Show that $\sum_{pq\leq x}\frac{1}{pq}$ = $(\ln \ln x)^2+O(\ln \ln x)$

Question

I know that $\sum_{pq\leq x}\frac{1}{pq}$=$\sum_{p\leq x}\frac{1}{p}\sum_{q\leq\frac{x}{p}} \frac{1}{q}$=$\sum_{p\leq x}\frac{1}{p}(\ln\ln(\frac{x}{p})+A+O(\frac{1}{\ln (\frac{x}{p})}))$. However, I'm stuck at this point. Any ideas?

Answer 1

$$\sum_{p \le x} \frac{1}{p} = \ln \ln x + O(1), \qquad \sum_{p \le x} \frac{1}{p}\frac{\ln p}{\ln x} = O( \ln\ln x)$$

$$\ln \ln (x/p) =\ln ((1-\frac{\ln p}{\ln x})\ln x)= \ln\ln x+O(\frac{\ln p}{\ln x})$$

$$\sum_{p \le x}\sum_{x/p<q\le x} \frac{1}{pq} = \sum_{p \le x} \frac{1}{p} (\ln \ln x - \ln\ln (x/p)+O(1)) = O(\sum_{p \le x} \frac{1}{p})+O(\sum_{p \le x} \frac{1}{p} \frac{\ln p}{\ln x})= O(\ln \ln x)$$

$$\sum_{p q\le x} \frac{1}{pq} =\sum_{p \le x}\sum_{ q \le x} \frac{1}{pq}-\sum_{p \le x}\sum_{ x/p<q\le x} \frac{1}{pq} = (\sum_{p\le x} \frac{1}{p})^2+O(\ln \ln x)$$

Answer 2

Here's a non-standard way to show it:

By Merten's estimates:

$$\sum_{pq \leq x} \frac{1}{pq} \leq \left( \sum_{p \leq x} \frac{1}{p} \right)^2 = (\ln \ln x)^2 + O(\ln \ln x).$$

We just have to show that the constant in the leading term is $1$. That is, for every $c < 1$, there is $x$ sufficiently large so that:

$$\sum_{pq \leq x} \frac{1}{pq} \nleq c (\ln \ln x)^2 + O(\ln \ln x).$$

By Merten's estimates again,

$$\left(\ln \ln \sqrt{ x } + A + O\left(\frac{1}{\ln \sqrt{ x } }\right)\right)^2 = \left( \sum_{p \leq \sqrt{x}} \frac{1}{p} \right)^2 \leq \sum_{pq \leq x} \frac{1}{pq}.$$

However, the following is true: for any $c < 1$, there is a sufficiently large $x$ such that: $$ \ln \ln \sqrt{ x } = \ln \frac{1}{2} \ln x \geq c \ln \ln x. $$ To see this, we replace the inner $\ln x$ with $x$. $$ \ln \frac{1}{2} x \geq c \ln x \iff \frac{1}{2} x \geq x^c \iff \frac{1}{2}\geq \frac{1}{x^{1-c}}. $$ As $c < 1$, the last inequality is clearly true for large $x$. So the constant of the leading term is at least $1$.