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From the prime number theorem we know that $\pi(x)\sim x/\log x$, i.e. $\dfrac{\pi(x)\log x}{x}\rightarrow 1$ as $x\rightarrow \infty$.

How can we use that to show that $\pi(x)\sim\int_2^x\dfrac{1}{\log t}dt$? The integral $\dfrac{1}{\log t}$ does not have a closed form, as far as I know.

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As for the question, show that $$\operatorname{Li}(x) = \int_2^x \frac{dt}{\log t} \sim \frac{x}{\log x},$$ for example by splitting the integral at something like $\dfrac{x}{(\log x)^2}$. –  Daniel Fischer Dec 14 '13 at 19:50
    
@DanielFischer What do you mean by splitting the integral? Like splitting into the intervals $[2,x/(\log x)^2]$ and $[x/(\log x)^2,x]$? I can't see how that makes it easier to evaluate/estimate things. –  PJ Miller Dec 14 '13 at 20:19
    
Yes, splitting like that. The first integral is easily estimated as $O\left(\frac{x}{(\log x)^2}\right)$, and the second as $\sim \frac{x}{\log x}$, since the integrand is "nearly constant" $\frac{1}{\log x}$ there. –  Daniel Fischer Dec 14 '13 at 21:10
    
@DanielFischer The first integral is fine, but I'm having trouble with the second one. I know the integrand is "nearly constant", but how can I argue formally that it's $\sum\frac{x}{\log x}$? –  PJ Miller Dec 15 '13 at 1:53
    
The second integral is bounded $$x\left(1 - \frac{1}{(\log x)^2}\right) \cdot \frac{1}{\log x} < \int_{x/(\log x)^2}^x \frac{dt}{\log t} < x\left(1-\frac{1}{(\log x)^2}\right)\cdot \frac{1}{\log x - 2 \log \log x}.$$ Both sides are $$\frac{x}{\log x} + o\left(\frac{x}{\log x}\right).$$ –  Daniel Fischer Dec 15 '13 at 11:48

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You actually have it backwards. The Prime Number Theorem shows that $\pi(x) \sim \int_2^x \frac{1}{\log t} dt,$ which is a much better approximation than $\pi(x) \sim \frac{x}{\log x}.$ To get the "standard form" from the first one, just integrate by parts.

To answer Daniel Fischer's comment, see this part of the wiki article on the PNT.

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That depends on what proof you read. The common complex analysis proof shows $\pi(x) \sim \dfrac{x}{\log x}$ (via $\vartheta(x)$). –  Daniel Fischer Dec 14 '13 at 19:51
    
@DanielFischer you are kidding, right? notice that the version you mention has a horrible error term whether or not you know the Riemann hypothesis, so I am sure you can prove this theorem, but it is not the prime number theorem. –  Igor Rivin Dec 14 '13 at 19:54
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I'm not kidding. Yes, you don't get good error bounds. Still, the asymptotic is right, and it is commonly called the prime number theorem. –  Daniel Fischer Dec 14 '13 at 19:56
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By the way, my name is not David. –  Daniel Fischer Dec 14 '13 at 20:00
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I didn't say "standard". I said "common". The really good thing about Newman's proof is that it's so easy that you can show it to students with not very much background in complex analysis, that's why it's commonly given in complex analysis courses. If you want to do number theory, you need the better result quite often, I'd think. –  Daniel Fischer Dec 14 '13 at 20:11

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