Integral Representation of Dirichlet Beta Function in terms of Gamma Function

Theorem

 * $\displaystyle \map \beta s = \frac 1 {\map \Gamma s} \int_0^\infty \frac {x^{s - 1} e^{-x} } {1 + e^{-2 x} } \rd x$

where:
 * $\beta$ denotes the Dirichlet beta function
 * $\Gamma$ denotes the gamma function
 * $s$ is a complex number with $\map \Re s > 0$.

Proof
We have, by Definite Integral to Infinity of $x^{s - 1} e^{-a x}$:


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

for $\map \Re s > 0$.

Summing, we have:

We have:

Therefore:

Giving:


 * $\displaystyle \map \beta s = \frac 1 {\map \Gamma s} \int_0^\infty \frac {x^{s - 1} e^{-x} } {1 + e^{-2 x} } \rd x$