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.