Timeline for Error bounds for $\pi(x)-Li(x)$ and convergence of the associated Dirichlet integral

8 events

when toggle format	what		by	license	comment
Sep 13, 2019 at 19:37	comment	added	reuns		If $\Theta=1/2$ or if there is a zero on $\Re(s) =\Theta$ then it diverges, if $\Theta=1$ then it converges.
Sep 13, 2019 at 17:13	comment	added	GH from MO		I think that both "yes" and "no" are consistent with our current knowledge. If, say, there is no zero with $\Re(s)\geq\Theta$ and the zeros with $\Re(s)\approx\Theta$ are sparsely located, then I think that $I_\Theta$ can converge. If, say, there is a single zero with $\Re(s)=\Theta$ and there is no zero with $1/2<\Re(s)<\Theta$, then surely $I_\Theta$ diverges. I have not verified these claims, but this was my immediate thought!
Sep 13, 2019 at 17:12	comment	added	Peter Humphries		@GregMartin, Landau's lemma is for integrals $\int_{1}^{\infty} f(x) x^{-s - 1} \, dx$ for which $f(x)$ is of constant sign for all sufficiently large $x$. Theorem 15.2, however, shows that $\pi(x) - \mathrm{Li}(x) = \Omega_{\pm}(x^{\Theta - \varepsilon})$.
Sep 13, 2019 at 16:41	comment	added	Greg Martin		Presumably no, by Landau's theorem? See Section 15.1 of Montgomery/Vaughan's Multiplicative Number Theory I.
Sep 13, 2019 at 14:17	history	edited	Q_p	CC BY-SA 4.0	deleted 16 characters in body
Sep 13, 2019 at 14:10	history	edited	Q_p	CC BY-SA 4.0	added 5 characters in body
Sep 13, 2019 at 14:05	review	First posts
Sep 13, 2019 at 14:09
Sep 13, 2019 at 14:04	history	asked	Q_p	CC BY-SA 4.0