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The truncated explicit formula has the shape
\begin{equation}\label{1} \psi(x) =x-\frac{\zeta^{\prime}(0)}{\zeta(0)}-\sum_{|\rho|\leq T}\frac{x^{\rho}}{\rho}+\sum_{n=1}^{\infty}\frac{x^{-2n}}{2n}+O\left(\frac{x\log (Tx)^2}{T} \right) , \end{equation} where $\rho=\beta+it$ are the nontrivial zeros of the zeta function, [1, Explicit Formula], and there are other versions of it.

The final paragraph on the proof of Theorem 6.9 in [2, page 181] states the following:

"... On combining these estimates we conclude that \begin{equation}\label{2} \psi(x) =x+O\left(x(\log x^2)\left( \frac{1}{T}+x^{-c/\log T} \right) \right) . \end{equation} We choose $T$ so that the two terms in the last factor of the error term are equal, $T = \exp(\sqrt{c\log x})$. With this choice of $T$, the error term above is \begin{equation}\label{3} \ll x(\log x)^2\exp(-\sqrt{c\log x})\ll x\exp(-c\sqrt{\log x}) \end{equation}
since we may suppose that $0 < c < 1$. Thus the proof of (6.12) is complete."

Question. Why does the proof of Theorem 6.9, based on the zero free region $1-c/\log T$, or any other known zero free region, uses fewer zeros $\rho=\beta+it$ in the approximation to $\psi(x)$. Specifically, $t\leq T< \exp(\sqrt{c\log x})<x$.

We should expect that increasing the number of zeta zeros to $t\leq T=x$, should have a smaller error term $E(x)$ since the approximation to $\psi(x)=\sum_{n\leq x}\Lambda(n)=x+E(x)$ is better. But, it does not, in fact, the error term of the truncated explicit formula (2) exceeds the main term. It has the shape \begin{equation}\label{4} \psi(x)=\sum_{n\leq x}\Lambda(n)=x+O(x(\log x)^2).\end{equation}

Please edit as required to improve the question. Many other paradoxes in the theory of the zeta function are documented in [3].

[1] Harold Davenport, Multiplicative number theory, third ed., Graduate Texts in Mathematics, vol. 74, Springer-Verlag, New York, 2000.

[2] Hugh L. Montgomery and Robert C. Vaughan, Multiplicative number theory. I. Classical theory, Cambridge Studies in Advanced Mathematics, vol. 97, Cambridge University Press, Cambridge, 2007.

[3] David W. Farmer, Currently there are no reasons to doubt the Riemann Hypothesis, arXiv:2211.11671.

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  • $\begingroup$ When you plug in $T=\exp(\sqrt{c\log t})$, then the error term will be significantly smaller than $x(\log x)^2$. $\endgroup$
    – Wojowu
    Nov 23, 2022 at 20:58
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    $\begingroup$ If you like my answer, please accept it officially (so that it turns green). Also, please do not add content to my answer (or any answer); fixing typos or links is fine. Thanks in advance! $\endgroup$
    – GH from MO
    Nov 25, 2022 at 23:51

1 Answer 1

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As you can see from the proof, with our current knowledge of the distribution of zeros, your expectation "We should expect that [...] the approximation to $\psi(x)$ [...] is better" is false. The problem is that, as far as we know, higher up there can be many zeros with real part very close to $1$, which makes the approximation worse, not better. For example, if we did not know that all zeros have real part less than $1$, then we could not even prove that $\psi(x)$ is asymptotically $x$.

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    $\begingroup$ If you want to accept an answer, m.o.q, the way to do it is to click in the check mark next to the answer. Editing the answer is not the way to go. $\endgroup$ Nov 24, 2022 at 22:10

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