Sum of Reciprocals of Squares of Odd Integers as Double Integral/Proof 1

Theorem

$\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2} = \int_0^1 \int_0^1 \frac 1 {1 - x^2 y^2} \rd x \rd y$

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

$=$

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

$\ds $

$=$

$\ds \sum_{n \mathop = 1}^\infty \int_0^1 x^{2 n - 2} \rd x \int_0^1 y^{2 n - 2} \rd y$

$\ds $

$=$

$\ds \sum_{n \mathop = 1}^\infty \int_0^1 \int_0^1 \paren {x^2 y^2}^{n - 1} \rd x \rd y$

$\ds $

$=$

$\ds \int_0^1 \int_0^1 \sum_{n \mathop = 0}^\infty \paren {x^2 y^2}^n \rd x \rd y$

Fubini's Theorem, reindexing the sum

$\ds $

$=$

$\ds \int_0^1 \int_0^1 \frac 1 {1 - x^2 y^2} \rd x \rd y$

$\blacksquare$