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The prime number theorem states that

$$\pi(x) \sim \frac x {\ln(x)}$$

Morally, this seems to suggest that there is a fundamental connection between primes and the natural logarithm. But since we're just dealing with asymptotics here, the choice of right hand side is not unique. In fact, we have

$$\pi(x) \sim \frac x {\ln(x)} + f(x)$$

For any $f\in o(\frac x {\ln(x)})$. What if the appearance of $\ln$ here is just an artefact of the specific method used to prove the PNT? Maybe for a suitable choice of $f$ you could make it looks like the primes have a deep connection to any function you like. Is there any argument that could in some sense show that $\ln$ is not arbitrary here?

Jack M
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3 Answers3

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The prime number theorem can also be formulated with the integral logarithm, i.e., as $\pi(x)\sim li(x)$. So we do not necessarily have $x/\log(x)$ directly. Still we can say that the logarithm appears naturally in connection with asymptotic results on primes, e.g., $$ \sum_{p\le x}\frac{1}{p}=\log(\log(x))+c+O(1/x). $$

Perhaps more convincing for you than asymptotic results are actual inequalities, like $$ \frac {x} {\ln x} < \pi(x) < 1.25506 \frac {x} {\ln x} \! \text{ for all } x ≥ 17. $$ This shows that $x/\log(x)$ does not only appear "just as an artefact of the specific method".

Dietrich Burde
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There are heuristic arguments that suggest that the probability that a random integer $n$ is prime is on the order of $1/\log n$. That's implicit in some of Euler's work. It's why the natural logarithm appears in the prime number theorem.

See this question and some of the answers there:

"Probability" of a large integer being prime

Community
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Ethan Bolker
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The prime number theorem is equivalent to the asymptotic estimate $p_n \sim n\ln n$, which is a direct link between logarithms and primes (not just counting primes). So $p_n/n\sim \ln n$.

KCd
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