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In Hardy & Wright's Theory of Numbers (p. 494f in 6th ed.) there's a little discussion following the proof of the prime number theorem.

We have $$ \pi(2x) - \pi(x) = \frac{x}{\log x} + o\left(\frac{x}{\log x}\right) \sim \pi(x). \tag{1} $$ Thus, to a first approximation, the number of primes between $x$ and $2x$ is the same as the number less than $x$. At first sight this is surprising, since we know that primes near $x$ 'thin out' (in some vague sense) as $x$ increases. In fact, $\pi(2x) - 2\pi(x) \to \infty$ as $x \to \infty$ (though we cannot prove this here), but this is not inconsistent with (1), which is equivalent to $$ \pi(2x) - 2\pi(x) = O(\pi(x)). \tag{2} $$

Isn't this just plain wrong? First of all, (1) is not equivalent to (2) but rather to $$ \pi(2x) - 2\pi(x) = o(\pi(x)). \tag{2'}$$ More importantly, how can $\pi(2x) - 2\pi(x)$ go to infinity if $\pi(2x) < 2\pi(x)$ for $x \ge 11$?

Thus the question is, as $x \to \infty$, what is $\pi(2x) - 2\pi(x)$ actually doing?

Community
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Unit
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    See my answer here: https://math.stackexchange.com/questions/1426666/a-particular-cases-of-second-hardy-littlewood-conjecture/1427143#1427143 – Peter Humphries Jul 15 '16 at 20:15
  • @Peter, if you post even just the asymptotic formula and a link to your proof, I will gladly accept the answer. – Unit Jul 22 '16 at 18:31

2 Answers2

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One can show via the prime number theorem that \[2\pi(x) - \pi(2x) \sim 2 \log 2 \frac{x}{(\log x)^2},\] so that $2\pi(x) \geq \pi(2x)$ for all sufficiently large $x$. See this answer.

Peter Humphries
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It is known that for any $\varepsilon > 0$ there is a point $x_0$ such that $$\frac{x}{\log x - 1 + \varepsilon} < \pi(x) < \frac{x}{\log x - 1 - \varepsilon}, \quad x \ge x_0$$ (e.g. Rosser (1941) Theorem 29A gives $x_0 = 55$ for $\varepsilon = 3$).

From this we deduce the upper bound $$\pi(2x) - 2\pi(x) < \frac{2x(2\varepsilon - \log 2)}{(\log 2x - 1 - \varepsilon)(\log x - 1 + \varepsilon)}$$ which holds eventually for any positive $\varepsilon$. Fixing $\varepsilon = \frac{1}{3} < \frac{1}{2} \log 2$ (because $e^2 < 2.75^2 = 8 - \frac{7}{16}$) we see that $\lim_{x \to \infty} \pi(2x) - 2\pi(x) = -\infty$, not $+\infty$. So there was a typo in the book.

Unit
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  • I think they talk instead of $\psi(2x) - 2 \psi(x)$ with $\psi(x) = \sum_{p^k < x} \ln p$ (and $\frac{\zeta'(s)}{\zeta(s)} = s\int_1^\infty \psi(x) x^{-s-1}dx$ ) and the PNT is $\psi(x) \sim x$ that is more or less direct from "$\zeta(s)$ has no zero on $Re(s) = 1$" – reuns Jul 15 '16 at 19:56
  • @user1952009 No, they are specifically talking about $\pi(x)$. – Unit Jul 15 '16 at 23:17
  • You said it is a typo, and if it is about the PNT then it is what I said – reuns Jul 15 '16 at 23:29
  • @user1952009 The "typo" I'm referring to is their claim that $\pi(2x) - 2\pi(x) \to \infty$ which appears in the blockquote in the question. – Unit Jul 15 '16 at 23:46