Definite Integral from 0 to 1 of Logarithm of x by Logarithm of One minus x

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Theorem

$\ds \int_0^1 \ln x \map \ln {1 - x} \rd x = 2 - \frac {\pi^2} 6$


Proof

\(\ds \int_0^1 \ln x \map \ln {1 - x} \rd x\) \(=\) \(\ds -\int_0^1 \ln x \paren {\sum_{n \mathop = 1}^\infty \frac {x^n} n} \rd x\) Power Series Expansion for $\map \ln {1 + x}$: Corollary
\(\ds \) \(=\) \(\ds -\sum_{n \mathop = 1}^\infty \frac 1 n \paren {\int_0^1 x^n \ln x \rd x}\) Fubini's Theorem
\(\ds \) \(=\) \(\ds \paren {-1}^2 \sum_{n \mathop = 1}^\infty \frac 1 n \paren {\frac {\map \Gamma 2} {\paren {n + 1}^2} }\) Definite Integral from $0$ to $1$ of $x^m \paren {\ln x}^n$
\(\ds \) \(=\) \(\ds 1! \sum_{n \mathop = 1}^\infty \frac 1 {n \paren {n + 1}^2}\) Gamma Function Extends Factorial
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {n + 1} } - \sum_{n \mathop = 1}^\infty \paren {\frac 1 {\paren {n + 1}^2} }\) partial fraction expansion
\(\ds \) \(=\) \(\ds 1 - \paren {\sum_{n \mathop = 1}^\infty \frac 1 {n^2} - 1}\) Telescoping Series: Example 1
\(\ds \) \(=\) \(\ds 2 - \frac {\pi^2} 6\) Basel Problem

$\blacksquare$


Sources