Difference between revisions of "Dirichlet beta in terms of Lerch transcendent"

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==Proof==
 
==Proof==
−
Starting from the Hadamard Fractional Integral representations of the Lerch Transcendent and the Dirichlet Beta functions, namely,
+
Starting from the series representations of the Lerch Transcendent and the Dirichlet Beta functions, namely,
  
−
$$\Phi (z,\alpha ,y) = \tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha -1}\tfrac{e^{-yu}}{1-z e^{-u}}\, du ,$$
+
$$\Phi (z,\alpha ,y) = \sum_{k=0}^\infty\tfrac{z^k}{(y+k)^\alpha},$$
  
 
and
 
and
  
−
$$\beta (\alpha ) = \tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha -1}\tfrac{e^{-u}}{1+e^{-2u}}\, du$$
+
$$\beta (\alpha ) = \sum_{k=0}^\infty \tfrac{(-1)^k}{(2k+1)^\alpha}.$$
  
−
$$2^{-\alpha } \Phi \left(-1,\alpha ,\dfrac{1}{2} \right) = 2^{-\alpha } \tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha -1}\tfrac{e^{-\dfrac{1}{2}u}}{1+ e^{-u}}\, du = 2^{-\alpha } \tfrac{1}{\Gamma (\alpha )}\cdot 2^{\alpha }\int_0^\infty \omega ^{\alpha -1}\tfrac{e^{-\omega }}{1+ e^{-2\omega }}\, du = \beta(\alpha )$$,
+
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.
 
and the proof is demonstrated.

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.

References

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