Two Representations of the Prime Counting Function 
The bounty  for the best work out of Greg's answer, especially the
"solving for $\pi^*(x;q,a)$ in terms of all $\Pi^*$ functions (tedious but possible)"
part is over. Since Raymond's contributions might be very helpful to recall the necessary math, upvoting his answers is highly appreciated...
I posted my attempt of a workout here. Comments welcome...


Original question:
I have two representations of $\pi(x)$:

*

*

The Prime Counting Function $\pi(x)$ is given
$$
\pi(x) = \operatorname{R}(x^1) - \sum_{\rho}\operatorname{R}(x^{\rho}) \tag{1}
$$
with $    \operatorname{R}(z) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} \operatorname{li}(z^{1/n})$ and $\rho$ running over all the zeros of $\zeta$ function.



*

This formula, while
widely believed to be correct, has not yet been proved.
$$
\pi(x) \approx \int\limits_2^x{\frac{dt}{\ln t}} - \frac{\sqrt x}{\ln x}\left( 1 + 2\sum_{\gamma} \ \ \frac{\sin(\gamma\ln x)}{\gamma}\right) \tag{2}, 
$$
with $\gamma=\text{Im}({\rho})$ being imaginary part of the roots  of the $\zeta$ function.

Now I have two questions:


*Does the truth of $(2)$ depend on Riemann's Hypothesis or is it "just" what Wikipedia says, that The amplitude of the "noisy" part is heuristically about $\sqrt x/\ln x$?

*How to show the equivalence between $(1)$ and $(2)$? The integral logarithm is easily found in both representations, but how do the $\rho$-parts fit together? How do I get $\sin$s from $\text{li}(z^{1/n})$s? Does this invoke Gram's series:
$$
    \operatorname{R}(z) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} \operatorname{li}(z^{1/n}) = 1 + \sum_{k=1}^\infty \frac{(\ln z)^k}{k! k \zeta(k+1)} ?
$$
We can rewrite $\displaystyle \frac{\sin(\gamma\ln x)}{\gamma}=\frac{x^{i\gamma}-x^{-i\gamma}}{i\gamma}$ and I remember that I've seen a similar expression at Wikipedia:
$$
    \psi_0(x) = x - \sum_\rho \frac{x^\rho}{\rho} - \ln 2\pi - \frac12 \ln(1-x^{-2}) ,
$$
But could this help, if at all? ($\psi_0(x)$ is the normalization of the Chebyshev function, see here)

 A: You probably know that
$$
\Pi^*(x;q,a) = \sum^*_{\substack{n\le x \\ n\equiv a\pmod q}} \frac{\Lambda(n)}{\log n} = \frac1{\phi(q)} \sum_{\chi\pmod q} \overline{\chi(a)}\Pi^*(x,\chi).
$$
I find that the easiest way to convert from counting prime powers (with weights) to counting only primes is to use the fact that
\begin{align*}
\Pi^*(x;q,a) &= \sum^*_{\substack{p\le x \\ p\equiv a\pmod q}} 1 + \sum^*_{\substack{p^2\le x \\ p^2\equiv a\pmod q}} \tfrac12 + \sum^*_{\substack{p^3\le x \\ p^3\equiv a\pmod q}} \tfrac13 + \cdots \\
&= \pi^*(x;q,a) + \tfrac12 \sum_{\substack{b\pmod q \\ b^2\equiv a\pmod q}} \pi^*(x^{1/2};q,b) + \tfrac13 \sum_{\substack{c\pmod q \\ c^3\equiv a\pmod q}} \pi^*(x^{1/3};q,c) + \cdots
\end{align*}
and solving for $\pi^*(x;q,a)$ in terms of all $\Pi^*$ functions (tedious but possible).
A: Let's start with the nice paper 'Prime Number Races' of Granville and Martin from your comments. It contains the sentence :  

"The precise formula he (Riemann) proposed is a bit too technical for
  this article, but we can get a good sense of it from the following
  approximation when x is large. This formula, while widely believed to
  be correct, has not yet been proved."

