Sum of Reciprocals of Powers of Odd Integers Alternating in Sign

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Theorem

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

where:

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


Corollary

$\ds \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s} = \frac 1 {\map \Gamma s} \int_1^\infty \frac {\ln^{s - 1} x} {x^2 + 1} \rd x$


Proof

\(\ds \int_0^\infty x^{s - 1} \map \sech x \rd x\) \(=\) \(\ds 2 \int_0^\infty \frac { x^{s - 1} } {e^x + e^{-x} } \rd x\) Definition of Hyperbolic Secant
\(\ds \) \(=\) \(\ds 2 \int_0^\infty \frac {x^{s - 1} e^{-x} } {1 - \paren {- e^{-2 x} } } \rd x\)
\(\ds \) \(=\) \(\ds 2 \int_0^\infty x^{s - 1} e^{-x} \sum_{n \mathop = 0}^\infty \paren {-1}^n e^{-2 n x} \rd x\) Sum of Infinite Geometric Sequence
\(\ds \) \(=\) \(\ds 2 \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^\infty x^{s - 1} e^{-\paren {2 n + 1} x} \rd x\) Fubini's Theorem
\(\ds \) \(=\) \(\ds 2 \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^\infty \paren {\frac t {2 n + 1} }^{s - 1} e^{-t} \frac {\rd t} {2 n + 1}\) substituting $t = \paren {2 n + 1} x$
\(\ds \) \(=\) \(\ds 2 \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s} \int_0^\infty t^{s - 1} e^{-t} \rd t\)
\(\ds \) \(=\) \(\ds 2 \map \Gamma s \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s}\) Definition of Gamma Function

So:

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

$\blacksquare$