Asymptotics of the sum $\sum_{n\leq x}\frac{\omega(n)}{n}$ as $x \to \infty$.

Question

If $\omega(n)$ denotes the number of distinct prime factors of $n$, it is known ( e.g. wikipedia) that

$$ \sum_{n\leq x}\omega(n)=x \log \log x+O(x) $$ as $x \to \infty$. Is it possible to use this formula to estimate the sum in question?

Thanks

Answer 1

Using Abel Summation Formula we have $\sum_{n\leq x}\frac{\omega(n)}{n}=\sum_{n\leq x} \omega(n).\frac{1}{x}+O(1)+\int_{1}^{x}\sum_{n \leq x}\omega(t)\frac{1}{t^2}dt = \\ \log\log(x)+O(1)+\int_{1}^{x}\frac{\log\log(t)}{t}dt+O(\int_{1}^{x}\frac{1}{t}dt) = \\ \log\log(x) + O(1) +\log(x)(\log\log(x)-1) +O(1)+O(log(x)) = \\\log\log(x)+\log(x)\log\log(x)+O(\log(x))$