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- Jan 31, 2012

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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.

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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