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

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Theorem

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


Proof

\(\displaystyle \int_0^1 \frac {\map \ln {1 - x} } x \rd x\) \(=\) \(\displaystyle -\int_0^1 \frac 1 x \paren {\sum_{n \mathop = 1}^\infty \frac {x^n} n} \rd x\) Power Series Expansion for $\map \ln {1 + x}$: Corollary
\(\displaystyle \) \(=\) \(\displaystyle -\sum_{n \mathop = 1}^\infty \paren {\frac 1 n \int_0^1 x^{n - 1} \rd x}\) Power Series is Termwise Integrable within Radius of Convergence
\(\displaystyle \) \(=\) \(\displaystyle -\sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) Primitive of Power
\(\displaystyle \) \(=\) \(\displaystyle -\frac {\pi^2} 6\) Basel Problem

$\blacksquare$


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