# Trace over the zeros with real part 1/2 Only

If RH is not true, we have that Weil's explicit formula still holds:

$$\sum_{\gamma} h(\gamma) = h(i/2)+h(-i/2)-2 \sum_{n=1}^{\infty} \frac{ \Lambda(n)}{ \sqrt n}g(logn)+\frac{1}{2\pi} \int_{-\infty}^{\infty}h(r)\Psi(1/4+ir/2)dr.$$

My question is if there are formulae that take into account ONLY the Riemann zeros with real part 1/2 and not the others.. thanks.

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Absent a proof of RH (or something close to RH), there are going to be some significant obstructions to the existence of such a formula, due to the possibility of two zeroes on the critical line that are very close together. A small perturbation of the zeta function could then move these two zeroes off of the critical line (much as perturbing the polynomial $z^2-\varepsilon^2$ to $z^2+\varepsilon^2$ moves two nearby real zeroes $\pm \varepsilon$ into two complex zeroes $\pm i \varepsilon$). The zeta function and its perturbation would be more or less indistinguishable from each other by the sort of contour integration operations that are traditionally used in Weil-type explicit formulae. As a consequence, such formulae cannot distinguish between two nearby zeroes on the critical line, and two nearby zeroes slightly away from the critical line, which already rules out a lot of options for a formula that only captures the critical line behaviour. Indeed, if one restricts attention to formulae which depend analytically with respect to analytic perturbations of the underlying complex function, I would imagine that there are no interesting formulae which only involve the real zeroes.
A model problem would be to try to locate an analytic formula relating the coefficients of a real polynomial with the purely real zeroes of that polynomial. Even in the quadratic case, it seems clear that no non-trivial formula exists, because one side of the formula would develop some sort of singularity (such as failure of analyticity) as one transitioned from $z^2-\varepsilon^2$ to $z^2+\varepsilon^2$, and the other side of the formula would not.