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$$\operatorname{Log}\biggl(\prod_p \Bigl(1-\frac{1}{p^{s}}\Bigr)^{-1}\biggr)=\prod_p \operatorname{Log}\Bigl(1-\frac{1}{p^{s}}\Bigr)^{-1} $$ where $Re(s)>1$ and $p$ is a prime. If I take the principal branch for logarithm, we need to verify that the sum of arguments in the logarithm on the right-hand side doesn't pass the branch line. I'm not sure if this is completely trivial. Could somebody help?

able20
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  • $\Pi_p$ is the product over primes $p$? That equation seems unbelievable to me. More plausible would be$$\operatorname{Log}(\prod_p (1-\frac{1}{p^{s}})^{-1})=\sum_p\operatorname{Log}(1-\frac{1}{p^{s}})^{-1}$$ – GEdgar Jan 26 '21 at 18:49
  • I'm so sorry, I needed to edit the post because I wrote the equation inrorrectly. – able20 Jan 26 '21 at 18:50
  • And $p$ is a prime and we have $Re(s)>1$ – able20 Jan 26 '21 at 18:50
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    The product $\prod_p(1-p^{-s})^{-1}$ is also called the Riemann zeta function $\zeta(s)$. Now look at https://math.stackexchange.com/questions/321602/how-to-understand-log-zetas-riemann-zeta-function, which can help you to answer your question. – jojobo Jan 26 '21 at 19:14

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