# The sum over zeros in the explicit formula for $\zeta(s)$

The explicit formula for $\zeta(s)$ is: $$\psi(x)=x-\sum_{|\operatorname{Im}\rho|<T}\frac{x^\rho}{\rho}-\log(2\pi)-\log\left(1-\frac{1}{x^2}\right)+O\left(\frac{x\log^2T}{T}\right),$$ where $\psi(x)=\sum_{p^k<x}\log p$, $x>1$ is a non-integer, $\rho$ is a nontrivial zero of $\zeta(s)$, and the sum over $\rho$ is taken with multiplicities.

Letting $T\rightarrow\infty$, we can conclude that the sum over all nontrivial zeros $\rho$ is convergent (albeit conditionally) since the other terms in the formula are finite. Is there a way to directly show that this sum is convergent?

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oops, yes I just fixed it. –  B.W. Apr 15 '14 at 21:41

Failed argument: The argument principle and an application of Jensen's lemma from complex variables, and Stirling's asymptotic for Gamma, via the functional equation for zeta, show that the number of zeros up to height $T$ is bounded by a constant times $T\log T$ (in fact giving asymptotics...). This, together with the symmetry of zeros from the functional equation and so on, gives the conditional convergence.
I think this shows clearly why say $\sum 1/\rho$ converges conditionally, but it is not clear to me why this implies the conditional convergence of $\sum x^{\rho}/\rho$. Just information about $N(T)$ does not preclude some very bad distribution of the ordinates. Also, where above do you see the difference between $x$ being a prime power, and not a prime power? –  Lucia Apr 13 '14 at 3:51
I don't think what you say is correct. What we need is the convergence of $\sum_{\gamma >0}\frac{\gamma e^{t\beta}}{\beta^2+\gamma^2}\sin(\gamma t)$ (I've combined conjugate zeros, and the cos part is fine). Here $t=\log x$ in the notation of the question. What you say works fine if $t=0$ as the sine term disappears. In general, just the zero count only localizes zeros in intervals of length $1$. What if the zeros align themselves to always satisfy $\sin (\gamma t) >0.5$ say. How do you rule this out? –  Lucia Apr 13 '14 at 15:14
If you mean the explicit formula of Weil, the test function should be bounded by $(1+|\im z|))^{2 + \epsilon}$ and we also have absolute convergence there. So I am not sure if your argument can be made rigorous, because it would imply an explicit formula with more relaxed conditions on the allowed test functions. –  Marc Palm Apr 14 '14 at 11:55