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Let $s=σ+it$. Assume that $ζ(s)-1/(s-1)$ has an analytic continuation to the half plane $σ>0$. Show that if $s = 1 + it$, with $t≠0$, and $ζ(s) = 0$ then $s$ is at most a simple zero of $ζ$.

I have no idea how to go about this. My initial idea was to differentiate, but I don't know much about the differential of $ζ$. I have also seen that no such $s$ exists in the first place, so Google has not helped much.

If it helps, the first part of the question was to show that $\zeta(s)^{-1}= \sum^{\infty}_{n=1}\mu(n)/n^{s}$ for $σ>1$, which I was able to do.

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    $\begingroup$ did you try mertens' trigonometric inequality ? it might even give a zero of order zero. $\endgroup$
    – Mr. No
    May 14 '15 at 10:46
  • $\begingroup$ @CaptainDarling That's the first time I've heard of it but I've looked it up and it does solve my problem. Thank you. $\endgroup$ May 14 '15 at 19:57
  • $\begingroup$ There are nowadays too many proofs that no zeros on the line $\Re(s)=1$ exist, the Mertens identity is maybe the most accessible. $\endgroup$
    – Mr. No
    May 14 '15 at 22:07
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theorem : if $F(s) = \int_1^\infty x^{-s-1} f(x) dx$ and $|f(x)| = \mathcal{O}(x)$ then the integral converges for $\Re(s) > 1$ and $|F(1+\epsilon+it)| < \frac{C}{\epsilon} $ (if it has a pole on $\Re(s) = 1$ then it is of order $\le 1$)

this is because $\int_1^{\infty} x^{-s-1} (C x^{\rho}) dx $ is a pole of order $1$ at $s = \rho$. thus $x^{1+it}$ is a pole of order $1$ at $s=1+it$.

application and demonstration of the theorem : because $\frac{1}{\zeta(s)} = \sum \mu_n n^{-s} = s \int_1^\infty M(x) x^{-s-1}dx$ and $M(x) = \sum_{n \le x} \mu_n = \mathcal{O}(x)$ you got that $$\left|\frac{1}{s \zeta(s)}\right| < \int_1^\infty x x^{-\sigma-1}dx = \frac{1}{\sigma-1}$$

which is a pole of order $1$ (and not more) at $Re(s) = 1$.

the moral of that exercise is that looking at how $f(x)$ is small or not when $x \to \infty$ (and also $x\to 0$ if the integral goes to $0$) tells us how poles of $F(s)$ are distributed and of what order they are.

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