$$
\tag{1}\boxed{\displaystyle\frac{\int\limits_2^x{\frac{dt}{\ln t}}-\pi(x)}{\sqrt x/\ln x} \approx 1 + 2\sum_{\gamma} \ \ \frac{\sin(\gamma\ln x)}{\gamma}} 
$$
I can only conjecture that the formula referenced in their last sentence is your formula $(1)$ (I don't know a proof of convergence of this formula either) derived from Riemann's explicit formula (proved by von Mangoldt) and not the approximate formula I reproduce above equivalent to your $(2)$ . Note that they add that this approximation is valid 'when $x$ is large'.
Let's add that this approximation doesn't include the trivial zeros from your other thread.
Concerning the $\displaystyle \frac{\sin(\gamma\ln x)}{\gamma}$ term it is a good approximation of Riesel's approximation $(2.30)$ from his fine book 'Prime Numbers and Computer Methods for Factorization' :
$$\tag{2}\frac{\cos(\gamma\ln x-\arg \rho)}{|\rho|},\quad\text{with}\ \ \rho=\frac 12+i\gamma$$
since $\ \gamma\gg 1 $ we have indeed $\ |\rho|\approx \gamma\ $ and $\ \arg\,\rho\approx \frac{\pi}2$
(note that $\gamma > 14.13\ $ so that $\ 1<\frac {|\rho|}{\gamma}<1.00063$)
Riesel's method is to approximate the oscillating term (see the next page from Riesel's book) :
$$\tag{3}\operatorname{R}(x^{\rho}) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n}\operatorname{li}(x^{\rho/n})$$
with the first term :
$$\tag{4}\operatorname{R}(x^{\rho}) \approx\left[\operatorname{li}(x^{\rho})=\operatorname{li}(e^{\rho\;\ln x})=\operatorname{Ei}(\rho\;\ln x)\right]$$
and to use the asymptotic expansion for $\operatorname{Ei}$ :
$$\tag{5}\operatorname{Ei}(z)\sim \frac{e^z}{z}\left(1+\frac {1!}z+\frac{2!}{z^2}+\cdots\right)$$
to get :
$$\operatorname{Ei}\left(\left(\frac 12+i\gamma\right)\;\ln x\right)\approx \frac{x^{1/2+i\gamma}}{(1/2+i\gamma)\;\ln x}$$
so that combining the two 'mirror' zeros : 
$$\operatorname{Ei}\left(\left(\frac 12+i\gamma\right)+\;\ln x\right)+\operatorname{Ei}\left(\left(\frac 12-i\gamma\right)+\;\ln x\right)\approx \frac{\sqrt{x}}{\ln x}\left(\frac{e^{i\gamma\ln x}}{1/2+i\gamma}+\frac{e^{-i\gamma\ln x}}{1/2-i\gamma}\right)$$
$$\approx\frac{\sqrt{x}}{\ln x}\frac{e^{i\gamma\ln x-i\arg{\rho}}+e^{-i\gamma\ln x+i\arg{\rho}}}{|\rho|}\quad\text{since}\ \ \rho=\frac 12+i\gamma$$
we get Riesel's approximation :
$$\tag{6}\operatorname{R}(x^{\rho})+\operatorname{R}(x^{\overline{\rho}})\approx\frac{2\;\sqrt{x}\;\cos(\gamma\ln x-\arg{\rho})}{|\rho|\;\ln x}$$
while $\operatorname{R}(x)$ was approximated by (using $(3), (4), (5)$) :
$$\tag{7}\operatorname{R}(x)\approx \operatorname{li}(x)-\frac 12\operatorname{li}\bigl(x^{1/2}\bigr)\quad \text{with}\ \operatorname{li}\bigl(x^{1/2}\bigr)\approx \frac{2\sqrt{x}}{\ln x}$$
this allows too to get better approximations if wished ($\operatorname{R}(x)\approx \operatorname{li}(x)-\frac 12\operatorname{li}\bigl(x^{1/2}\bigr)-\frac 13\operatorname{li}\bigl(x^{1/3}\bigr) $ for example, adding the trivial zeros contribution and so on...)
Now to answer your questions :


*

*We have to suppose the R.H. to be able to write $\rho=\frac 12+i\gamma\ $ for any root $\rho$ so that the simplifications proposed apply (in practice we don't need R.H. since $(2)$ requires precomputed zeros and since all the zeros known satisfy the R.H. ... to this date...)

*The implication from $(1)\to (2)$ was provided without using the Gram series


Hoping this clarified things,
