Relationship between $\pi (x)$ and $\pi _0 (x)$

Question

Let $p\in\mathbb{P}$, $\pi (x)=\sum_{p\le x}1$ and $\pi _0 (x)=\sum_{p\lt x}1$. Then $$\pi _0 (x)=\sum_{n\gt 0}\frac{\mu (n)}{n}J_0 \left(x^{\frac{1}{n}}\right)$$ and $$\begin{align}J_0 (x)&=\operatorname{li}x-\sum_{\rho}\operatorname{li}x^{\rho}-\ln 2+\int_{x}^{\infty} \frac{dt}{t\left(t^2-1\right)\ln t}\\&=\frac{1}{2}\left(\sum_{p^n\lt x}\frac{1}{n}+\sum_{p^n\le x}\frac{1}{n}\right).\end{align}$$ What is the relationship between $\pi (x)$ and $\pi _0 (x)$? Both $\pi (x)$ and $\pi _0 (x)$ map only to non-negative integers. So I think it's certainly not the case that $$\pi _0 (x)=\begin{cases}\pi (x) &\mbox{ }\text{for}\, x\, \text{composite}\\ \pi (x)-\frac{1}{2} &\mbox{ }\text{for}\, x\, \text{prime}\end{cases}$$ as this would imply that two integers can differ by $\frac{1}{2}$.

Answer 1

The prime counting $\pi(x)$ and Riemanns prime-power counting function $J(x)$ are defined as follows where $p$ is a prime and $k$ is a positive integer.

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(1) $\quad\pi(x)=\sum\limits_{p\le x} 1$

(2) $\quad J(x)=\sum\limits_{p^k\le x}\frac{1}{k}$

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The relationships between $\pi(x)$ and $J(x)$ are as follows. The sums over $n$ below can be terminated at the finite upper evaluation limit $n=\lfloor\log_2(x)\rfloor$ since $\pi(x)=J(x)=0$ for $x<2$.

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(3) $\quad J(x)=\sum\limits_{n=1}^\infty\frac{1}{n}\ \pi\left(x^{1/n}\right)=\sum\limits_{n=1}^{\lfloor\log_2(x)\rfloor}\frac{1}{n}\ \pi\left(x^{1/n}\right)$

(4) $\quad \pi(x)=\sum\limits_{n=1}^\infty\frac{\mu(n)}{n}\ J\left(x^{1/n}\right)=\sum\limits_{n=1}^{\lfloor\log_2(x)\rfloor}\frac{\mu(n)}{n}\ J\left(x^{1/n}\right)$

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Riemann's explicit formula for the prime-power counting function $J(x)$ needs to be evaluated as shown in (5) or (6) below. Formula (6) below evaluates faster than formula (5) below when using a modest finite upper evaluation limit $M$ for the sum over the trivial zeta zeros. Both formulas below are valid for all $x>1$.

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(5) $\quad J_o(x)=li(x)-\underset{T\to\infty}{\text{lim}}\left(\sum\limits_{|\Im(\rho)|\le T} Ei(\log(x)\,\rho)\right)-\log(2)+\int_x^\infty\frac{1}{t\,\left(t^2-1\right)\log(t)}\,dt$

(6) $\quad J_o(x)=li(x)-\underset{T\to\infty}{\text{lim}}\left(\sum\limits_{|\Im(\rho)|\le T} Ei(\log(x)\,\rho)\right)-\log(2)-\underset{M\to\infty}{\text{lim}}\left(\sum\limits_{m=1}^M Ei(-2\,m\,\log(x))\right)$

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When evaluated properly, $J_o(x)=\underset{\epsilon\to 0}{\text{lim}}\frac{J(x-\epsilon)+J(x+\epsilon)}{2}$ for all $x>1$.

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The following figure illustrates $J_o(x)$ defined in (6) above in orange overlaid on the reference function $J(x)$ in blue where formula (6) is evaluated over the first $100$ pairs of non-trivial zeta zeros and first $100$ trivial zeta zeros. The red discrete evaluation points illustrate the evaluation of formula (6) for $J_o(x)$ at the first 7 prime powers.

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Figure (1): Illustration of formula (6) for $J_o(x)$

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Relationship (4) above leads to the following explicit formula for $\pi(x)$ which is valid for all $x>1$.

(7) $\quad\pi_o(x)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{\mu(n)}{n}\ J_o\left(x^{1/n}\right)\right)$

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When evaluated properly, $\pi_o(x)=\underset{\epsilon\to 0}{\text{lim}}\frac{\pi(x-\epsilon)+\pi(x+\epsilon)}{2}$ for all $x>1$ which is equivalent to $\pi_o(x)=\pi(x)-\frac{1}{2}$ when $x$ is a prime-power and $\pi_o(x)=\pi(x)$ when $x>1$ and $x$ is not a prime-power.

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The following figure illustrates formula (7) for $\pi_o(x)$ in orange overlaid on the reference function $\pi(x)$ in blue where formula (6) for $J_o(x)$ is evaluated over the first $100$ pairs of non-trivial zeta zeros and first $100$ trivial zeta zeros and the sum for $\pi_o(x)$ in formula (7) is evaluated at $N=\lfloor \log_2(12)\rfloor=3$. The red discrete evaluation points illustrate the evaluation of the formula (7) for $\pi_o(x)$ at the first 5 prime integers.

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Figure (2): Illustration of formula (7) for $\pi_o(x)$