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An integral representation of the Hurwitz zeta function

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
For $ \text{Re} (a) >0$ and $\text{Re} (s)>1$, the Hurwitz zeta function is defined as $ \displaystyle \zeta(s,a) = \sum_{n=0}^{\infty} \frac{1}{(a+n)^{s}} $.

Notice that $\zeta(s) = \zeta(s,1)$.

So the Hurwitz zeta function is a generalization of the Riemann zeta function.

And just like the Riemann zeta function, the Hurwitz zeta function can be continued analytically to all complex values of $s$ excluding $s=1$.

One way to see this is an integral representation that generalizes the one I recently posted for the Riemann zeta function.

$\displaystyle \zeta(s,a) = 2 \int_{0}^{\infty} \frac{\sin (s \arctan \frac{t}{a} )}{(a^{2}+t^{2})^{s/2} (e^{2 \pi t}-1)} \ dt + \frac{1}{2a^{s}} + \frac{a^{1-s}}{s-1} $

The derivation of this integral representation shouldn't be that much different.

But since this representation is stated almost nowhere, I thought it would be something interesting to post.


EDIT: It actually is stated on Wolfram MathWorld.
 
Last edited:

DreamWeaver

Well-known member
Sep 16, 2013
337

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
Using contour integration makes the derivation/proof so much easier.
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
You can derive a simple yet exotic-looking representation of the gamma function from this integral representation of the Hurwitz zeta function.


$$ \frac{\partial }{\partial s} \zeta(s,a) \Big|_{s=0} = \zeta'(0,s) = 2 \int_{0}^{\infty} \frac{\arctan (\frac{t}{a})}{e^{2 \pi t}-1} \ dt - \frac{\log a}{2} + a \log a -a $$


Binet's integral formula once again states


$$ \int_{0}^{\infty} \frac{\arctan \left( \frac{x}{z} \right)}{e^{2 \pi x} -1} \ dx = \ln \Gamma(z) - \left( z- \frac{1}{2} \right) \ln z + z - \frac{\ln (2 \pi)}{2} $$


So

$$ \zeta'(0,a) = \log \Gamma(a) - a \log a + \frac{\log a}{2} + a - \frac{\log (2 \pi)}{2} - \frac{\log a}{2} + a \log a -a = \ln \Gamma(a) - \frac{\log (2 \pi)}{2}$$

$$ \implies \Gamma(a) = \sqrt{2 \pi} e^{\zeta'(0,a)} $$


(Speechless)

I think my brain just exploded a little bit.
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Nicely done, RV!! (Clapping)

I think I might have recommended this link before, but just in case, the following paper of Adamchik contains quite a few integrals analogous to the one above...

http://arxiv.org/pdf/math/0308086v1.pdf