To show a function is bounded by a function when x is large

Question

I want to show $\sum_{p^a\leq x}\log p = O(\sqrt{x}\log^2 x)$,where sum runs over $a\geq2$.

I only know that $\sum_{\sqrt{x}<p \leq x}\log p \leq 2x\log x$.

I tried using above property but I am not getting anything interesting which can help.

Answer 1

It's clear that the largest $\log p$ can be with $p \leq x$ is $\log x$, so we get that $$ \sum_{p^a \leq x} \log p \ll \log x \sum_{p^a \leq x} 1. \tag{1}$$ Now we just need to count the number of prime powers up to $x$.

By the Prime Number Theorem, there are $O(x / \log x)$ primes up to $x$. You are restricting to $a \geq 2$, so we use the equivalent fact that there are $O(\sqrt x / \log x)$ primes up to $\sqrt x$.

The largest power of $a$ that appears is from $2$, which occurrs $\log_2(x) = O(\log x)$ many times.

Together, this means that there are at most $O(\sqrt x / \log x)$ primes satisfying $p^2 \leq x$, and each occurs at most $O(\log x)$ times (from powers of $a$). Thus there are at most $O(\sqrt x)$ primes satisfying $p^a \leq x$ with $a \geq 2$.

Inserting into the estimate in $(1)$, we see that

$$ \sum_{p^a \leq x} \log p \ll x \log x,$$

which is slightly stronger than you needed.