frac{x^n}{n}\Big)$ uniformly equals to zero for $m\to \infty$ , for some big enough $m$

$\displaystyle\Bigg|\int\limits_{\delta}^{1-\delta} \ln{x} \Big(\ln{(1-x)} + \sum^{m}_{n=1} \frac{x^n}{n}\Big) dx\Bigg|< \varepsilon/2.$

Adding up this inequality and $(4)$ , we get required inequality $(3)$ for all $m \geq m_0$ .

Now, we can integrate $(1)$ , we find the value of the integral:

$\displaystyle\int\limits_0^1 \ln{x} \cdot \ln{(1-x)} dx = \sum^{\infty}_{n=1} \frac{1}{n(n+1)^2} = \sum^{\infty}_{n=1} \Big(\frac{1}{n(n+1)} - \frac{1}{(n+1)^2}\Big) =$

$\displaystyle = \sum^{\infty}_{n=1} \Big(\frac{1}{n} - \frac{1}{(n+1)}\Big) - \sum^{\infty}_{n=2} \frac{1}{n^2} = 1 - \Big(\sum^{\infty}_{n=1} \frac{1}{n^2} - 1\Big) = 2 - \frac{\pi^2}{6} = \boxed{-0.355}$ .