# What is the relationship between simple prime-power counting function and $\log\zeta(s)$?

This question assumes the following definitions of prime-counting functions where $$p$$ denotes a prime number, $$k$$ denotes a positive integer, and $$\theta(y)$$ is the Heaviside step function which takes a unit step at $$y=0$$.

(1) $$\quad\pi(x)=\sum\limits_{p}\theta(x-p)\,\qquad\quad\text{(fundamental prime counting function)}$$

(2) $$\quad\Pi(x)=\sum\limits_{n=p^k}\frac{1}{k}\,\theta(x-n)\,\quad\text{(Riemann's prime-power counting function)}$$

(3) $$\quad k(x)=\sum\limits_{n=p^k}\theta(x-n)\,\qquad\text{(simple prime-power counting function)}$$

The following plot illustrates $$\pi(x)$$, $$\Pi(x)$$, and $$k(x)$$ in blue, orange, and green respectively. Note that $$\pi(x)$$ takes a step of $$1$$ at each prime and $$k(x)$$ takes a step of $$1$$ at each prime-power. $$\Pi(x)$$ is more complicated in that it takes a step of $$\frac{1}{k}$$ at each prime-power $$p^k$$.

The $$\pi(x)$$ and $$\Pi(x)$$ functions defined above are related to $$\log\zeta(s)$$ as defined below.

(4) $$\quad\log\zeta(s)=s\int\limits_0^\infty\Pi(x)\,x^{-s-1}\,dx\,,\quad\Re(s)>1$$

(5) $$\quad\log\zeta(s)=s\int\limits_0^\infty\frac{\pi(x)}{x\,\left(x^s-1\right)}\,dx\,,\qquad\quad\Re(s)>1$$

Question: What is the relationship between $$k(x)$$ and $$\log\zeta(s)$$? More specifically, what is the definition of the function $$f(x)$$ consistent with (6) below?

(6) $$\quad\log\zeta(s)=s\int\limits_0^\infty k(x)\,f(x)\,dx\,,\qquad\quad\Re(s)>1$$

The following relationships between $$\pi(x)$$, $$\Pi(x)$$, and $$k(x)$$ may provide some insight where $$rad(n)$$ is the greatest square-free divisor of $$n$$ also referred to as the square-free kernel of $$n$$.

(7) $$\quad\Pi(x)=\sum\limits_{n=1}^{\log_2(x)}\frac{1}{n}\,\pi(x^{1/n})$$

(8) $$\quad\pi(x)=\sum\limits_{n=1}^{\log_2(x)}\frac{\mu(n)}{n}\,\Pi(x^{1/n})$$

(9) $$\quad k(x)=\sum\limits_{n=1}^{\log_2(x)}\pi(x^{1/n})$$

(10) $$\quad\pi(x)=\sum\limits_{n=1}^{\log_2(x)}\mu(n)\,k(x^{1/n})$$

(11) $$\quad k(x)=\sum\limits_{n=1}^{\log_2(x)}\frac{\phi(n)}{n}\,\Pi(x^{1/n})$$

(12) $$\quad\Pi(x)=\sum\limits_{n=1}^{\log_2(x)}\frac{\mu(rad(n))\,\phi(rad(n))}{n}\,k(x^{1/n})$$

• Comments are not for extended discussion; this conversation has been moved to chat. – Aloizio Macedo Jan 14 at 19:07