Estimation of sum of prime power product

Question

In the book 'Probabilistic Number Theory I Mean- Value Theorems' by P.D.T.A. Elliott the writer mentioned the following fact without proof. Fact: $$ \sum_{\substack{p\neq q\\ p^k q^l\leq x}} \!\! p^k q^l \ll \frac{x^2 \ln \ln x}{\ln x}. $$ Could someone please help me in proving this. Any help or hints would be appreciated. Thanks in advance.

Answer 1

The argument should go something like this I think, but I'm guessing something is wrong because I'm not using $p\not =q$.

From Chebyshev's upper bound for primes we have for any $X>1$

\[ \sum _{p^k\leq X}p^k\leq X\left (\sum _{p\leq X}1+\sum _{k\geq 2\atop {k\leq \log X/\log 2}}\sum _{p\leq X^{1/k}}\right )\ll \frac {X^2}{\log X}+X^{3/2}\log X.\]

In your sum, we must have that one of $p^k$ or $q^l$ is $\leq \sqrt x$, let's bound the ones with $q^l\leq \sqrt x$, i.e. $\log x/q^l\gg \log x$. These contribute to your sum

\[ \leq \sum _{q^l\leq x\atop {q^l\leq \sqrt x}}q^l\cdot \frac {x^2}{q^{2l}\log (x/q^{l})}\leq \frac {x^2}{\log x}\sum _{q^l\leq x}\frac {1}{q^l}.\]

It's known that the sum of reciprocals of primes is asymptotically $\log \log x$, so the above sum is

\[ \leq \sum _{q\leq x}\frac {1}{q}+\sum _{n=1}^\infty \frac {1}{n^2}\ll \log \log x.\]