Sum of Reciprocals of Squares of Odd Integers/Proof 4

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Theorem

\(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2}\) \(=\) \(\ds 1 + \dfrac 1 {3^2} + \dfrac 1 {5^2} + \dfrac 1 {7^2} + \dfrac 1 {9^2} + \cdots\)
\(\ds \) \(=\) \(\ds \dfrac {\pi^2} 8\)


Proof




\(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2n - 1}^2}\) \(=\) \(\ds \int_0^1 \int_0^1 \frac 1 {1 - x^2y^2} \rd x \rd y\) Sum of Reciprocals of Squares of Odd Integers as Double Integral


Applying the substitution:

$\tuple {x, y} = \tuple {\dfrac {\sin u} {\cos v}, \dfrac {\sin v} {\cos u} }$

the Jacobian matrix is:

$\mathbf J_{\mathbf f} := \begin{pmatrix} {\dfrac \partial {\partial u} \dfrac {\sin u} {\cos v} } & {\dfrac \partial {\partial v} \dfrac {\sin u} {\cos v} } \\ {\dfrac \partial {\partial u} \dfrac {\sin v} {\cos u} } & {\dfrac \partial {\partial v} \dfrac {\sin v} {\cos u} } \end{pmatrix}$

and the Jacobian determinant is:

\(\ds \size {\frac {\partial \tuple {x, y} } {\partial \tuple {u, v} } }\) \(=\) \(\ds \frac \partial {\partial u} \frac {\sin u} {\cos v} \frac \partial {\partial v} \frac {\sin v} {\cos u} - \frac \partial {\partial v} \frac {\sin u} {\cos v} \frac \partial {\partial u} \frac {\sin v} {\cos u}\)
\(\ds \) \(=\) \(\ds \frac {\cos u \, \cos v} {\cos v \, \cos u} - \frac {\sin^2 u \, \sin^2 v} {\cos^2 v \, \cos^2 u}\) Derivative of Secant Function, Derivative of Sine Function
\(\ds \) \(=\) \(\ds 1 - \tan^2 u \tan^2 v\)


Under this substitution, the image of the region $\closedint 0 1^2$, that is the unit square, is an isosceles triangle $\bigtriangleup$ with:

base and height $\dfrac \pi 2$
vertices: $\tuple {0, 0}; \tuple {0, \dfrac \pi 2}; \tuple {\dfrac \pi 2, 0}$


We have:

\(\ds 0\) \(\le\) \(\ds \dfrac {\sin u} {\cos v}\) \(\ds \le 1\)
\(\ds \leadsto \ \ \) \(\ds u\) \(\ge\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \sin u\) \(\le\) \(\ds \cos v\)

and we have:

\(\ds 0\) \(\le\) \(\ds \dfrac {\sin v} {\cos u}\) \(\ds \le 1\)
\(\ds \leadsto \ \ \) \(\ds v\) \(\ge\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \sin v\) \(\le\) \(\ds \cos u\)


This gives us the region:

$\set {\tuple {u, v}: u, v \ge 0 \land \cos u \ge \sin v \land \cos v \ge \sin u}$

which is equivalent to the region:

$\set {\tuple {u, v}: u, v \ge 0 \land v \le \dfrac \pi 2 - u}$


By Change of Variables Theorem (Multivariable Calculus):

\(\ds \int_0^1 \int_0^1 \frac 1 {1 - x^2 y^2} \rd x \rd y\) \(=\) \(\ds \iint_{\bigtriangleup} \frac {1 - \tan^2 u \, \tan^2 v} {1 - \paren {\tan u \, \tan v}^2} \rd u \rd v\)
\(\ds \) \(=\) \(\ds \iint_{\bigtriangleup} \rd u \rd v\)
\(\ds \) \(=\) \(\ds \frac 1 2 \int_0^{\frac \pi 2} \int_0^{\frac \pi 2} \rd u \rd v\) Area of Triangle in Terms of Side and Altitude
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac \pi 2}^2\) Primitive of Constant
\(\ds \) \(=\) \(\ds \frac {\pi^2} 8\)

$\blacksquare$