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All:

We all know that zeros of Zeta function determine (or predict) the locations of primes through explicit formula.

Is there anything similar that shows how primes determine the locations of zeros of Zeta functions ?

Thank you.

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The connection between primes and the zeroes of the Riemann zeta function is based on the fact that $$ -\frac{\zeta'(s)}{\zeta(s)} = \sum_{n \geq 1} \frac{\Lambda (n)}{n^s}, $$ where $\Lambda(n) = \log p$ if $n = p^k$ a prime power. Morally, if $\zeta(\sigma + it) = 0$, then there is a contributing term of size $x^\sigma$ in the asymptotic expansion of $\pi(x)\log x$. That is, the asymptotic should look something like $$ \pi(x) \log x = x + (\text{maybe some terms from other zeroes}) + C_\sigma x^\sigma + o(x^\sigma),$$ where this is to be interpreted very loosely. (But it can be stated very precisely).

What this means is that if there is a zero with real part $\sigma$, then we should expect the number of primes up to $x$ to eventually look at least as large as $$ \pi(x) \gg \frac{x}{\log x} + C_\sigma \frac{x^\sigma}{\log x} \tag{1}$$ for some constant $C_\sigma$. Similarly, if the number of primes looked like the right hand side of $(1)$, then we would expect a zero with real part $\sigma$.

More precisely, one could look at sums of the form $$ \sum_{n \geq 1} \Lambda(n) e^{-n/X},$$ which is morally a sum up to $X$, and try to detect second order asymptotics here. In this form, secondary terms from zeroes of the zeta function appear in a concrete, explicit, not at all hand-wavy way since convergence issues are no longer a problem.

A clear understanding of an analytic proof of the prime number theorem would make this more clear. Consider reading this answer of mine on another question, and perhaps a book on analytic number theory. I find Apostol's Introduction to Analytic Number Theory to be extremely friendly. Alternately, Montgomery and Vaughan's book is excellent, though a bit more sophisticated.

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    $\begingroup$ This is morally true but technically very false. The PNT states that $\pi(x) = \mathrm{Li}(x) + O(e^{-c\sqrt{\log x}})$, and so unconditionally we know that $\pi(x) \sim \frac{x}{\log x} + \frac{x}{(\log x)^2}$. You really have to use $\mathrm{Li}(x)$ to see that the existence of zeroes off the critical line implies a bad error term in the PNT. $\endgroup$ – Peter Humphries Sep 15 '15 at 15:24

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