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Can the prime number theorem be obtained from the explicit formula,

$\psi(x)=x-\sum_{\zeta(\rho)=0}\frac{x^\rho}{\rho}+O(1)$?

Here, $\psi(x)=\sum_{k=1}^\infty\sum_{p^k<x}\log p$

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    $\begingroup$ It does not follow solely from the explicit formula, because the Prime Number Theorem requires the nontrivial zeros $\rho$ to have real part $<1$. Once you have this, you can prove the PNT without the full explicit formula, which typically comes later in most textbooks. See Montgomery-Vaughan, Multiplicative Number Theory. $\endgroup$
    – Stopple
    Commented Jul 31, 2023 at 17:48
  • $\begingroup$ @Stopple I meant to say that does pnt follow from the explicit formula along with the non vanishing of zeta on the boundary of the critical strip? $\endgroup$
    – Mustafa Said
    Commented Jul 31, 2023 at 19:36
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    $\begingroup$ So there's two parts: $\psi(x)\sim x\Rightarrow$ PNT, which is in almost any book on the subject, and the Explicit Formula $\Rightarrow \psi(x)\sim x$. This latter requires more than just the real part of each individual $\rho<1$, one would need to understand the contribution of all of them. And, the series is only conditionally, not absolutely, convergent. $\endgroup$
    – Stopple
    Commented Aug 1, 2023 at 17:50

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Expanding on my comments above, most textbooks will show how the Prime Number Theorem follows from $\psi(x)\sim x$. This does not require the full strength of the Explicit Formula for $\psi(x)$, and most textbooks will prove the PNT before the Explicit Formula. One needs more than just the real part of each $\rho$ is $<1$; one needs to understand the contribution of all of them. The fact that the infinite series is not absolutely convergent complicates matters.

However, one can consider instead $$\psi_1(x)=\int_0^x\psi(u)\, du=\sum_{n\le x}(x-n)\Lambda(n)$$ This function also has an Explicit Formula $$\psi_1(x)=\frac{x^2}{2}-\sum_\rho \frac{x^{\rho+1}}{\rho(\rho+1)}-x\frac{\zeta^\prime}{\zeta}(0)+\frac{\zeta^\prime}{\zeta}(-1)-\frac{1}{2} x \log \left(1-\frac{1}{x^2}\right)-\coth ^{-1}(x)$$ and now the sum over $\rho$ is absolutely convergent. This is Theorem 28 in Ingham's "The Distribution of Prime Numbers". Ingham comments on p.74

A generalised form of Theorem 28 was the basis of de la Vallee Poussin's proof of the Prime Number Theorem...

(Updated to address the questions of Steven Clark below)

The explicit formula for $\psi_1(x)$ is not obtained by integrating the explicit formula for $\psi(x)$ term by term; again that's not allowed without absolute convergence. Instead the approach is via an inverse Mellin transform $$ I(m)=-\frac{1}{2\pi i}\int_{C(m)}\frac{x^{s+1}}{s(s+1)}\frac{\zeta^\prime}{\zeta}(s)\, ds, $$ where $C(m)$ is a large rectangle avoiding the zeros of $\zeta(s)$ The Residue Theorem, standard estimates and letting $m\to \infty$ give the explicit formula. (see Ingham for details.)

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  • $\begingroup$ Just to add some details: by monotonicity of $\psi(t)$, $\psi_1(x)-\psi_1(x-h) = \int_{x-h}^{x}\psi(t)dt\le h\psi(x) \le \int_{x}^{x+h}\psi(t)dt = \psi_1(x+h)-\psi_1(x)$ for any $h>0$. So if one proves $\psi_1(t) = t^2/2 + O(R(t))$ using the formula Stopple cites (and a zero-free region) one gets for free $\psi(x)=x +O(h + h^{-1}\max_{[x-h,x+h]} R(t))$. The most common $R$ is $R(t) = t^2 \exp(-c\sqrt{\log t})$ for which $\psi(x) = x + O(h +h^{-1}x^2 \exp(-c'\sqrt{\log x}))$, now take $h=x\exp(-c''\sqrt{\log x})$ to recover PNT. $\endgroup$
    – Ofir Gorodetsky
    Commented Aug 2, 2023 at 19:59
  • $\begingroup$ I also think some clarification is needed with respect to $\psi_1(x)=\int\limits_0^x \psi(u)\,du$ in the derivation of the explict formula for $\psi_1(x)$ from the explicit formula $$\psi(x)=x-\underset{T\to\infty}{\text{lim}}\left(\sum\limits_{|\Im(\rho)|<T}\frac{x^{\rho}}{\rho}\right)-\frac{\zeta'(0)}{\zeta(0)}-\frac{1}{2} \log\left(1-\frac{1}{x^2}\right).$$. $\endgroup$
    – Steven Clark
    Commented Aug 6, 2023 at 20:18
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    $\begingroup$ But $$\psi_1(x)=\frac{x^2}{2}-\sum_\rho \frac{x^{\rho+1}}{\rho(\rho+1)}-x\frac{\zeta^\prime}{\zeta}(0)-\frac{1}{2} x \log\left(1-\frac{1}{x^2}\right)-\coth^{-1}(x)$$ seems to evaluate with an offset, whereas $$\psi_1(x)=\frac{x^2}{2}-\sum_\rho \frac{x^{\rho+1}}{\rho(\rho+1)}-x\frac{\zeta^\prime}{\zeta}(0)+\frac{\zeta^\prime}{\zeta}(-1)-\frac{1}{2} x \log\left(1-\frac{1}{x^2}\right)-\coth^{-1}(x)$$ seems to evaluate more correctly. $\endgroup$
    – Steven Clark
    Commented Aug 6, 2023 at 21:37
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    $\begingroup$ @StevenClark See updated answer. $\endgroup$
    – Stopple
    Commented Aug 7, 2023 at 20:35
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    $\begingroup$ Very good. I think that your modification of the von Mangoldt version of R's explicit formula is not as well-known as it should be! :) $\endgroup$
    – paul garrett
    Commented Aug 7, 2023 at 23:44

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