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I was looking around for a simple formula that can describe Riemann Zeta $\zeta$ in one go, at $\mathbb C-\{1\}$, but I couldn't really find one. Could someone help me find one?

I know one way to do this, but it gives a ridiculously complicated formula:

From analytic continuation of $$\Xi(s)=\frac1{s-1}-\frac1s+\int_1^\infty(u^{-s/2-1/2}+u^{s/2-1})\sum_{n=1}^\infty e^{-\pi n^2u}du$$ which holds for $s\in\mathbb{C}-\{0,1\}$, we can deduce a general formula of $\zeta$: $$\begin{align} \zeta(s) &= \pi^{s/2}\frac{\Xi(s)}{\Gamma(s/2)}\\ &= \pi^{s/2}\left(\frac1{s-1}-\frac1s+\int_1^\infty(u^{-s/2-1/2}+u^{s/2-1})\sum_{n=1}^\infty e^{-\pi n^2u}du\right)\left(e^{\gamma s/2}\frac s2 \prod_{n=1}^\infty (1+\frac s{2n})e^{-s/(2n)}\right)\\ &=\frac12(\pi e^{\gamma})^{s/2}\left(\frac 1{s-1}+s\int_1^\infty(u^{-s/2-1/2}+u^{s/2-1})\sum_{n=1}^\infty e^{-\pi n^2u}du\right)\left(\prod_{n=1}^\infty \frac{1+\frac s{2n}}{e^{s/(2n)}}\right) \end{align}$$ Thus we can write an explicit formula of $\zeta$ that should hold for $\mathbb C-\{1\}$ as following: $$\zeta(s)=\frac12(\pi e^{\gamma})^{s/2}\left(\frac 1{s-1}+s\int_1^\infty(u^{-s/2-1/2}+u^{s/2-1})\sum_{n=1}^\infty e^{-\pi n^2u}du\right)\left(\prod_{n=1}^\infty \frac{1+\frac s{2n}}{e^{s/(2n)}}\right)$$

Note that infinite product of $\frac1{\Gamma(s/2)}$ was used to make the expression more direct.

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    There are some globally convergent series expressions for zeta on the MathWorld entry.2012-12-30
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    This question is probably not supposed to make sense to me. Nonetheless, could you give a more clear notion of what you're trying to find?2012-12-30
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    anon That formula looks pretty nice, thank you. However, if possible, I'd like to find an elementary formula that doesn't involve contour integral around a... nontrivial contour. @Limitless I'm looking for a simple expression of $\zeta$ that converges everywhere at $\mathbb C-\{1\}$.2012-12-30
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    @progressiveforest That makes more sense, thank you. I get a slight notion of what you're saying--enough to see where you're going, even if I can't make the journey myself.2012-12-30

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I think the formula that anon is referring to is $$ \forall s \in \mathbb{C} \setminus \{ 1 \}: \quad \zeta(s) = \frac{1}{1 - 2^{1-s}} \sum_{n=0}^{\infty} \left[ \frac{1}{2^{n+1}} \sum_{k=0}^{n} (-1)^{k} \binom{n}{k} (k + 1)^{-s} \right]. $$ I do not see how a contour integral is being used to compute each term. The relevant references are:

  • Havil, J. Gamma: Exploring Euler's Constant. Princeton, NJ: Princeton University Press, 2003.
  • Hasse, H. Ein Summierungsverfahren für die Riemannsche Zeta-Reihe. Math. Z. 32, 458-464, 1930.

I hope this helps. :)

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    Thanks. That's pretty direct. I think what anon meant though is formula at the bottom of this page: http://planetmath.org/encyclopedia/AnalyticContinuationOfRiemannZetaUsingIntegral.html2012-12-30
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    Note that this formula is not actually defined on all $s\in\Bbb C\setminus\{1\}$, but rather all $s$ such that $2^{1-s}\ne1$, due to the division at the outset. The correct statement is $$\zeta(s_0)=\lim_{s\to s_0}\frac1{1-2^{1-s}} \sum_{n=0}^\infty\left[\frac1{2^{n+1}}\sum_{k=0}^n(-1)^k\binom nk(k + 1)^{-s} \right].$$2014-07-21
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There is also an integral formula:

$$ \zeta \left( s \right) =\frac{1}{2}+ \frac{1}{s-1}+2\,\int _{0}^{ \infty }\!{\frac {\sin \left( s\arctan \left( t \right) \right) }{ \left( {t}^{2}+1 \right) ^{s/2} \left( -1+{{\rm e}^{2\,t\pi }} \right) }}{dt} $$ (according to Maple)

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    thank you, this is also very nice.2012-12-30