# $\pi(x)$ asymptotic as integral $1/\log t$

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

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.
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