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Let $\pi(x)$ denote the prime counting function and $$\pi_0(x) = \lim_{\epsilon \to 0} \frac{\pi(x+\epsilon)+\pi(x-\epsilon)}{2}.$$ I've seen noted in a few references the explicit formula $$\pi_0(x) = R(x)-\sum_\rho R(x^\rho), \quad x>1,$$ where the sum is over all zeros $\rho$ of the Riemann zeta function (the nontrivial zeros taken in conjugate pairs and in order of imaginary part), and where $$R(s) = \sum_{n = 1}^\infty \frac{\mu(n)}{n} \operatorname{li}(s^{1/n}).$$ See, for example, Watkins - the "encoding" of the distribution of prime numbers by the nontrivial zeros of the Riemann zeta function [common approach]. I am wondering if someone can help by providing a proof of this explicit formula for $\pi_0(x)$ (perhaps from some other known explicit formula, e.g., for $\Pi_0(x)$), or else supply a reference that proves it. I can't find a proof of it in the literature.

EDIT: The explicit formula for $\pi_0(x)$ above likely follows from the Riemann–von Mangoldt explicit formula for $\Pi_0(x)$ because, for all $x >1$, one has \begin{align*} \pi_0(x) & =\sum_{n=1}^\infty \frac{\mu(n)}{n}\Pi_0(x^{1/n}) \\ & = \sum_{n=1}^\infty \frac{\mu(n)}{n} \left( \operatorname{li}(x^{1/n}) - \sum_\rho \operatorname{Ei}\left(\frac{\rho \log x}{n}\right) - \log 2 \right) \\ & = R(x) - \sum_{n=1}^\infty \sum_\rho \frac{\mu(n)}{n}\operatorname{Ei}\left(\frac{\rho \log x}{n}\right), \end{align*} where we have used the fact that $\sum_{n = 1}^\infty \frac{\mu(n)}{n} = 0$, and then the explicit formula for $\pi_0(x)$ in question follows provided that the sums over $n$ and $\rho$ above can be interchanged with one another. Thus, my question is: can the two sums in fact be interchanged with one another?

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    $\begingroup$ the formula is due to Riemann, you can find a proof at math.stackexchange.com/a/282848/87355 $\endgroup$ Commented Apr 6, 2022 at 20:27
  • $\begingroup$ The end of the MathWorld article at mathworld.wolfram.com/RiemannPrimeCountingFunction.html states "However, no proof of the equality of (12) appears to exist in the literature (Borwein et al. 2000)", so the status seems somewhat controversial. $\endgroup$ Commented Apr 7, 2022 at 1:41
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    $\begingroup$ See empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/borwein1.pdf page 4 for the claim. $\endgroup$
    – Charles
    Commented Apr 7, 2022 at 2:57
  • $\begingroup$ @StevenClark. The Mathworld article claims it's true summing over the nontrivial zeros, but I think you have to sum over all zeros, including the real ones. $\endgroup$ Commented Apr 9, 2022 at 8:05
  • $\begingroup$ @CarloBeenakker. The formula isn't proved there. All it says is that it follows from (9) by Mobius inversion, but "with questionable convergence". $\endgroup$ Commented Apr 9, 2022 at 8:06

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