Improper integral

Evaluate the integral:
$$\int_0^1 \ln x \ln (1-x) \, {\rm d}x$$

Solution



By invoking the Taylor expansion of $\ln (1-x)$ the integral becomes successively:

$$\begin{aligned}
\int_{0}^{1}\ln x \ln (1-x)\, {\rm d}x &=-\int_{0}^{1} \ln x \sum_{n=1}^{\infty}\frac{x^n}{n}\, {\rm d}x \\
 &= - \int_{0}^{1}\sum_{n=1}^{\infty}\frac{x^n \ln x}{n}\, {\rm d}x\\
 &= - \sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{1}x^n \ln x \, {\rm d}x\\
 &= \sum_{n=1}^{\infty}\frac{1}{n (n+1)^2}\\
 &= \sum_{n=1}^{\infty}\left [ \frac{1}{n} - \frac{1}{n+1} - \frac{1}{(n+1)^2} \right ]\\
 &= \cancelto{1}{\sum_{n=1}^{\infty}\left [ \frac{1}{n} - \frac{1}{n+1} \right ]} - \sum_{n=2}^{\infty}\frac{1}{n^2} \\
 &= 1 - \sum_{n=1}^{\infty}\frac{1}{n^2}+1 \\
 &= 2 - \frac{\pi^2}{6}
\end{aligned}$$

Although known that the series $\displaystyle \sum_{n=1}^{\infty}\left [ \frac{1}{n} - \frac{1}{n+1} \right ]$ telescopes to $1$ here is an analytic solution.

$$\begin{aligned}
\sum_{k=1}^{n}\left [ \frac{1}{k} - \frac{1}{k+1} \right ] &= \left ( 1- \cancel{\frac{1}{2}} \right ) + \left ( \cancel{\frac{1}{2}} - \cancel{\frac{1}{3}} \right )+\cdots+ \left ( \cancel{\frac{1}{n}} - \frac{1}{n+1} \right )\\
 &= \cdots \\
 &= 1 - \frac{1}{n+1}
\end{aligned}$$

Hence:

$$\sum_{n=1}^{\infty}\left [ \frac{1}{n} - \frac{1}{n+1} \right ]= \lim_{n \rightarrow +\infty}\sum_{k=1}^{n}\left [ \frac{1}{k} - \frac{1}{k+1} \right ]= \lim_{n \rightarrow +\infty} \left [ 1 - \frac{1}{n+1} \right ]= 1$$

proving the result.