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

V.G
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A2011
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1 Answers1

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

Arun Bharadwaj
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  • Yes I just want the upper bound. I got stuck in the Abel Lemma. – A2011 Feb 15 '21 at 17:17
  • My previous comment was just straight up wrong, so I have deleted it. I have made an edit to the answer, the bound looks quite ugly but it is at least integrable now. – Arun Bharadwaj Feb 15 '21 at 17:47
  • Thanks Arun. But in your second integral it looks like the $t$ can be simplified. Is there anyway I can have a full answer? Thanks again – A2011 Feb 15 '21 at 17:56
  • Made edit 3, which should now be a full answer. Remember, in the version I gave you, $n$ starts at 1, and in your case, it starts at $3$. So be sure to subtract away the values of $\pi(\sqrt{n})$ for $n =1, 2$ from my answer. – Arun Bharadwaj Feb 15 '21 at 18:03
  • Thank you very much Arun. Yes I will subtract those from your answer. – A2011 Feb 15 '21 at 18:09
  • Of course! Happy to help. Just one last thing I wanted to mention is that this is only AN upper bound; and is by no means the the best one. Cleverer substitutions and simplification of the ugly integrand can give you better bounds. – Arun Bharadwaj Feb 15 '21 at 18:40
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    Note that the error committed by approximating the prime counting function will be $ \sum\limits_{p = 3}^x {\mathcal{O}!\left( {\frac{{\sqrt p }}{{\log ^2 p}}} \right)} = \mathcal{O}!\left( {\frac{{x^{3/2} }}{{\log ^2 x}}} \right)$. Thus, the final result should be $$\sum\limits_{p = 3}^x {\pi (\sqrt p )} = 2\frac{{x^{3/2} }}{{\log x}} + \mathcal{O}!\left( {\frac{{x^{3/2} }}{{\log ^2 x}}} \right).$$ – Gary Feb 16 '21 at 07:50
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    @Gary Thank you for the addition! I completely overlooked the error bound. – Arun Bharadwaj Feb 16 '21 at 18:21
  • Thanks to both of you Gary and Arun for your contributions. – A2011 Feb 17 '21 at 02:32