# hurwitz zeta function

• June 5th 2011, 02:12 PM
klw289
hurwitz zeta function
I'm trying to calculate $\zeta$(4, $\frac{1}{2}$)

I used $\zeta$(s,a) = $\Gamma$(1-s) I(s,a)

Am I right by saying $\Gamma$(1-s) is undefined for s=4 and
I(4, 1/2) = 0 as s is a positive whole number

So $\zeta$ (4, 1/2) =0

Am I doing this right or am I just making a mess of this

• June 5th 2011, 04:35 PM
chisigma
Quote:

Originally Posted by klw289
I'm trying to calculate $\zeta$(4, $\frac{1}{2}$)

I used $\zeta$(s,a) = $\Gamma$(1-s) I(s,a)

Am I right by saying $\Gamma$(1-s) is undefined for s=4 and
I(4, 1/2) = 0 as s is a positive whole number

So $\zeta$ (4, 1/2) =0

Am I doing this right or am I just making a mess of this

Better chance You have using the relation...

$\zeta(s,\frac{1}{2}) = \sum_{n=0}^{\infty} (n+\frac{1}{2})^{-s} = 2^{s}\ \sum_{n=0}^{\infty} (2n+1)^{-s} =$

$= 2^{s}\ (1-2^{-s})\ \zeta (s)= (2^{s}-1)\ \zeta(s)$ (1)

... so that is...

$\zeta(4,\frac{1}{2}) = 15\ \zeta(4) = \frac{\pi^{4}}{6}$

Kind regards

$\chi$ $\sigma$
• June 5th 2011, 04:59 PM
Renji Rodrigo
A little generalization xD

Calculate
$\zeta(2n,\frac{1}{2})$
where $n \in N$.

$\zeta(2n,\frac{1}{2})=\sum_{k=0}^{\infty}\frac{1}{ (k+\frac{1}{2})^{2n}} =2^{2n}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{2n}}$

if you know
$\sum^{\infty}_{k=0}\frac{1}{(2k+1)^{2n}}=\frac{B_{ 2n}(-1)^{n+1}.[4^{n}-1]\pi^{2n}}{2(2n)!}$

we have

$\zeta(2n,\frac{1}{2}) =\frac{4^nB_{2n}(-1)^{n+1}.[4^{n}-1]\pi^{2n}}{2(2n)!}$

where B_n is the n-th Bernoulli number .