Asymptotic expansion of Riemann's prime counting function

Question

Riemann's prime counting function is given as

$$J(n)=\sum_{k=1}^{\infty}\frac{\mu(k)}{k}\operatorname{li}(n^{1/k})$$

the approximations

\begin{align} \operatorname{li}(n)\sim J(n)\tag{1}\\ \operatorname{li}(n)-\sqrt{n}/\log n\sim J(n)\tag{2}\\ (1-\sqrt{n}/\log n)\operatorname{li}(n+\sqrt{n}\log n)\sim J(n)\tag{3}\\ \end{align}

get increasingly closer. Could these be the first few terms of an asymptotic expansion of $J(n),$ or can $(3)$ not be taken any further?

Answer 1

To make it clear $\sum_{k \ge 1} \frac{\mu(k)}{k} li(x^{1/k})$ is not a prime counting function.

The prime counting functions are $$\psi(x) = \sum_{p^k \le x} \log p, \quad J(x) = \sum_{p^k \le x} \frac{1}{k} = \int_{2-\epsilon}^x \frac{\psi'(y)}{\log y} dy = \sum_{k \ge 1} \frac{\pi(x^{1/k})}{k}, \quad \pi(x) = \sum_{p \le x} 1$$ Whose Mellin transforms are $\frac{-1}{s}\frac{\zeta'(s)}{\zeta(s)},\frac{1}{s}\log \zeta(s),\frac{1}{s}P(s) $.

The (Riemann) explicit formulas are $$\psi(x) = x - \sum_\rho \frac{x^\rho}{\rho} - \sum_{m\ge 1} \frac{x^{-2m}}{-2m}-\log 2 \pi, \\J(x) = \text{Li}(x)- \sum_\rho \text{Li}(x^\rho) - \sum_{m\ge 1} \text{Li}(x^{-2m})-A, \\\pi(x) = R(x) - \sum_\rho R(x^\rho) - \sum_{m\ge 1} R(x^{-2m})-B,\\ \text{Li}(x) = \int_2^x \frac{dy}{\log y}, \qquad R(x ) = \sum_{k \ge 1} \frac{\mu(k)}{k} \text{Li}(x^{1/k})$$