# Does this explicit formula for the prime-counting function $\pi(x)$ converge?

This question is related to an answer I posted earlier at the following link.

Explicit Formula for $$\pi(x)$$

A potential explicit formula for the fundamental prime-counting function $$\pi(x)$$ is typically discussed in terms of Riemann's explicit formula for the prime-power counting function $$\Pi(x)$$ and the relationship between $$\pi(x)$$ and $$\Pi(x)$$, but I've investigated several explicit formulas analogous to (1) below where $$\nu(k)$$ is the number of primes in the factorization of $$k$$ and $$M_o(x)$$ is the explicit formula for the Mertens function illustrated in (2) below.

(1) $$\quad\pi_o(x)=\sum\limits_{k=1}^K\nu(k)\,M_o\left(\frac{x}{k}\right)$$

(2) $$\quad M_o(x)=-2+\sum\limits_\rho\frac{x^{\rho}}{\rho\,\zeta'\left(\rho\right)}+\sum\limits_n\frac{x^{-2\,n}}{(-2\,n)\,\zeta'(-2 n)}$$

The following figure illustrates $$\pi(x)$$ in blue and formula (1) for $$\pi_o(x)$$ in orange where formula (1) is evaluated with an upper limit of $$K=x$$ and formula (2) is evaluated over the first $$200$$ trivial zeta-zeros and non-trivial zeta-zero pairs. The red discrete portion of the plot illustrates the evaluation of formula (1) at integer values of $$x$$.

Figure (1): Illustration of formula (1) for $$\pi_o(x)$$ (orange) evaluated with an upper limit $$K=x$$

I've noticed the spikes exhibited in formula (1) above near integer values of $$x$$ can be eliminated by evaluating formula (1) with an upper limit of $$K=x+1$$ instead of $$K=x$$ as illustrated in Figure (2) below.

Figure (2): Illustration of formula (1) for $$\pi_o(x)$$ (orange) evaluated with an upper limit $$K=x+1$$

Question (1): Does formula (1) truly converge when evaluated with an upper limit of $$K=x$$ and/or $$K=x+1$$ as the number of trivial zeta zeros and non-trivial zeta-zero pairs evaluated in formula (2) increases towards $$\infty$$?

The upper evaluation limit $$K$$ in formula (1) above for $$\pi_o(x)$$ can't be increased to be arbitrarily larger than $$x$$ because the explicit formula for $$M_o(x)$$ defined in (2) above doesn't converge all the way down to $$x=0$$ which is illustrated in the following figure where formula (2) for $$M_o(x)$$ is evaluated over the first $$200$$ trivial zeta-zeros and non-trivial zeta-zero pairs (orange curve). The underlying blue reference function is $$M(x)$$, and the red discrete portion of the figure illustrates the evaluation of formula (2) at integer values of $$x$$.

Figure (3): Illustration of formula (2) for $$M_o(x)$$ (orange)

Question (2): What is the lower convergence bound of formula (2) for $$M_o(x)$$? In other words, if $$M_o(x)$$ converges for $$x>a$$, what is $$a$$?

Additional explicit formulas analogous to (1) above are illustrated at the following links. Some, but not all, of these explicit formulas exhibit spikes near integer values of $$x$$ when evaluated with an upper limit of $$x$$ analogous to the spikes illustrated in Figure (1) above, and I believe these spikes can often be eliminated by evaluating these explicit formulas with an upper limit of $$x+1$$ instead of $$x$$ analogous to Figure (2) above.

Explicit Formula for $$\lfloor x\rfloor$$

Explicit Formulas for Divisor Summatory Functions

Explicit Formula for Divisor Summatory Function Derived from Explicit Formula for $$\psi(x)$$

Figures (4) and (5) below illustrate formula (1) for $$\pi_o(x)$$ evaluated with an upper limit $$K=x$$ and $$K=x+1$$ respectively (orange curves) where the underlying explicit formula (2) for $$M_o(x)$$ is evaluated over the first $$200$$ trivial zeta-zeros and non-trivial zeta-zero pairs. Note the discontinuity at $$x=6$$ in Figure (4) below seems to be eliminated in Figure (5) below.

Figure (4): Illustration of formula (1) for $$\pi_o(x)$$ (orange) evaluated with an upper limit $$K=x$$

Figure (5): Illustration of formula (1) for $$\pi_o(x)$$ (orange) evaluated with an upper limit $$K=x+1$$

• Why do you care of what happens at the discontinuities ? The question is if it converges in $L^1_{loc}$ and uniformly on $x < A, |x-n| > c$. See en.wikipedia.org/wiki/Gibbs_phenomenon – reuns Apr 23 at 1:05
• @reuns Are you referring to the discontinuities of $\pi(x)$ or the discontinuities of $M(x)$? For example, note the spike in Figure (1) at $x=6$ where $\pi(x)$ has no discontinuity. But $M(x)$ has discontinuities for $x\in \{1,2,3,6\}$ which are all divisors of $6$, so I can see where the Gibbs phenomenon of the underlying explicit formula for $M(x)$ is perhaps contributing to the spike at $x=6$ in the derived formula for $\pi(x)$. – Steven Clark Apr 24 at 18:08
• @reuns But I don't believe the spikes are totally related to the Gibbs phenomenon because I believe the piece-wise evaluation of the derived formula (where the evaluation limit is changed at integer values of $x$) also introduces discontinuities at integer values of $x$. My follow-on question at math.stackexchange.com/q/3200866 addresses this issue. – Steven Clark Apr 24 at 18:10
• The Gibbs phenomenom occurs at $M(n)$. You asked 30 times the same questions and I don't think you care of the answers. There are many books about the proof of the explicit formula for $\psi(x)$, follow them. – reuns Apr 24 at 18:26
• @reuns I don't know where you get 30 times as I believe this is the first time I asked a question about the particular approach illustrated in this question. The primary source of the spikes illustrated in Figure (1) above is discontinuities in the approach defined here at integer values of $x$, not the Gibbs phenomenon. My follow-on question is about a different approach that eliminates these discontinuities, and hence divergence at integer values of $x$ is limited to the Gibbs phenomenon. I think its valid to ask different questions about different approaches. – Steven Clark Apr 24 at 22:33