TITLE: Is there an intuitive reason why the natural logarithm shows up in the Prime Number Theorem?
QUESTION [7 upvotes]: I've always wondered if it has something to do with the idea that the probability that an integer is divisible by a prime $p$ is $1/p$.

REPLY [11 votes]: Here is a heuristic argument. As you say, heuristically the probability that an integer is divisible by a prime $p$ is $\frac{1}{p}$, hence the probability that it's not divisible by $p$ is $1 - \frac{1}{p}$. Making the simplifying assumption that these events are independent, we get that the probability that an integer $n$ is prime is the probability that it isn't divisible by any smaller primes, hence is
$$\prod_{p < n} \left( 1 - \frac{1}{p} \right).$$
So we want to explain why this grows like $\frac{1}{\ln n}$. Well, let's invert it, getting
$$\prod_{p < n} \frac{1}{1 - \frac{1}{p}} = \prod_{p < n} \left( 1 + \frac{1}{p} + \frac{1}{p^2} + \dots \right).$$
Expanding this out produces a sum over terms of the form $\frac{1}{k}$ where $k$ ranges over all positive integers whose prime divisors are all less than $n$. (This is closely related to the Euler product factorization of the Riemann zeta function.) The bulk of this sum is
$$\sum_{k < n} \frac{1}{k} \sim \ln n$$
by the usual Riemann sum argument, and we are going to blithely ignore the rest of the sum (this can be done a bit more carefully but eh). Done!