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The Riemann Hypothesis is the statement that the Riemann zeta function $\zeta(s)$ does not vanish for $1/2<\Re(s)<1$. $\zeta(s)$ can also be expressed by the Euler product over primes $$\zeta(s)=\prod_{p}(1-p^{-s})^{-1}$$ for $\Re(s)>1$. I have heard several times that this product should play a crucial role in any solution of the RH. But since this product is only defined for $\Re(s)>1$, how is it expected to say anything about the zeros of $\zeta$ in the critical strip $0<\Re(s)<1$ ?

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    $\begingroup$ Because there are variants of zeta function without Euler products, but otherwise sharing many of its properties, which do not satisfy the analogue of Riemann hypothesis. The simplest example is probably Hurwitz zeta function. $\endgroup$ – Wojowu Dec 20 '18 at 13:07
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    $\begingroup$ I don't think it is "reasonable" to say that the Euler product will play a crucial role for solving RH. The product is just the same as $\zeta(s)$ for $\Re(s)>1$. We could also say "the prime property will play a crucial role in solving the strong Goldbach conjecture, i.e., that every even number $n\ge 4$ is the sum of two primes". $\endgroup$ – Dietrich Burde Dec 20 '18 at 13:21
  • $\begingroup$ @Wojowu, yes, but still the fact remains that the Euler product is only valid for $\Re(s)>1$, hence is irrelevant as far as the critical strip is concerned. In any case, note that the Davernport-Heilbronn zeta function $f(s)$ (whose analogue of the RH is false) can be written as a linear combination of two L-functions which have Euler products (with a common factor) and this fact plays the crucial role in Karatsuba's proof of the lower bound result for the number of zeros of $f(s)$. See Aleksander Ivic's ''Some reasons to doubt the RH, '' p.6. $\endgroup$ – OneTwoOne Dec 20 '18 at 13:22
  • $\begingroup$ Then the usual zeta function is also "irrelevant" as fas as the critical line is concerned, because it is only defined for $\Re(s)>1$. So this is not true. We consider meromorphic continuations anyway. And RH concerns $\zeta(s)$, and not some other $f(s)$. $\endgroup$ – Dietrich Burde Dec 20 '18 at 13:24
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    $\begingroup$ The Euler product is the zeta function for $\Re(s)>1$. If you have a meromorphic continuation for one of two equal things, you have it for both. $\endgroup$ – Dietrich Burde Dec 20 '18 at 13:28
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  • The Euler product implies that for $\Re(s) > 1$ $$\log \zeta(s) = \sum_{p^k} \frac{(p^k)^{-s}}{k}$$ the sum being over prime powers.

  • The first answer to your question is that we can prove the RH is true if and only if the Dirichlet series $$\sum_{p^k} \frac{(p^k)^{-s}}{k} - \sum_{n \ge 2} \frac{n^{-s}}{\log n}$$ converges for $\Re(s) > 1/2$.

    Since for $\Re(s) > 1/2$, the Dirichlet series $\sum_{p^k} \frac{(p^k)^{-s}}{k}-\sum_p p^{-s}$ converges and is analytic, and $\int_2^x \frac{dt}{\log t} = \sum_{2 \le n \le x} \frac{1}{\log n} + O(x^{\epsilon})$ this is often stated as "the RH is true iff $\pi(x)=\sum_{p \le x} 1 = \int_2^x \frac{dt}{\log t}+O(x^{1/2+\epsilon})$".

    Note this connection to the behavior of $\pi(x)$ is the only reason why we care so much of the zeros of $\zeta(s)$.

    The proof is quite similar to the prime number theorem, and relies on many properties of $\zeta(s)$ (analytic continuation, functional equation, bounds on the critical strip, density of zeros, non-vanishing on $\Re(s) = 1$ due to non-negativity of the coefficients of $\log \zeta(s)$...) all those properties being needed to deduce that if $\zeta(s)$ has no zeros for $\Re(s) \ge c$ then $\log \zeta(s) =O(\log(s-1))$ which is enough to obtain the absolute convergence of the inverse Mellin transform integral $$\int_1^x( \pi(y)-\int_2^y \frac{dt}{\log t}) dy \approx \frac{1}{2i \pi}\int_{c-i \infty}^{c+i \infty} \log ((s-1) \zeta(s)) \frac{x^{s+1}}{s(s+1)} ds = O(x^{c+1})$$

  • Now there is another question in your question which is "can we conjecture the RH for some Dirichlet not in the Selberg class, in particular with no Euler product ?" which is indeed very complicated.

    The intuitive answer is that for a Dirichlet series $F(s)$ not converging everywhere then both $F(s),\log F(s)$ have "non-messy" coefficients implies that $\log F(s)$ is almost a series over prime powers and $F(s)$ almost has an Euler product.

    For Dirichlet series with functional equation and meromorphic continuation with finitely many poles, this can be made rigorous and there are proofs (see this and this) that no Euler product implies (infinitely many) zeros in the neighborhood of $\Re(s)=1$.

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  • $\begingroup$ This is definitely more than what i expected, thanks ! $\endgroup$ – OneTwoOne Dec 23 '18 at 7:37

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