Engineering Mathematics: Sep 8, 2012

DBHZ.ID.1: Dirichlet Beta - Hurwitz zeta relation

$\beta (s) = \frac{1}{4^{s}}(\zeta (s,\frac{1}{4}) - \zeta (s,\frac{3}{4}))$

Proof

From DBG.ID.1 , the

$\beta (s)\Gamma (s)= \int _{0}^{\infty }\frac{x^{s-1} }{e^{x}+e^{-x}}dx$

$\beta (s) = \frac{1}{\Gamma (s)}\int _{0 }^{\infty}\frac{x^{s-1}}{e^{x}+e^{-x}} dx$

Multiply both numerator and denominator of integral by $e^{-x}$

$\beta (s) = \frac{1}{\Gamma (s)}\int _{0 }^{\infty}\frac{x^{s-1}e^{-x}}{1+e^{-2x}} dx$

(1)

The integral representation of Hurwitz zeta function:

$\zeta (s,a) = \frac{1}{\Gamma (s)}\int _{0 }^{\infty}\frac{x^{s-1}e^{-ax}}{1-e^{-x}} dx$

(2)

$\zeta (s,\frac{1}{4}) = \frac{1}{\Gamma (s)}\int _{0 }^{\infty}\frac{x^{s-1}e^{-\frac{x}{4}}}{1-e^{-x}} dx$

$\zeta (s,\frac{3}{4}) = \frac{1}{\Gamma (s)}\int _{0 }^{\infty}\frac{x^{s-1}e^{-\frac{3x}{4}}}{1-e^{-x}} dx$

change of variable x=2t , dx=2dt

$\zeta (s,\frac{3}{4}) = \frac{1}{\Gamma (s)}\int _{0 }^{\infty}\frac{(2t)^{s-1}e^{-\frac{3t}{2}}}{1-e^{-2t}} 2dt$

$\zeta (s,\frac{3}{4}) = \frac{2^{s}}{\Gamma (s)}\int _{0 }^{\infty}\frac{t^{s-1}e^{-\frac{3t}{2}}}{1-e^{-2t}} dt$

(3)

similarly

$\zeta (s,\frac{1}{4}) = \frac{2^{s}}{\Gamma (s)}\int _{0 }^{\infty}\frac{t^{s-1}e^{-\frac{t}{2}}}{1-e^{-2t}} dt$

(4)

$\zeta (s,\frac{1}{4}) - \zeta (s,\frac{3}{4}) = \frac{2^{s}}{\Gamma (s)}\int _{0 }^{\infty}\frac{t^{s-1}(e^{-\frac{t}{2}}-e^{-\frac{3t}{2}})}{1-e^{-2t}} dt$

(5)

$1-e^{-2t}= (1-e^{-t})(1+e^{-t})$

(6)

substituting (6) into (5)

$\zeta (s,\frac{1}{4}) - \zeta (s,\frac{3}{4}) = \frac{2^{s}}{\Gamma (s)}\int _{0 }^{\infty}\frac{t^{s-1}e^{-\frac{t}{2}}(1-e^{-t})}{(1-e^{-t})(1+e^{-t})} dt$

(7)