What's the asymptotic distribution of $p^n$ (powers of primes)?

Question

We know by the prime number theorem that $\lim_{n\to\infty}\frac{\pi(n)}{n\,/\ln n} = 1$ An even better approximation is $\lim_{n\to\infty}\frac{\pi(n)}{\int_2^n\frac{1}{\ln t}\mathrm{d}t} = 1$.

Is there a similar formula that approximates the distribution of natural numbers of the form $p^n$ where $p$ is a prime? That is, an approximation of $$\pi'(x)=\left|\,\Pi'\cap\{1,\ldots,x\}\,\right| \qquad\mbox{ where }\quad \Pi'=\{p^n\;|\;p\mbox{ is prime}, n\in\mathbb{N}\}$$

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(There is already a question that asks if $\lim_{n\to\infty}\frac{\pi'(n)}{n}=0$, but I'm interested in a more precise approximation.)

Answer 1

The same asymptotics hold, and the intuitive explanation for that is that the number of prime powers up to $x$ are negligible in comparison to the number of primes up to $x$. In fact, the usual proofs of the prime number theorem actually prove the asymptotics for $\pi'$ first, and then one shows that these coincide with the asymptotics of $\pi$.

Since $\pi'(x)$ for a real $x$ counts the number of primes $p$ up to $x$, the number of prime squares $p^2$ up to $x$, the number of prime cubes $p^3$ up to $x$, and so on, we can write $$\pi'(x)=\pi(x)+\pi(x^{1/2})+\pi(x^{1/3})+\cdots+\pi(x^{1/m})$$ where $m=\log(x)+1$. Now, using the prime number theorem, we get $$\pi'(x)-\pi(x)\sim x^{1/2}/\log(x^{1/2})+\cdots+x^{1/m}/\log(x^{1/m})$$ which can be bounded above by $x^{1/2}/\log(x^{1/2})\cdot (\log x+1)\leq x^{1/2}/\log(x^{1/2})\cdot 2\log x=4 x^{1/2},$ and thus $\pi'(x)=\pi(x)+O(x^{1/2})$, so $\pi'(x)$ is asymptotic to $x/\log x$.