Let $S(x)=\sum_{p\le x,\; q\le x,\; pq\gt x}\frac{1}{pq}$, where p and q are primes. Find the limit of this function.

Question

Let

$$S(x)=\sum_{p\le x,\; q\le x,\; pq\gt x}\frac{1}{pq},$$

where $p$ and $q$ denote prime numbers. Show that as $x\to\infty$,$S(x)$ converges to a constant, and find the value of that constant.

First, I think I need to manipulate the sum to something more convenient to deal with, but I'm really struggling to make any progress with this problem. I'm trying to learn Analytic Number Theory on my own, and I would greatly appreciate it if anyone could show me some genius way to work through this.

Answer 1

As Eric Naslund's answer below shows, the calculation here in fact doesn't work. One wonders where the error is.

Roughly speaking, $\sum_{p \leq x} \frac{1}{p} \approx \log\log x$. Therefore $$ \sum_{p,q \leq x, pq > x} \frac{1}{pq} = \sum_{p \leq x} \frac{1}{p} \sum_{x/p < q \leq x} \frac{1}{q} \approx \sum_{p \leq x} \frac{1}{p} \left(\log\log x - \log\log \frac{x}{p}\right). $$ Using $(d/dt) \log t = 1/t$, $$ \log\log x - \log\log \frac{x}{p} = \log(\log x) - \log(\log x - \log p) \approx \frac{\log p}{\log x}, $$ and so $$ \sum_{p,q \leq x, pq > x} \frac{1}{pq} \approx \frac{1}{\log x} \sum_{p \leq x} \frac{\log p}{p}. $$ Again roughly speaking, $\sum_{p \leq x} \frac{\log p}{p} \approx \log x$, and we conclude that $$ \sum_{p,q \leq x, pq > x} \frac{1}{pq} \approx 1. $$

(In order to get the estimates on $\sum_{p \leq x} \frac{1}{p}$ and $\sum_{p \leq x} \frac{\log p}{p}$, use $\sum_{p \leq x} f(p) \approx \int^x \frac{f(p)}{\log p} \, dp$.)

Answer 2

In this answer I will prove that:

$$\sum_{\begin{array}{c} p\leq x,q\leq x\\ pq>x \end{array}}\frac{1}{pq}=\frac{\pi^{2}}{6}+O\left(\frac{1}{\log x}\right).$$

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To handle the error terms correctly we need to use the hyperbola method and split the sum into ranges according to the geometry of the hyperbola. Doing so we have that $$\sum_{\begin{array}{c} p\leq x,q\leq x\\ pq>x \end{array}}\frac{1}{pq}=2\sum_{\sqrt{x}<p\leq x}\frac{1}{p}\sum_{\frac{x}{p}<q\leq x}\frac{1}{q}-\left(\sum_{\sqrt{x}<p\leq x}\frac{1}{p}\right)^{2}.$$ Merten's second theorem says that $$\sum_{p\leq x}\frac{1}{p}=\log\log x+B+O\left(\frac{1}{\log x}\right),$$ where $B$ is Merten's constant, and so the second term equals $$\left(\sum_{\sqrt{x}<p\leq x}\frac{1}{p}\right)^{2}=\left(\log\log x-\log\log\sqrt{x}+O\left(1/\log\sqrt{x}\right)\right)=\left(\log2\right)^{2}+O\left(\frac{1}{\log x}\right).$$ Now, using Merten's theorem again, the first double sum equals $$\sum_{\sqrt{x}<p\leq x}\frac{1}{p}\sum_{\frac{x}{p}<q\leq x}\frac{1}{q}=\sum_{\sqrt{x}<p\leq x}\frac{1}{p}\left(\log\left(1-\frac{\log p}{\log x}\right)^{-1}+O\left(\frac{1}{\log\sqrt{x}}\right)\right) =\sum_{\sqrt{x}<p\leq x}\frac{1}{p}\log\left(1-\frac{\log p}{\log x}\right)^{-1}+O\left(\frac{1}{\log\sqrt{x}}\right).$$ Now converting to a Riemann Stieltjes integral we have that $$\sum_{\sqrt{x}<p\leq x}\frac{1}{p}\log\left(1-\frac{\log p}{\log x}\right)^{-1}=\int_{\sqrt{x}}^{x}\frac{1}{t}\log\left(1-\frac{\log t}{\log x}\right)^{-1}d\pi(t),$$ and by writing $\pi(x)=\int_{2}^{x}\frac{1}{\log t}dt+E(x)$, this equals $$\int_{\sqrt{x}}^{x}\frac{1}{t\log t}\log\left(1-\frac{\log t}{\log x}\right)^{-1}dt+\int_{\sqrt{x}}^{x}\frac{1}{t\log t}\log\left(1-\frac{\log t}{\log x}\right)^{-1}dE(t).$$ The second term above can be handled due to the Prime Number Theorem. Note: Here it is critical that our range is from $\sqrt{x}$ to $x$ in order to obtain cancellation from the PNT, as the integral from $2$ to $x$ of the above function equals a specific constant. This is why we split the sum earlier according to the hyperbola method. Now, letting $t=x^{u}$, our integral above equals $$-\int_{1/2}^{1}\frac{\log\left(1-u\right)}{u}du=\text{Li}_{2}(1)-\text{Li}_{2}\left(\frac{1}{2}\right)=\frac{\pi^{2}}{12}-\frac{1}{2}\left(\log(2)\right)^{2}$$ where $\text{Li}_{2}(x)=\sum_{k=0}^{\infty}\frac{z^{k}}{k^{2}}$ is the dilogarithm. Thus we obtain equation $(1)$ above.