# Sum of Reciprocals of Squares of Odd Integers as Double Integral

Jump to navigation
Jump to search

## Theorem

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

## Proof

\(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\) | \(=\) | \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1} \paren {2 n - 1} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{n \mathop = 1}^\infty \int_0^1 x^{2 n - 2} \rd x \int_0^1 y^{2 n - 2} \rd y\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{n \mathop = 1}^\infty \int_0^1 \int_0^1 \paren {x^2 y^2}^{n - 1} \rd x \rd y\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \int_0^1 \int_0^1 \sum_{n \mathop = 0}^\infty \paren {x^2 y^2}^n \rd x \rd y\) | Fubini's Theorem | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \int_0^1 \int_0^1 \frac 1 {1 - x^2 y^2} \rd x \rd y\) | Sum of Infinite Geometric Progression |

$\blacksquare$