This question is related to my previous question at the following link.
Does this explicit formula for the prime-counting function $\pi(x)$ converge?
My previous question utilized an explicit formula for Mertens function to derive an explicit formula for the prime-counting function $\pi(x)$. I haven't been able to find a definitive statement as to whether the explicit formula for Mertens function has been proven, and even if so it doesn't seem to be valid all the way down to $x=0$. This limits the number of terms that can be evaluated in the derived explicit formula for $\pi(x)$ requiring the derived formula to be evaluated piece-wise where the evaluation limit changes at integer values of $x$ resulting in discontinuities at integer values of $x$.
This motivated me to investigate an approach based on well-known proven explicit formulas which allow the evaluation limit for the derived explicit formula for $\pi(x)$ to be increased towards infinity. This allows a single evaluation limit to be used over an entire range of $x$ values under evaluation which avoids discontinuities associated with piece-wise evaluation resulting in a continuous function across the entire range of $x$ values under evaluation.
The approach outlined in this question can be applied to many functions of the form $f(x)=\sum_{n=1}^x a(n)$, but this question uses the prime-counting function $\pi(x)$ as an example of $f(x)$. Two explicit formulas for $\pi(x)$ are defined and illustrated. The first explicit formula for $\pi(x)$ is based on von Mangoldt's explicit formula for the second Chebyshev function $\psi(x)$, and the second explicit formula for $\pi(x)$ is based on Riemann's explicit formula for the prime-power counting function $\Pi(x)$.
This question assumes the following definitions. The offset of $6$ below was chosen to avoid a transition of $g_o(x)$ and $h_o(x)$ at $x=0$ which I believe would have adversely affected the convergence of the derived formulas for $\pi_o(x)$ in (10) and (18) below. The offset of $6$ also guarantees a non-zero value of $r(1)$ and $u(1)$ which is necessary in order to derive the Dirichlet inverses below.
(1) $\quad\pi(x)=\sum\limits_{n=1}^x a(n)\,,\quad a(n)=\begin{array}{cc} \{ & \begin{array}{cc} 1 & n\in \mathbb{P} \\ 0 & \text{True} \\ \end{array} \\ \end{array}$
(2) $\quad b(n)=\sum\limits_{d|n} a(d)\,\mu\left(\frac{n}{d}\right)\qquad\qquad\qquad\qquad\qquad\text{(Moebius Transform)}$
(3) $\quad \psi(x)=\sum\limits_{n=1}^x\Lambda(n)$
(4) $\quad g(x)=\psi(x+6)-\log(60)=\sum\limits_{n=1}^x\Lambda(n+6)$
(5) $\quad\psi_o(x)=x-\sum\limits_\rho\frac{x^\rho}{\rho}-\log(2\,\pi)+\sum\limits_n\frac{x^{-2\,n}}{2\,n}\\$ $\qquad\qquad\quad\,\,=x-\sum\limits_\rho\frac{x^\rho}{\rho}-\log(2\,\pi)-\frac{1}{2}\,\log\left(1-\frac{1}{x^2}\right)$
(6) $\quad g_o(x)=\psi_o(x+6)-\log(60)$
(7) $\quad r(n)=\sum\limits_{d|n}\Lambda(d+6)\,\mu\left(\frac{n}{d}\right)\qquad\qquad\qquad\qquad\text{(Moebius Transform)}$
(8) $\quad s(n)=\begin{array}{cc} \{ & \begin{array}{cc} \frac{1}{r(1)} & n=1 \\ \frac{1}{r(1)}\sum\limits_{d|n\land d<n} r\left(\frac{n}{d}\right)\,s(d) & \text{True} \\ \end{array} \\ \end{array}\quad\text{(Dirichlet Inverse)}$
