# Asymptotic expansion of Riemann's prime counting function

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?

• Small typo in the definition of prime counting function: $li(n^{1/k})$. Sep 3, 2017 at 15:15
• In what sense is the question not answered in your MathWorld link? They write it as the "Gram Series". Sep 3, 2017 at 18:03
• @IgorRivin I don't think the Gram series is the same as the form as given above - if it is, I can't see it - am I missing something? Sep 3, 2017 at 18:41
• It is not the same, but it IS an asymptotic expansion. Sep 3, 2017 at 18:50

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 $$\psi(x) = x - \sum_\rho \frac{x^\rho}{\rho} - \sum_{m\ge 1} \frac{x^{-2m}}{-2m}-\log 2 \pi, \\J(x) = \text{Li}(x)- \sum_\rho \text{Li}(x^\rho) - \sum_{m\ge 1} \text{Li}(x^{-2m})-A, \\\pi(x) = R(x) - \sum_\rho R(x^\rho) - \sum_{m\ge 1} R(x^{-2m})-B,\\ \text{Li}(x) = \int_2^x \frac{dy}{\log y}, \qquad R(x ) = \sum_{k \ge 1} \frac{\mu(k)}{k} \text{Li}(x^{1/k})$$
• @martin what is your question then ? Do you see why $\text{Li}(x)$ is easier than $\text{li}(x) = \int_0^x \text{pv.}(\frac{1}{\log y}) dy$ ? Sep 5, 2017 at 21:13