What is $\sum_{3\leq p\leq x} \pi(\sqrt{p})$?

Question

What is the $\displaystyle \sum_{3\leq p\leq x} \pi(\sqrt{p})$? I thought about starting from $\displaystyle 2\sum_{3\leq p\leq x}\frac{\sqrt{p}}{\log p}$.

Answer 1

You may want to look at Abel's lemma. For a sequence of complex numbers $\{a_n\}$ and $f(t)$ a differentiable function for $t\geq 0$, $$\sum_{n\leq x} a_nf(x) = A(x)f(x) - \int_{1}^x A(t)f'(t)dt$$
where $A(t) = \sum_{n\leq x} a_n$. In your case, $a_n = 1$ for all $n$ (so, $A(x) = [x]$ where [$\cdot$] is the greatest integer function of $x$), and $f(t) = \frac{2\sqrt{t}}{\log(t)}$

Of course, this only gives the Asymptotic behavior, because $\pi(t)$ itself is not differentiable for all $t\geq 0$

Edit: Since we know that this is only going to give us the asymptotic behavior anyway (based on our approximation of $\pi$ by $f$) you might as well take $A(x) = x$ to simplify the integral computation.

Edit 2: In this case, your $A(x) = x$ if you are looking for the upper bound, you can bound the second ugly integral by $$-\int_1^x \frac{t(\log(t)-2)}{2\sqrt{t}\log^2(t)} \leq -\int_{1}^x t\frac{(\log(1)-2)}{t\sqrt{t}}dt$$ Because for $t \geq 1$, $\log(t) \geq \log(1)$ and $\frac{1}{2\sqrt{t}\log^2(t)} \geq \frac{1}{t\sqrt{t}}$ as this plot shows

Edit 3: $-\int_{1}^x t\frac{(\log(1)-2)}{t\sqrt{t}}dt = \int_{1}^x \frac{(2)}{\sqrt{t}}dt = \int_{1}^x 2t^{-1/2} = 4t^{1/2}]_1^x = 4\sqrt{x} - 4$

So, by Abel, $\sum_{n\leq x} a_nf(x) = x\frac{2\sqrt{x}}{\log(x)} + 4\sqrt{x} - 4$