We have for $x>1$:
$$\tag{1}\pi^*(x)=R(x)-\sum_{\rho} R(x^{\rho})$$
$\rho$-s are zeros of Riemann zeta function (trivial and nontrivial). I know how to derive $(1)$ and also know how to derive: $$\tag{2}\psi^*(x)=x-\sum_{\rho} \frac {x^{\rho}}{\rho}-\frac{\zeta'(0)}{\zeta(0)}$$ This question concerns only about $(1)$ so please do not answer like "instead of $(1)$ look at $(2)$". (I know it is somehow easier in understanding and evaluation.)
Back to $(1)$...
$R(x)$ is Rieamann R-function and is defined as:
$$\tag{3}R(x)=1+\sum _{n=1}^{\infty } \frac{\big(\log(x)\big)^n}{nn! \zeta (n+1)}$$ Now, whether I use $$\tag{3.1}R(x^\rho)=1+\sum _{n=1}^{\infty } \frac{\big(\log(x^\rho)\big)^n}{nn! \zeta (n+1)}$$ or $$\tag{3.2}R(x^\rho)=1+\sum _{n=1}^{\infty } \frac{\big(\rho \log(x)\big)^n}{nn! \zeta (n+1)}$$ in $(1)$, in both cases I do not get $\pi^*(x)$. (Series seems to diverge/oscillate.)
Can you tell me what a definition of $R(x)$ to use to get the correct results? I am interested only in exact formulas, approximations are out of interest!
I have seen approximations using Möbius function $\mu(x)$ and exponential integral $Ei\big(\rho \log(x)\big)$.
There is the easiest way to obtain convergent explicit formulas :
Since $$\psi(x) =\sum_{p^k \le x} \log p=( x-\sum_\rho \frac{x^\rho}{\rho}-\log 2\pi)1_{x > 2}$$ (sum over trivial and non-trivial zeros)
Letting $\Pi(x) = \sum_{p^k \le x} \frac{1}{k}= \int_{2-\epsilon}^x \frac{\psi'(x)}{\log x}dx$ and $1_{x > 2}\int_2^x ( \frac{d}{dt}\frac{t^\rho}{\rho}) \frac{dx}{\log x} = 1_{x > 2}\int_2^x \frac{\rho t^{\rho-1}}{\rho\log t} dt=1_{x > 2}\int_{2^\rho}^{x^\rho} \frac{u}{\log u} du= (\text{li}(x^\rho)-\text{li}(2^\rho))1_{x > 2}$
then $$\Pi(x) = (\text{li}(x)-\text{li}(2))1_{x > 2}-\sum_\rho (\text{li}(x^\rho)-\text{li}(2^\rho))1_{x > 2}$$
Now let $\pi(x) =\sum_{n=1}^\infty \frac{\mu(n)}{n} \Pi(x^{1/n})=\sum_{n=1}^{\log_2(x)} \frac{\mu(n)}{n} \Pi(x^{1/n})1_{x^{1/n}> 2}$ and $\text{R}_\rho(x) = \sum_{n=1}^\infty \frac{\mu(n)}{n} (\text{li}(x^{\rho/n})-\text{li}(2^{\rho/n}))1_{x^{1/n} > 2}$ (those are finite sums)
therefore $$\pi(x) = \text{R}_1(x)-\sum_\rho \text{R}_\rho(x)$$