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Let $\psi(x)=\sum_{p^{k}\leq x} \log p$, $k\in \mathbb{N}$. If i recall correctly, the convergence of the integral $s\int_{1}^{\infty} (\psi(x)-x)x^{-s-1} \mathrm{d}x$ at $s=1$ is equivalent to the Prime number theorem.

Can anyone provide a reference/proof of this result ?

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  • $\begingroup$ Okay i found nice explanations here google.com/… $\endgroup$
    – user146617
    Commented Oct 1, 2019 at 10:53

1 Answer 1

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The link to Baker's notes gives one direction, but the other direction relies on what you mean by "the prime number theorem". In its weakest form, the prime number theorem asserts that $\psi(x)\sim x$. Note that if $\psi(x)\sim x$ but $|\psi(x) - x|\gg x/\log x$, then the integral diverges. Indeed, the integral diverges when $\psi(x)\sim x$ but

$|\psi(x) - x|\gg\frac{x}{(\log x)(\log_2 x)\cdots(\log_{k-1} x)(\log_k x)}$,

where $\log_2 x = \log (\log x)$, and $\log_{j+1} x=\log_j(\log x)$ (increasing the lower limit of the integral to a sufficiently large fixed constant as needed).

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