Simple Formula to Describe $\zeta$ in One Go

Question

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.

Answer 1

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. :)

Answer 2

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)