MHB An integral representation of the Hurwitz zeta function

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The Hurwitz zeta function, defined for Re(a) > 0 and Re(s) > 1, generalizes the Riemann zeta function, with the relationship zeta(s) = zeta(s,1). It can be analytically continued to all complex values of s except s = 1. An integral representation of the Hurwitz zeta function is presented, which simplifies its derivation using contour integration. This representation leads to a connection with the gamma function, revealing that Gamma(a) can be expressed in terms of the Hurwitz zeta function's derivative. The discussion highlights the significance of this representation, noting its rarity in literature.
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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.
 
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Using contour integration makes the derivation/proof so much easier.
 
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.
 
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
 
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