Sum of Reciprocals of Powers of Odd Integers Alternating in Sign

Theorem

 * $\displaystyle \sum_{n \mathop = 0}^\infty \frac {\left({-1}\right)^n} {\left({2 n + 1}\right)^s} = \frac 1 {2 \Gamma\left({s}\right)} \int_0^\infty x^{s - 1} \operatorname{sech} \left({x}\right) \rd x$

where:


 * $\operatorname{Re}\left({s}\right) > 0$
 * $\Gamma$ is the gamma function
 * $\operatorname{sech}$ is the hyperbolic secant function.

Proof
So:


 * $\displaystyle \frac 1 {2 \Gamma \left({s}\right)} \int_0^\infty x^{s - 1} \operatorname{sech} \left({x}\right) \rd x = \sum_{n \mathop = 0}^\infty \frac {\left({-1}\right)^n} {\left({2 n + 1}\right)^s}$