Between Mertens' theorems

Question

It is well-known that $$ \sum_{p\le x}\frac{\log p}{p}=\log x+O(1) $$ and $$ \sum_{p\le x}\frac1p=\log\log x+M+o(1). $$

What is the order of $$ \sum_{p\le x}\frac{\sqrt{\log p}}{p} $$ ?

Answer 1

Let's calculate:

$$\begin{align} \sum_{p\leqslant x} \frac{\sqrt{\log p}}{p} &= \sum_{n \leqslant x} \frac{\sqrt{\log n}}{n}\left(\pi(n) - \pi(n-1)\right)\\ &= \sum_{n \leqslant x}\frac{\sqrt{\log n}}{n}\pi(n) - \sum_{n \leqslant x-1}\frac{\sqrt{\log (n+1)}}{n+1}\pi(n)\\ &= \sum_{n \leqslant x} \left(\frac{\sqrt{\log n}}{n} - \frac{\sqrt{\log (n+1)}}{n+1}\right)\pi(n) - \frac{\sqrt{\log x}}{x}\pi(x) + O(1)\\ &= \sum_{n\leqslant x} \frac{\sqrt{\log n} - n\left(\sqrt{\log (n+1)} - \sqrt{\log n}\right)}{n(n+1)}\pi(n) + O(1). \end{align}$$

The first $O(1)$ comes from replacing $\lfloor x\rfloor$ with $x$ for the term after the sum. Since $\frac{\sqrt{\log x}}{x}\pi(x) \sim \frac{1}{\sqrt{\log x}}$, we can absorb it in the $O(1)$ in the next line.

Now we split the sum, the second part is bounded, we have

$$n\left(\sqrt{\log (n+1)} - \sqrt{\log n}\right) = \frac{n(\log (n+1) - \log n)}{\sqrt{\log (n+1)} + \sqrt{\log n}} = \frac{n\log (1 + \frac1n)}{\sqrt{\log (n+1)} + \sqrt{\log n}} \approx \frac{1}{2\sqrt{\log n}},$$

and thus

$$\begin{align} \sum_{n\leqslant x} \frac{n\left(\sqrt{\log (n+1)} - \sqrt{\log n}\right)}{n(n+1)}\pi(n) &\leqslant \sum_{n\leqslant x} \frac{\pi(n)}{2n(n+1)\sqrt{\log n}}\\ &\leqslant \sum_{n\leqslant x} \frac{1}{(n+1)(\log n)^{3/2}}\\ &= O(1). \end{align}$$

So it remains to look at

$$\sum_{n \leqslant x} \frac{\sqrt{\log n}}{n(n+1)}\pi(n).$$

Using $\displaystyle\pi(x) = \frac{x}{\log x} + O\left(\frac{x}{(\log x)^2}\right)$, we observe that the second term again yields a sum of $1/\left(n(\log n)^{3/2}\right)$ and hence remains bounded. For the remaining part, we find

$$ \sum_{2 \leqslant n \leqslant x} \frac{\sqrt{\log n}}{n(n+1)}\frac{n}{\log n} = \sum_{2 \leqslant n \leqslant x} \frac{1}{(n+1)\sqrt{\log n}} $$

sandwiched:

$$\int_3^{x+1} \frac{dt}{(t+1)\sqrt{\log t}} \leqslant \sum_{2 \leqslant n \leqslant x} \frac{1}{(n+1)\sqrt{\log n}} \leqslant \int_2^{x} \frac{dt}{(t+1)\sqrt{\log t}}$$

Approximating,

$$\int_a^b \frac{dt}{t\sqrt{\log t}} = \int_{\log a}^{\log b} \frac{du}{\sqrt{u}} = 2\left(\sqrt{\log b} - \sqrt{\log a}\right),$$

with the difference to the original integrand being of the order $t^{-2}(\log t)^{-1}$ with a finite integral.

Altogether,

$$\sum_{p\leqslant x} \frac{\sqrt{\log p}}{p} = 2\sqrt{\log x} + O(1).$$