Let $\psi(x)=\sum_{n\leq x} \Lambda(n)$, where $\Lambda(n)$ is the von Mangoldt function. Then as Chebyshev showed, the following equality holds $$\sum_{n\leq x} \psi(x/n)=x\log(x)-x+O(\log(x)).$$ My question is, how far can one go towards proving the prime number theorem by only using the above estimate and the fact that $\psi$ is increasing? Alternatively, is there a well-known example of an increasing function $\psi(x)$ for which the above equality holds, but $\lim_{x\to \infty}\frac{\psi(x)}{x}\neq 1?$ Thank you very much.

  • $\begingroup$ David Speyer has described interesting sets of natural numbers for which the (analogue of the) prime number theorem fails but that share some of the density properties of the set of primes. You might try to concoct a suitable $\psi$ function out of his "fake primes." $\endgroup$ Apr 27, 2020 at 16:56

1 Answer 1


Theorem 1 in the following paper of Ingham shows that the stated estimate, together with $\psi$ being positive and nondecreasing, is 'enough' to deduce that $\psi(x)/x \to 1$:

A. E. Ingham: Some Tauberian theorems connected with the prime number theorem. J. London Math. Soc. 20, 171–180 (1945). Full text (paywalled)

In fact, this remains true even if one only has the weaker error term $o(x)$ in place of $O(\log{x})$. Of course, one has to be careful about what `enough' means here; Ingham's proof still uses the nonvanishing of $\zeta(s)$ on $\Re(s)=1$, which is the key input in the usual proofs of the prime number theorem.

  • $\begingroup$ Thank you very much! $\endgroup$
    – Zamanyan
    Apr 26, 2020 at 19:21

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