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

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

Revision as of 12:01, 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 Hadamard Fractional Integral 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

$$\beta (\alpha ) = \sum_{k=0}^\infty \tfrac{(-1)^k}{(2k+1)^\alpha}$$

$$2^{-\alpha } \Phi \left(-1,\alpha ,\dfrac{1}{2} \right) = 2^{-\alpha } \sum_{k=0}^\infty\tfrac{(-1)^k}{\left( \dfrac{1}{2}+k\right) ^\alpha} = \beta(\alpha )$$,

and the proof is demonstrated.

References

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