# Prime number theorem lemma: prove that $\psi(x)\sim\pi(x)\log(x)$

I'm trying to follow the proof in Wikipedia that the PNT is equivalent to the assertion $\psi(x)\sim x$, by proving that $\psi(x)\sim\pi(x)\log x$, which it claims is a very simple proof. One direction of inequality is an actual bound, $\psi(x)\le\pi(x)\log x$, but the other inequality has a fuzz factor:

$$\psi(x) \ge \sum_{x^{1-\epsilon}\le p\le x} \log p\ge\sum_{x^{1-\epsilon}\le p\le x}(1-\epsilon)\log x=(1-\epsilon)(\pi(x)+O(x^{1-\epsilon}))\log x.$$

But this doesn't actually complete the proof, because we want $\psi(x)\ge(1-\epsilon)\pi(x)\log x$ without the fuzz factor. If we take large enough $x$ and use $\epsilon/2$ in the above equation we get

$$\psi(x)\ge(1-\epsilon/2)\pi(x)\log x+Ax^{1-\epsilon/2}\log x,$$ so it is sufficient to prove that $Ax^{1-\epsilon/2}\le\frac{\epsilon}2\pi(x)$ for sufficiently large $x$, i.e. $x^{1-\epsilon/2}\in o(\pi(x)),$ and although I am sure there is a proof of this, it's not so simple that the proof can be completely omitted, at least as far as I can see. Is there an easy proof to be found here? The only one I am seeing is Chebyshev's weak version of the PNT, $\frac x{\log x}\in O(\pi(x))$, which takes some significant work to prove.

• IMO, the lower bounds are far more elegantly obtained by using $\frac{x}{(\log x)^k}$ as the cut-off point than with $x^{1-\epsilon}$. – Daniel Fischer Nov 25 '17 at 22:23
• @DanielFischer Could you elaborate? It seems that this would only make the fuzz factor worse in the equation. – Mario Carneiro Nov 26 '17 at 2:35
• $$\pi(x)\log x\leqslant \sum_{p \leqslant \frac{x}{(\log x)^2}} \log x+ \log x \sum_{\frac{x}{(\log x)^2} <p \leqslant x} \frac{\log p}{\log p}\leqslant \frac{x}{(\log x)^2}\log x+ \frac{\log x}{\log x - 2\log \log x} \sum \log p \leqslant \frac{x}{\log x} + \frac{1}{1 - 2\frac{\log \log x}{\log x}}\vartheta(x)$$ for $x \geqslant 3$, using the trivial $\pi(y) \leqslant y$ and $\vartheta(x)- \vartheta(y) \leqslant \vartheta(x)$, and $\log p > \log \frac{x}{(\log x)^2} = \log x - 2\log \log x$ for $p > \frac{x}{(\log x)^2}$. You get explicit bounds (which one could make sharper), no fuzz factor. – Daniel Fischer Nov 26 '17 at 14:15

You have: $$\psi(x)\geq (1-\varepsilon)\sum_{x^{1-\varepsilon}<p<x}\log x \geq (1-\varepsilon)(\pi(x)-\pi(x^{1-\varepsilon}))\log x$$ but, by the Chebyshev bound: $$\frac{\pi(x^{1-\varepsilon})}{\pi(x)}\leq\frac{1.11\frac{x^{1-\varepsilon}}{(1-\varepsilon)\log x}}{0.92\frac{x}{\log x}}\leq \frac{1.21}{1-\varepsilon}\, x^{-\varepsilon}\tag{1}$$ is enough to prove that: $$\psi(x)\geq (1-\tilde{\varepsilon})\,\pi(x)\log x.$$ To replace $(1)$ with something easier to prove than the Chebyshev bound, you can use $\pi(x^{1-\varepsilon})\leq x^{1-\varepsilon}$ and any suitable lower bound for $\pi(x)$.
• @MarioCarneiro: it is not necessary, but it is necessary to prove that the density of the primes in $[1,x]$ does not drop to zero too fast. Any bound like $\pi(x)\geq\frac{x}{(\log x)^{\eta}}$ for a positive $\eta$ does the job. – Jack D'Aurizio Sep 18 '14 at 17:54
• For instance, it is not difficult to prove that $$\pi(x)\geq \frac{x}{\log^2 x}$$ with elementary sieve methods and the Buchstab identity. – Jack D'Aurizio Sep 18 '14 at 17:59