Difference between revisions of "Dirichlet beta in terms of Lerch transcendent"
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$$\beta (\alpha ) = \tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha -1}\tfrac{e^{-u}}{1+e^{-2u}}\, du$$ | $$\beta (\alpha ) = \tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha -1}\tfrac{e^{-u}}{1+e^{-2u}}\, du$$ | ||
| − | $$2^{-\alpha } \Phi \left(-1,\alpha ,\dfrac{1}{2} \right) = 2^{-\alpha } \tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha -1}\ | + | $$2^{-\alpha } \Phi \left(-1,\alpha ,\dfrac{1}{2} \right) = 2^{-\alpha } \tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha -1}\frac{e^{-\tfrac{u}{2}}}{1+ e^{-u}}\, du = 2^{-\alpha } \tfrac{1}{\Gamma (\alpha )}\cdot 2^{\alpha }\int_0^\infty \omega ^{\alpha -1}\frac{e^{-\omega }}{1+ e^{-2\omega }}\, du = \beta(\alpha )$$, |
and the proof is demonstrated. | and the proof is demonstrated. | ||
Revision as of 20:27, 20 March 2022
Theorem
The following formula holds: $$\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}\frac{e^{-\tfrac{u}{2}}}{1+ e^{-u}}\, du = 2^{-\alpha } \tfrac{1}{\Gamma (\alpha )}\cdot 2^{\alpha }\int_0^\infty \omega ^{\alpha -1}\frac{e^{-\omega }}{1+ e^{-2\omega }}\, du = \beta(\alpha )$$,
and the proof is demonstrated.
