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

## 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

$\blacksquare$