Sum of Reciprocals of Powers of Odd Integers Alternating in Sign

Theorem

 * $\displaystyle \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s} = \frac 1 {2 \map \Gamma s} \int_0^\infty x^{s - 1} \map \sech x \rd x$

where:


 * $\map \Re s > 0$
 * $\Gamma$ is the gamma function
 * $\sech$ is the hyperbolic secant function.

Proof
So:
 * $\displaystyle \frac 1 {2 \map \Gamma s} \int_0^\infty x^{s - 1} \map \sech x \rd x = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s}$