Dirichlet beta in terms of Lerch transcendent

Theorem

The following formula holds: $$\beta(x) = 2^{-x} \Phi \left(-1,x,\dfrac{1}{2} \right),$$ where $\beta$ denotes Dirichlet beta and $\Phi$ denotes the Lerch transcendent.

Proof

Starting from the Hadamard Fractional Integral representations of the Lerch Transcendent and the Dirichlet Beta functions, namely,

$$\Phi (z,\alpha ,y) = \tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha -1}\tfrac{e^{-yu}}{1-z e^{-u}}\, du ,$$

and

$$\beta (\alpha ) = \tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha -1}\tfrac{e^{-u}}{1+e^{-2u}}\, du$$

$$2^{-\alpha } \Phi \left(-1,\alpha ,\dfrac{1}{2} \right) = 2^{-\alpha } \tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha -1}\tfrac{e^{-\dfrac{1}{2}u}}{1+ e^{-u}}\, du = 2^{-\alpha } \tfrac{1}{\Gamma (\alpha )}\cdot 2^{\alpha }\int_0^\infty \omega ^{\alpha -1}\tfrac{e^{-\omega }}{1+ e^{-2\omega }}\, du = \beta(\alpha )$$,

and the proof is demonstrated.