Timeline for Integral of the error estimate in the prime number theorem
Current License: CC BY-SA 2.5
3 events
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| Nov 29, 2009 at 22:52 | comment | added | Terry Tao | Well, from the explicit formula for $\psi(x)$, this is equivalent to having no zeroes with real part greater than $1 - \delta$, so the answer depends on what you mean by "without the Riemann hypothesis". Basically, a single exceptional zero $\rho$ (and its conjugates, of course) is already close to the worst-case scenario, and in this scenario there is no distinction between average-case and worst-case behaviour of $\psi(x)$. | |
| Nov 27, 2009 at 0:05 | comment | added | Mark Lewko | Thanks. I would be interested to know if this can be improved (without the Riemann Hypothesis) to $\int_X^{2X} |\psi(x) - x|^2dx << X^{3-\delta}$ for $\delta>0$. | |
| Nov 26, 2009 at 22:50 | history | answered | engelbrekt | CC BY-SA 2.5 |