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

Theorem

$\ds \int_0^1 \dfrac {\ln x \map \ln {1 - x} } {\paren {1 - x} } \rd x = \map \zeta 3$

$\ds \int_0^1 \dfrac {\ln x \map \ln {1 - x} } {\paren {1 - x} } \rd x$

$=$

$\ds \int_1^0 \dfrac {\map \ln {1 - u} \ln u } u \paren {-\rd u}$

$x \to \paren {1 - u}$ and $\rd x \to -\rd u$

$\ds $

$=$

$\ds \int_0^1 \dfrac {\map \ln {1 - u} \ln u } u \rd u$

$\ds $

$=$

$\ds -\int_0^1 \frac {\ln u} u \paren {\sum_{n \mathop = 1}^\infty \frac {u^n} n}$

$\ds $

$=$

$\ds -\int_0^1 \ln u \paren {\sum_{n \mathop = 1}^\infty \frac {u^{n - 1} } n}$

$\ds $

$=$

$\ds -\sum_{n \mathop = 1}^\infty \frac 1 n \paren {\int_0^1 u^{n - 1} \ln u \rd u}$

$\ds $

$=$

$\ds \paren {-1}^2 \sum_{n \mathop = 1}^\infty \frac 1 n \paren {\frac {\map \Gamma 2} {\paren {\paren {n - 1} + 1}^2} }$

$\ds $

$=$

$\ds 1! \sum_{n \mathop = 1}^\infty \frac 1 {n^3}$

$\ds $

$=$

$\ds \map \zeta {3}$

$\blacksquare$