# Does the analytic continuation of $\zeta(s)$ for $s\in\mathbb{C} \ \backslash \ \{1\}$ require Cauchy's integral formula?

I've seen Cauchy's integral formula used in proofs for the analytic continuation of the Riemann zeta function to the entire complex plane ($$s\neq1$$). My question is whether a counterexample can be given that does not require the use of Cauchy's integral formula at all to achieve the above analytic continuation for the zeta function.

• Actually, I don't think Cauchy's integral formula is at all an essential item is such a proof... because not everything does have an analytic continuation, so something substantially more would be needed. If you could expand on the literal argument you're meaning to refer to, that would help people understand the substance of your question (and your previous one, commented-upon by @reuns). Apr 27, 2021 at 21:52
• My interest is with regard to how Cauchy's integral formula effectively defines each value of a function non-locally via the arithmetic mean of values on an enclosing boundary. I am wondering if this mean-value property could have any bearing on the unique behavior of the zeta function in its analytically-continued domain. Apr 27, 2021 at 21:58
• Ah... thanks for the clarification. I think that Cauchy's formula itself cannot readily distinguish zeta, because linear combinations of $L$-functions (e.g., studied by Bombieri and Hejhal) are known to have infinitely-many zeros off the line. And zeta functions of quaternion algebras provably have $100\%$ of their zeros off the line. Apr 27, 2021 at 22:32

A proof using zero theorem:

For $$s>1$$

$$\zeta(s)=\sum_{n\ge 1}n^{-s}=\sum_{n\ge 1}s\int_n^\infty x^{-s-1}dx$$ $$=s\int_1^\infty \lfloor x\rfloor x^{-s-1}dx= \frac{s}{s-1}+s\int_1^\infty f_1(x)x^{-s-1}dx$$ where $$f_1(x)=x-\lfloor x\rfloor$$ is one-periodic.

Given $$f_n$$ one-periodic let $$c_n=\int_0^1 f_n(x)dx$$ and $$f_{n+1}(x)=\int_1^x (f_n(y)-c_n)dy$$ which is one-periodic again then $$(s+n-1)\int_1^\infty f_n(x)x^{-s-n}dx= c_n+(s+n-1)(s+n)\int_1^\infty f_{n+1}(x)x^{-s-n-1}dx$$ is analytic for $$\Re(s) > -n$$.

Whence by induction $$\zeta(s)-\frac{s}{s-1}$$ is entire.

• What do you mean by "one-periodic?" Apr 27, 2021 at 22:04
• With period 1 and mean value $c_n$ Apr 27, 2021 at 22:06
• Anything unclear Apr 28, 2021 at 4:01
• Can you provide any sources to this proof? Apr 28, 2021 at 13:05
• A source for what, which step do you not understand? Apr 28, 2021 at 13:08