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Jul 29, 2019 at 20:52 comment added H A Helfgott In practice, though, you would use a verification of RH up to a certain height $T$, and so the zero-free region would become relevant only for $|\gamma|$ larger than $T$.
Jul 29, 2019 at 7:34 comment added 2734364041 Take $\zeta(s)$ in the PNT, for example. The sum over zeros in the explicit formula would be $\sum_{\beta+i\gamma} x^{\beta+i\gamma} \Gamma(\beta+i\gamma)$. Using $|x^{\beta+i\gamma}|\leq x^{\beta}$, and applying the standard zero-free region, the sum over zeros is bounded by $\sum_{\beta+i\gamma} x^{1-c/\log(3+|\gamma|)} |\Gamma(\beta+i\gamma)|$. Rewrite the sum as an integral using partial summation, and then apply Stirling's formula. The integral is roughly $\int_{0}^{\infty} x^{1-c/\log(3+t)} e^{-\pi t/2}\log(3+t)dt$, a stationary phase integral.
Jul 29, 2019 at 2:58 comment added H A Helfgott Again, I don't understand. Where would the stationary phase integral be coming from? Are you thinking of what happens when you apply an explicit formula to estimate an exponential sum?
Jul 29, 2019 at 0:01 comment added 2734364041 @HAHelfgott My mistake, the convergence is indeed exponential in vertical strips. But since your concern is for explicit estimates, if you are not assuming GRH (this is the probably the situation in which such smoothing is typically most advantageous), the sum over zeros will become a stationary phase integral once you apply the Stirling bound. This integral becomes difficult to explicitly estimate with decent constants using existing zero-free regions (though it's easy to estimate asymptotically).
Jul 27, 2019 at 18:23 comment added H A Helfgott But how does $\Gamma(s) = \Gamma(s+1)/s$ imply that one can achieve at most a polynomial rate of convergence in the integral? $\Gamma(s)$ decreases exponentially.
Jul 17, 2019 at 17:46 history answered 2734364041 CC BY-SA 4.0