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Soft question: Where is the connection between the zeros of Riemann's $\zeta$-funciton and the density of prime numbers?

Is there a short answer to this question, to get the overview? I once had a lecture about this topic, but it took several weeks to proof the prime number theorem using $\zeta$, so I lost track of the general idea.

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1 Answer 1

The Riemann hypothesis, about the non-trivial zeros of $\zeta(s)$, is equivalent to the following statement: For any real number $x\ge 1$ the number of prime numbers less than $x$ is approximately $Li(x)$ and this approximation is essentially square root accurate. More precisely, $$ \pi(x)=Li(x)+O(\sqrt{x}\log(x)). $$ Here $\pi(x)=\sum_{p\le x}1$ is the number of primes up to $x$. This says that the distribution of primes has the best possible error term if and only if the Riemann hypothesis holds true.

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This is about as much as I already knew. So what is the idea to proof this equivalence? –  Leif Sabellek Nov 26 '13 at 19:26
Well, more or less the same ideas as for the PNT, which took several weeks, as you said. Elkies has a relatively short proof, see page $4$ in math.harvard.edu/~elkies/M259.02/pnt.pdf. –  Dietrich Burde Nov 26 '13 at 19:39

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