0

Show there exists a $K > 0$ such that for all $x> K$ the interval $]x,2x]$ contains fewer primes than the interval $[0,x]$.

My approach so far has been to try to show $\pi(2x)-2\pi(x)$ is negative for all $x>K$ as this would be the desired conclusion. My idea is to apply the Prime Number Theorem, but so far I have not been able to do it in a fruitful way.

I would appreciate any help or hints for this problem. Thanks!

user114158
  • 680
  • 1
    Not that it helps, but the lowest such $K$ is $10$ – Henry Dec 14 '16 at 13:54
  • @Henry You can become a little more helpful if you provide a reference for that :) – Wojowu Dec 14 '16 at 14:13
  • I think that you need to use the lowest and highest bounds of the prime-counting function, i.e., prove that $H_{\pi(2x)}-2L_{\pi(x)}$ is negative. – barak manos Dec 14 '16 at 14:13
  • @Wojowu: the original question is obviously true with the difference tending to widen as $x$ increases, with "obvious" meaning "I will leave it to others to prove". My statement comes from looking at the differences for $x$ up to $50000$, by which time it is well over $600$ – Henry Dec 14 '16 at 14:26
  • See this answer for a more precise asymptotic: https://math.stackexchange.com/questions/1426666/a-particular-cases-of-second-hardy-littlewood-conjecture/1427143#1427143 – Peter Humphries Dec 14 '16 at 18:45

1 Answers1

2

Use the prime number theorem in the form $$\pi(y) = \int_2^y \frac{1}{\log t}dt + O_A(y (\log y)^{-A}).$$ We have $$2\pi(x) - \pi(2x) = \int_2^x \frac{1}{\log t}dt - \int_x^{2x}\frac{1}{\log t}dt + O_A(x (\log x)^{-A}),$$ and we easily see that $$\int_x^{2x}\frac{1}{\log t}dt \leq \frac{x}{\log x}.$$ If we now integrate by parts once, we get $$\int_2^x \frac{1}{\log t}dt = O(1) + \frac{x}{\log x} + \int_2^x \frac{1}{\log^2 t}dt,$$ so $$2\pi(x) - \pi(2x) \geq (1+o(1))\int_2^x \frac{1}{\log^2 t}dt \gg \frac{x}{\log^2 x}$$ as $x$ tends to infinity.

user397701
  • 683
  • Again not helping much, but in fact $2\pi(x) - \pi(2x) \geq \dfrac{x}{(\log_e x)^2}$ for $x \ge 101$ – Henry Dec 14 '16 at 14:08