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 1

\(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1} \paren {2 n - 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \int_0^1 x^{2 n - 2} \rd x \int_0^1 y^{2 n - 2} \rd y\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \int_0^1 \int_0^1 \paren {x^2 y^2}^{n - 1} \rd x \rd y\)
\(\displaystyle \) \(=\) \(\displaystyle \int_0^1 \int_0^1 \sum_{n \mathop = 0}^\infty \paren {x^2 y^2}^n \rd x \rd y\) 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 Sequence

$\blacksquare$



Proof 2

\(\displaystyle \int_0^1 \int_0^1 \frac 1 {1 - x^2 y^2} \rd x \rd y\) \(=\) \(\displaystyle \int_0^1 \int_0^1 \paren {1 + x^2 y^2 + x^4 y^4 + x^6 y^6 + \cdots} \rd x \rd y\) Sum of Infinite Geometric Sequence
\(\displaystyle \) \(=\) \(\displaystyle \int_0^1 \intlimits {y + \frac {x^2y^3} 3 + \frac {x^4y^5} 5 + \frac {x^6y^7} 7 + \cdots } {y \mathop = 0} {y \mathop = 1} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \int_0^1 \paren {1 + \frac 1 3 x^2 + \frac 1 5 x^4 + \frac 1 7 x^6 + \cdots} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \intlimits {x + \frac {x^3} 9 + \frac {x^5} {25} + \frac {x^7} {49} + \cdots } {x \mathop = 0} {x \mathop = 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {1 + \frac 1 9 + \frac 1 {25} + \frac 1 {49} + \cdots }\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\)

$\blacksquare$


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