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I try to prove, that the logarithm of the $\zeta$-function has the following representation for $z \in \mathbb{C}$ with $\text{Re}(z) > 1$ $$\log\zeta(z)=-\sum_p\log\left(1-\frac{1}{p^z}\right).$$ My idea was to show this identity for real argument and then to use the analytic continuation theorem, but I don't know, how to show that the right side is analytic. I have thought on the Weierstrass M-test and I hoped that I can try something like $$\left|\log\left(1-\frac{1}{p^z}\right)\right| \leq (1+\varepsilon)\frac{1}{p^{1+\delta}} $$ for all $z \in \mathbb{C}$ with $\text{Re}(z) \geq 1 + \delta$ for any fixed $\delta > 0$, but I failed. I will be grateful, if you could help me to show that the right side is analytic.

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    $\begingroup$ This approach looks perfect to me! When you say you failed, what exactly is the nature of the obstacle? $\endgroup$ – Greg Martin Jul 10 at 16:44
  • $\begingroup$ @GregMartin, I showed that $\lim_{z \to 0} \frac{\log(1+z)}{z} = 1$. Hence, for fixed $z$ and for every $\varepsilon > 0$ there is some $N \in \mathbb{N}$ such that for all $p \geq N$: $\left|\log\left(1-\frac{1}{p^z}\right)\right| \leq (1+\varepsilon) \frac{1}{p^z}$. But I don't know, why it should work for every $z$ with $Re(z) \geq 1 + \delta$. Because $N$ depends on both $\varepsilon$ and $z$. $\endgroup$ – Imperio Jul 10 at 17:46
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    $\begingroup$ Using the Taylor series of $\log(1+w)$ near $0$ and taking absolute values, you can easily show $|\log(1+w)| \le 2|w|$ if $|w|$ small enough and that should be enough for normal convergence in your case as primes are at least $2$... $\endgroup$ – Conrad Jul 10 at 17:54
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    $\begingroup$ $-\log(1-p^{-s}) \sim p^{-s}$ so the series converges absolutely and (locally) uniformly, for the derivative it is $-\frac{p^{-s} \log p}{1-p^{-s}} \sim -p^{-s} \log p$ so the series for $(\log \zeta(s))'$ converges absolutely and (locally) uniformly, thus we can integrate $(\log \zeta(s))'$ term by term to recover $\log \zeta(s)$ so that $\log \zeta(s)$ is holomorphic and hence analytic (by the Cauchy integral formula). The same kind of argument works for any Dirichlet series and Laplace transform (in their domain of convergence) $\endgroup$ – reuns Jul 10 at 19:05
  • $\begingroup$ @Conrad, thank you very much, I have finished my proof! $\endgroup$ – Imperio Jul 12 at 20:22

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