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Define $\theta(x)=\sum_{p\leq x} \log p $, where $p>1$ denotes a prime. Nicolas proved that if the Riemann zeta function $\zeta(s)$ vanishes for some $s$ with $\Re(s)\leq 1/2 + b$, where $b\in(0, 1/2)$, then

$$\log\Big(e^{\gamma}(\log \theta(x))\prod_{p\leq x} (1-p^{-1})\Big)=\Omega_{\pm }(x^{-b}).$$

Is the reverse necessarily true ? It appears so to me, since

$$\theta(x)=x +O(\sqrt{x})-\sum_{\rho} \frac{x^{\rho}}{\rho},$$

where the sum is over the complex zeros of $\zeta$ and $\prod_{p\leq x} (1-p^{-1})\sim e^{-\gamma}(\log x)^{-1} ?$

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  • $\begingroup$ Do you have a reference? $\endgroup$ – András Bátkai Mar 10 at 18:41
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    $\begingroup$ @AndrásBátkai Nicolas puts most of his stuff on his site. math.univ-lyon1.fr/~nicolas and publications at math.univ-lyon1.fr/~nicolas/publications.html $\endgroup$ – Will Jagy Mar 10 at 22:31
  • $\begingroup$ @WillJagy: thanks, I also found this. I meant a bit more precise reference for the result recalled above. $\endgroup$ – András Bátkai Mar 11 at 8:07
  • $\begingroup$ @AndrásBátkai, I see what you mean. Maybe the OP will loh at th Nicolas page and indicate which publication is relevant $\endgroup$ – Will Jagy Mar 11 at 17:06
  • $\begingroup$ The explicit formula doesn't help a lot because it doesn't converge absolutely. Under the RH $\theta(x) = x+O(\sqrt{x} \log^2x)$ you can't improve it much. It is easy to show if $\theta(x)-x = o(x^{\sigma})$ then $\zeta(s)$ has no zeros for $\Re(s) \ge \sigma$, the converse is much harder $\endgroup$ – reuns Mar 11 at 21:08
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The result Nicolas proved (Theorem 3, Jean-Louis Nicolas. Petites valeurs de la fonction d'Euler, J. Number Theory, 17, 1983, 375--388, paper linked HERE) is actually not quite what is claimed in this post, but:

$\log\Big(e^{\gamma}(\log \theta(x))\prod_{p\leq x} (1-p^{-1})\Big)=\Omega_{\pm }(x^{b-\frac{1}{2}-\epsilon})$, for all $\epsilon > 0$, where $0< b < \frac{1}{2}$ is such that RZ has a zero with real part $\frac{1}{2} + b$, so it is in the classical spirit as the worse RH fails (the higher the supremum of the real part of the critical zeros), the stronger the claimed $\Omega$.

The estimate claimed in the post above is wrong (obviously because it is stronger for a weaker claim, not to speak that since RZ always vanishes on the critical line the claim of the post would imply the estimate to be true for $b$ going to zero and that contradicts the later claim when RZ is true). Using the zeros symmetry around the critical line, one can rephrase the $\Omega$ result to:

$\log\Big(e^{\gamma}(\log \theta(x))\prod_{p\leq x} (1-p^{-1})\Big)=\Omega_{\pm }(x^{-\beta-\epsilon})$, for all $\epsilon > 0$, where $0< \beta < \frac{1}{2}$ is such that RZ has a zero with real part $\beta$

He also proved that if RH is true $\log\Big(e^{\gamma}(\log \theta(x))\prod_{p\leq x} (1-p^{-1})\Big)x^{\frac{1}{2}}\log x$ is always negative for $x>2$, has inferior limit $\log{4\pi}-4-\gamma$ and superior limit less or equal to $\gamma - \log{4\pi}$ which are both close to $-2$, so obviously any $\Omega$ result as above with a power $x^{-\beta}, 0 < \beta < \frac{1}{2}$ dispproves RH, with the lower the $\beta$ the worse RH fails, while any $O$ result gives bounds on the supremum of the real parts of critical zeros, with the higher the $\beta$ the better the bounds

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