Sum of Reciprocals of Squares of Odd Integers as Double Integral/Proof 2

Theorem

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

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

$=$

$\ds \int_0^1 \int_0^1 \paren {1 + x^2 y^2 + x^4 y^4 + x^6 y^6 + \cdots} \rd x \rd y$

$\ds $

$=$

$\ds \int_0^1 \intlimits {y + \frac {x^2y^3} 3 + \frac {x^4y^5} 5 + \frac {x^6y^7} 7 + \cdots } {y \mathop = 0} {y \mathop = 1} \rd x$

$\ds $

$=$

$\ds \int_0^1 \paren {1 + \frac 1 3 x^2 + \frac 1 5 x^4 + \frac 1 7 x^6 + \cdots} \rd x$

$\ds $

$=$

$\ds \intlimits {x + \frac {x^3} 9 + \frac {x^5} {25} + \frac {x^7} {49} + \cdots } {x \mathop = 0} {x \mathop = 1}$

$\ds $

$=$

$\ds \paren {1 + \frac 1 9 + \frac 1 {25} + \frac 1 {49} + \cdots }$

$\ds $

$=$

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

$\blacksquare$