Engineering Mathematics

Saturday, October 20, 2012

Numerical value of Riemann zeta function (Euler Product)

NVRZ.ID.1: Numerical value of Euler Product

$\prod \left ( \frac{1+p^{-2s}}{1-p^{-2s}} \right )= \frac{5}{2}\pi^{2}$

refer to RS.ID.2

$\frac{\zeta ^{2}(s)}{\zeta (2s)}= \prod \left ( \frac{1+p^{-s}}{1-p^{-s}} \right )$

(1)

if s=2 , then

$\frac{\zeta ^{2}(2)}{\zeta (4)} = \prod \left ( \frac{1+p^{-2s}}{1-p^{-2s}} \right )$

(2)

$\zeta (2)= \frac{\pi ^{2}}{6}$

$\zeta (4)= \frac{\pi ^{2}}{90}$

(3)

From (2) and (3) ,

$\frac{\frac{\pi ^{2}}{6}\frac{\pi ^{2}}{6}}{\frac{\pi ^{2}}{90}} = \prod \left ( \frac{1+p^{-2s}}{1-p^{-2s}} \right )$

(4)

$\frac{5}{2}\pi^{2} = \prod \left ( \frac{1+p^{-2s}}{1-p^{-2s}} \right )$

(5)

Sunday, September 9, 2012

Hurwitz Zeta function

HZ.ID.1: Hurwitz Zeta- Gamma relation

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

Proof

$\zeta (s,a)= \sum _{n=0}^{\infty } \frac{1}{(n+a)^{s}}$

(1)

From (), the multiplication of (1) with gamma integral definition

$\zeta (s,a)\Gamma (s)= \sum _{n=0}^{\infty }\int_{0}^{\infty } \frac{t^{s-1}e^{-t}}{(n+a)^{s}} dt$

(2)

change of variable

$t=(n+a)x , dt = (n+a)dx$

(3)

substituting (3) into (2)

$\zeta (s,a)\Gamma (s)= \sum _{n=0}^{\infty }\int_{0}^{\infty } \frac{(n+a)^{s-1}x^{s-1} e^{-(n+a)x}}{(n+a)^{s}} (n+a)dx$

(4)

$\zeta (s,a)\Gamma (s)= \sum _{n=0}^{\infty }\int_{0}^{\infty } x^{s-1} e^{-(n+a)x} dx$

(5)

$\zeta (s,a)\Gamma (s)= \sum _{n=0}^{\infty }e^{-nx}\int_{0}^{\infty } x^{s-1} e^{-ax} dx$

(6)

using geometric series

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

(7)

Saturday, September 8, 2012

Dirichlet Beta - Hurwitz zeta relation

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)

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

(8)

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

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

(9)

from (1) and (9) , Dirichlet Beta - Hurwitz zeta relation is given as by:

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

(10)

Saturday, June 2, 2012

Dirichlet Beta - Gamma

DBG.ID.1: Integral representation

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

given

Dirchelet Beta :

$\beta (s)= \sum _{0}^{\infty }\frac{(-1)^{n}}{(2n+1)^{s}}$

(1)

$\beta (s)\Gamma (s)= \sum _{0}^{\infty }(-1)^{n} \int _{0}^{\infty }\frac{t^{s-1}e^{-t}}{(2n+1)^{s}} dt$

(2)

$t=(2n+1)x \, , \, dt= (2n+1)dx$

(3)

replacing (2) with (3)

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

(4)

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

(5)

using geometric series

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

(6)

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

(7)

DBG.ID.2: Integral representation

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

proof

multiply both sides by s

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

(8)

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

(9)

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

(10)

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

(11)

DBG.ID.3:Recurrence Relation

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

given

from DBG.ID.1

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

(1)

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

(2)

from DBG.ID.2

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

(3)

(3) - (2) results in :

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

(4)

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

(5)

DBG.ID.5: Integral representation by natural log

$\beta (s)\Gamma (s)=\int _{0}^{1}\frac{In(\frac{1}{t})^{s-1}}{1+t^{2}} dt$

from DBG.ID.1

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

(1)

Change of variable

$x= In(\frac{1}{t}) \: ,\: dx = -\frac{1}{t}dt$

Therefore,

$\beta (s)\Gamma (s)=-\int _{1}^{0}\frac{In(\frac{1}{t})^{s-1}e^{-In(\frac{1}{t})}}{1+e^{-2In(\frac{1}{t})}t} dt$

(2)

$\beta (s)\Gamma (s)=\int _{0}^{1}\frac{In(\frac{1}{t})^{s-1}}{1+t^{2}} dt$

(3)

if s=2

$\beta (2)=\int _{0}^{1}\frac{In(\frac{1}{t})}{1+t^{2}} dt$

$\beta (2)=-\int _{0}^{1}\frac{In(t)}{1+t^{2}} dt$

(4)