Sum of Reciprocals of Squares of Odd Integers as Double Integral

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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

$\blacksquare$


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