(9) $\quad t(n)=\sum\limits_{d|n} s(d)\,b\left(\frac{n}{d}\right)\qquad\qquad\qquad\qquad\qquad\text{(Dirichlet Convolution)}$
(10) $\quad\pi_o(x)=\sum\limits_{j=1}^J t(j)\,g_o(\frac{x}{j})$
(11) $\quad \Pi(x)=\sum\limits_{n=2}^x\frac{\Lambda(n)}{\log(n)}$
(12) $\quad h(x)=\Pi(x+6)-\frac{7}{2}=\sum\limits_{n=1}^x\frac{\Lambda(n+6)}{\log(n+6)}$
(13) $\quad\Pi_o(x)=li(x)-\sum\limits_\rho Ei(\log\,(x)\,\rho)-\log(2)-\sum\limits_n Ei(-2\,n\,\log(x))\\$ $\qquad\qquad\qquad=li(x)-\sum\limits_\rho Ei(\log(x)\,\rho)-\log (2)+\int_x^\infty\frac{1}{t\,\left(t^2-1\right)\log(t)}\,dt$
(14) $\quad h_o(x)=\Pi_o(x+6)-\frac{7}{2}$
(15) $\quad u(n)=\sum\limits_{d|n}\frac{\Lambda(d+6)}{\log(d+6)}\,\mu\left(\frac{n}{d}\right)\qquad\qquad\qquad\qquad\quad\text{(Moebius Transform)}$
(16) $\quad v(n)=\begin{array}{cc} \{ & \begin{array}{cc} \frac{1}{u(1)} & n=1 \\ \frac{1}{u(1)}\sum\limits_{d|n\land d<n} u\left(\frac{n}{d}\right)\,v(d) & \text{True} \\ \end{array} \\ \end{array}\quad\text{(Dirichlet Inverse)}$
(17) $\quad w(n)=\sum\limits_{d|n} v(d)\,b\left(\frac{n}{d}\right)\qquad\qquad\qquad\qquad\qquad\text{(Dirichlet Convolution)}$
(18) $\quad\pi_o(x)=\sum\limits_{j=1}^J w(j)\,h_o(\frac{x}{j})$
The following two figures illustrate the formulas for $\pi_o(x)$ defined in (10) and (18) above (orange curves) where the formulas for $\pi_o(x)$ are evaluated with an upper limit of $J=25$ and the underlying explicit formulas for $\psi_o(x)$ and $\Pi_o(x)$ defined in (5) and (13) above are both evaluated over the first $1,000$ pairs of non-trivial zeta-zero pairs. The underlying blue reference function corresponds to $\pi(x)$ defined in (1) above. The red discrete portions of the plots in the figures below represent the evaluation of the formulas for $\pi_o(x)$ at integer values of $x$.
Figure(1): Illustration of Formula (10) for $\pi_o(x)$ (orange curve)
Figure(2): Illustration of Formula (18) for $\pi_o(x)$ (orange curve)
Question (1): Do the formulas for $\pi_o(x)$ defined in (10) and (18) above truly converge as $J\to\infty$ and as the number of non-trivial zeta-zero pairs evaluated in the underlying explicit formulas is increased towards $\infty$?
The following three figures illustrate formulas (10) and (18) above for $\pi_o(x)$ eliminate the discontinuities of the previous formula for $\pi_o(x)$ (which was derived from an explicit formula for the Mertens function $M(x)$) at integer values of $x$. Formulas (10) and (18) above are evaluated at an evaluation limit of $J=10$, and the previous formula is evaluated at an evaluation limit of $K=x$ (see formula (1) here). All underlying explicit formulas are evaluated over over the first $200$ pairs of non-trivial zeta zeros. The explicit formulas are illustrated in orange, and the reference function $\pi(x)$ is illustrated in blue.
Figure (3): Illustration of Previous Formula for $\pi_o(x)$ Derived from $M_o(x)$
Figure (4): Illustration of Formula (10) for $\pi_o(x)$ Derived from $\psi_o(x)$
Figure (5): Illustration of Formula (18) for $\pi_o(x)$ Derived from $\Pi_o(x)$