Sum of Reciprocals of Squares of Odd Integers as Double Integral

Theorem

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

Also see

 * Sum of Reciprocals of Squares of Odd Integers
 * Riemann Zeta Function as a Multiple Integral