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, http://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/encoding1.htm. 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?

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?

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, http://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/encoding1.htm. 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.

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, http://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/encoding1.htm. 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?