Difference between revisions of "Dirichlet beta in terms of Lerch transcendent"
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==Proof== | ==Proof== | ||
| − | Starting from the | + | Starting from the series representations of the Lerch Transcendent and the Dirichlet Beta functions, namely, |
$$\Phi (z,\alpha ,y) = \sum_{k=0}^\infty\tfrac{z^k}{(y+k)^\alpha},$$ | $$\Phi (z,\alpha ,y) = \sum_{k=0}^\infty\tfrac{z^k}{(y+k)^\alpha},$$ | ||
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and | and | ||
| − | $$\beta (\alpha ) = \sum_{k=0}^\infty \tfrac{(-1)^k}{(2k+1)^\alpha}$$ | + | $$\beta (\alpha ) = \sum_{k=0}^\infty \tfrac{(-1)^k}{(2k+1)^\alpha}.$$ |
| + | |||
| + | We have | ||
$$2^{-\alpha } \Phi \left(-1,\alpha ,\dfrac{1}{2} \right) = 2^{-\alpha } \sum_{k=0}^\infty\tfrac{(-1)^k}{\left( \dfrac{1}{2}+k\right) ^\alpha} = \beta(\alpha )$$, | $$2^{-\alpha } \Phi \left(-1,\alpha ,\dfrac{1}{2} \right) = 2^{-\alpha } \sum_{k=0}^\infty\tfrac{(-1)^k}{\left( \dfrac{1}{2}+k\right) ^\alpha} = \beta(\alpha )$$, | ||
Latest revision as of 12:03, 30 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 series representations of the Lerch Transcendent and the Dirichlet Beta functions, namely,
$$\Phi (z,\alpha ,y) = \sum_{k=0}^\infty\tfrac{z^k}{(y+k)^\alpha},$$
and
$$\beta (\alpha ) = \sum_{k=0}^\infty \tfrac{(-1)^k}{(2k+1)^\alpha}.$$
We have
$$2^{-\alpha } \Phi \left(-1,\alpha ,\dfrac{1}{2} \right) = 2^{-\alpha } \sum_{k=0}^\infty\tfrac{(-1)^k}{\left( \dfrac{1}{2}+k\right) ^\alpha} = \beta(\alpha )$$,
and the proof is demonstrated.
