# On an oscillation Theorem involving the Chebyshev function and the zeros of the Riemann zeta function

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 $$\prod_{p\leq x} (1-p^{-1})\sim e^{-\gamma}(\log x)^{-1} ?$$

• Do you have a reference? – András Bátkai Mar 10 at 18:41
• @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 – Will Jagy Mar 10 at 22:31
• @WillJagy: thanks, I also found this. I meant a bit more precise reference for the result recalled above. – András Bátkai Mar 11 at 8:07
• @AndrásBátkai, I see what you mean. Maybe the OP will loh at th Nicolas page and indicate which publication is relevant – Will Jagy Mar 11 at 17:06
• 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 – reuns Mar 11 at 21:08

$$\log\Big(e^{\gamma}(\log \theta(x))\prod_{p\leq x} (1-p^{-1})\Big)=\Omega_{\pm }(x^{b-\frac{1}{2}-\epsilon})$$, 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