How to show Riemann zeta function $\zeta(s)$ is holomorphic (except at s=1)? The Riemann zeta function $\zeta(s)$ is known to be holomorphic, except at $s=1$.
I am not trained to university maths, but I interpret this to mean:

*

*it is complex differentiable at every point (smoothly changing values, no discontinuities)


*and is equal to Taylor series developed around any point in its domain
Question: How do we show $\zeta(s)$ is holomorphic?
Is it enough to show that the series which represent it, $\zeta(s)=\sum 1/n^s$ valid for $\Re(s)>1$ for example, are infinitely differentiable, except at the known pole at $s=1$?
Is it also correct to say that by the principle of analytic continuation, any other series, such as $\zeta(s)=(1-s^{1-s})^{-1}\eta(s)$ valid over larger domains, are also holomorphic, except at any new poles. By separate analysis, there are no new poles in $0<\Re(s)\leq 1$.
Apologies if this question seems naive, I am self-teaching.
 A: Let $U=\{s\in\Bbb{C}\,:\, \text{Re}(s)>1\}$, and consider first $\zeta_0:U\to\Bbb{C}$ defined as
\begin{align}
\zeta_0(s)&:=\sum_{n=1}^{\infty}\frac{1}{n^s}
\end{align}
For each $\sigma>1$, if $\text{Re}(s)\geq \sigma$, then
\begin{align}
\sum_{n=1}^{\infty}\left|\frac{1}{n^s}\right|&=\sum_{n=1}^{\infty}\frac{1}{n^{\text{Re}(s)}}\leq \sum_{n=1}^{\infty}\frac{1}{n^{\sigma}}<\infty
\end{align}
This implies local uniform convergence of a series of holomorphic functions on $U$, and thus $\zeta_0$ is a holomorphic function on $U$. The only reason I use the notation $\zeta_0$ is to emphasize that we have not done any analytic continuation yet, so it's not the full Riemann zeta function yet. One has the following theorem:

Theorem.
There exists an analytic/holomorphic function $\zeta: \Bbb{C}\setminus \{1\}\to\Bbb{C}$ such that $\zeta|_U=\zeta_0$, and such that $s=1$ is a simple pole of $\zeta$ (with residue $1$).

We call the full mapping $\zeta$ the Riemann zeta function. $\zeta_0$ is merely a part of the restriction. As mentioned in the theorem, one has to prove that such an analytic/holomorphic function $\zeta:\Bbb{C}\setminus\{1\}\to\Bbb{C}$ exists. This is not a trivial thing (in general, existence of analytic continuations/extensions is a non-trivial issue), but in this case there are several well-known proofs of this theorem.
In the first part of my answer, showing the local uniform convergence of the series only shows that $\zeta_0$ is holomorphic on $U$. This says nothing about the existence of an analytic/holomorphic extension of $\zeta_0$ from the initial domain $U$ to $\Bbb{C}\setminus\{1\}$.

As I mentioned, there are several well-known proofs of the theorem. Here are two I remember off the top of my head:

*

*The first method can be found in Ahlfors. Start by observing that for $\text{Re}(s)>1$ (i.e $s\in U$) we have $\Gamma(s)\zeta_0(s)=\int_0^{\infty}\frac{x^{s-1}}{e^x-1}\,dx$. One can then deform the integration to get it as an integral over a certain contour $C$ which avoids the non-negative real axis, and then with a bit of work, one can show
\begin{align}
\zeta_0(s)&=-\frac{\Gamma(1-s)}{2\pi i}\int_C\frac{(-z)^{s-1}}{e^z-1}\,dz
\end{align}
(see Ahlfors for the details). Now, one observes that on the RHS, the integral is an entire function of $s$ (this is not trivial, but also not too hard if we have things like Fubini's theorem and Morera's theorem available). $\Gamma(1-s)$ has poles at $s=1,2,3,\dots$. In short, the RHS is a mermomorphic function on $\Bbb{C}$; but actually on the LHS, we already know $\zeta_0$ is holomorphic at $s=2,3,4,\dots$. In summary, the RHS provides us with a meromorphic extension of $\zeta_0$ to $\Bbb{C}$ with only a simple pole at $s=1$.


*Another approach for showing the existence of an analytic continuation involves using Mellin transforms and the Jacobi theta identity. Wikipedia gives a formula for $\zeta_0$. You can find more details in the delightful little book A Brief Introduction to Theta Functions (there are sections on Jacobi's identity, the Mellin Transform and the zeta function).
