Questions related to Moebius Transform of Characteristic Function of the Primes Consider the function $f(x)$ defined in (2) below related to the fundamental prime counting function $\pi(x)$ defined in (1) below. Note $b(n)$ is the Möbius transform of $a(n)$.
(1) $\quad \pi(x)=\sum\limits_{n=1}^{x}a(n)\,,\quad a(n)=\begin{array}{cc}
 \{ & 
\begin{array}{cc}
 1 & n\in\mathbb{P} \\
 0 & n\notin\mathbb{P} \\
\end{array}
 \\
\end{array}\qquad$ (see https://oeis.org/A010051)
(2) $\quad f(x)=\sum\limits_{n=1}^{x}b(n)\,,\quad b(n)=\sum\limits_{d|n} a(d)\,\mu\left(\frac{n}{d}\right)\qquad\,\,$ (see https://oeis.org/A143519)

The following plot illustrates $f(x)$ defined in formula (2) above.


Figure (1): Illustration of $f(x)$ defined in formula (2)

The integer zeros of $f(x)$ for $x\le 10,000$ are listed in (3) below.
(3) $\quad${1,6,9,12,19,30,79,80,81,116,193,201,287,288,291,668,673,679,680,685,686,1109}

The zero crossings of $f(x)$ for $x\le 10,000$ where $f(x)$ doesn't settle at zero are listed in (4) below.
(4) $\quad${14,21,33,114,115,118,195,286,290,295,442,445,665,667,670,671,678,682}

Question (1): Does $f(x)$ have a finite number of integer zeros, and if so what is the largest integer zero of $f(x)$?
Question (2): Does $f(x)$ have an finite number of zero crossings, and if so what is the largest zero crossing of $f(x)$?
Question (3): What is the asymptotic for the long term growth of $f(x)$? What are the associated error bounds predicted by the Prime Number Theorem and the Riemann Hypothesis?

The Dirichlet transforms of $a(n)$ and $b(n)$ are defined in (5) and (6) below where $P(s)$ is the prime zeta function.
(5) $\quad\sum\limits_{n=1}^\infty\frac{a(n)}{n^s}=P(s),\quad\Re(s)>1$
(6) $\quad\sum\limits_{n=1}^\infty\frac{b(n)}{n^s}=\frac{P(s)}{\zeta(s)},\quad\Re(s)>1$

The following figure illustrates the Dirichlet series for $\frac{P(s)}{\zeta(s)}$ defined in (6) above in orange where formula (6) is evaluated over the first $10,000$ terms. The underlying blue reference function is $\frac{P(s)}{\zeta(s)}$.


Figure (2): Illustration of formula (6) for $\frac{P(s)}{\zeta(s)}$ (orange curve) and reference function (blue curve)

The following four figures illustrate formula (6) for $\frac{P(s)}{\zeta(s)}$ evaluated along the line $s=1+i\,t$ in orange where formula (6) is evaluated over the first $1,000$ terms. The underlying blue reference function is $\frac{P(s)}{\zeta(s)}$. The red discrete portions of the plots illustrate the evaluation of formula (6) for $\frac{P(1+i\,t)}{\zeta(1+i\,t)}$ where $t$ equals the imaginary part of a non-trivial zeta zero.


Figure (3): Illustration of formula (6) for $\left|\frac{P(1+i\,t)}{\zeta(1+i\,t)}\right|$


Figure (4): Illustration of formula (6) for $\Re\left(\frac{P(1+i\,t)}{\zeta(1+i\,t)}\right)$


Figure (5): Illustration of formula (6) for $\Im\left(\frac{P(1+i\,t)}{\zeta(1+i\,t)}\right)$


Figure (6): Illustration of formula (6) for $Arg\left(\frac{P(1+i\,t)}{\zeta(1+i\,t)}\right)$

Question (4): What is the range of convergence of the Dirichlet series for $\frac{P(s)}{\zeta(s)}$ defined in (6) above? Does it converge only for $\Re(s)>1$, or does it also converge for $\Re(s)=1\land\Im(s)\ne 0$?

Note $\frac{P(s)}{\zeta(s)}$ has a pole at each non-trivial zeta zero.

Question (5): Are there explicit formulas for $f(x)$ and $\frac{P(s)}{\zeta(s)}$ expressed in terms of the non-trivial zeta zeros?
 A: Proof of (4): We have that $f(x) := \sum_{n \leq x} f_0(n)$ is the summatory function of the sequence $f_0(n) := \sum_{d|n} \mu(n/d) (\pi(d) - \pi(d-1))$. So the Dirichlet series of $f_0$ is given by 
$$D_{f_0}(s) = \sum_{n \geq 1} \frac{f_0(n)}{n^s} = \frac{1}{\zeta(s)} \times \sum_{p\mathrm{\ prime}} p^{-s} = \frac{P(s)}{\zeta(s)},$$ where $P(s)$ is the prime zeta function. 
Partial solution to (5): Now by Perron's formula, we obtain that for integers $x \geq 1$, 
$$f(x) - \frac{1}{2}f_0(x) = {\sum_{n \leq x}}^{\prime} f_0(n) = \frac{1}{2\pi\imath} \int_{c-\imath\infty}^{c+\imath\infty} D_{f_0}(s) \frac{x^s}{s} ds$$, where $c > \sigma_{f_0}$ makes the Dirichlet series absolutely convergent. The integrand of the previous equation has poles at 1) $s := 0$; 2) $s := \frac{1}{k}$ for all integers $k \geq 1$ (by the representation of $P(s)$ by a series over $\log \zeta(s)$); and 3) at all of the zeros of $\zeta(s)$. Thus, if we were able to apply the residue theorem (see my request for an answer), AND there were only singularities of the integrand at the negative even integers (corresponding to the trivial zeros of $\zeta(s)$), then we might get a formula that looks something like this:
$$\require{cancel}{\sum_{n \leq x}}^{\prime} f_0(n) = -\lim_{s \rightarrow 0} 2 P(s) + \sum_{\rho} \frac{P(\rho) x^{\rho}}{\rho \cdot \zeta^{\prime}(\rho)} - \sum_{n \geq 1} \cancel{\frac{P^{\ast}(-2n)}{2n \cdot x^{2n} \zeta^{\prime}(-2n)}} + \sum_{k \geq 1} \operatorname{Res}_{s=\frac{1}{k}}\left[\frac{P(s)}{\zeta(s)}\right].$$
