# The nontrivial zeros of the zeta function and the prime counting function

The truncated explicit formula has the shape
$$$$\label{1} \psi(x) =x-\frac{\zeta^{\prime}(0)}{\zeta(0)}-\sum_{|\rho|\leq T}\frac{x^{\rho}}{\rho}+\sum_{n=1}^{\infty}\frac{x^{-2n}}{2n}+O\left(\frac{x\log (Tx)^2}{T} \right) ,$$$$ where $$\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 $$$$\label{2} \psi(x) =x+O\left(x(\log x^2)\left( \frac{1}{T}+x^{-c/\log T} \right) \right) .$$$$ We choose $$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$$, the error term above is $$$$\label{3} \ll x(\log x)^2\exp(-\sqrt{c\log x})\ll x\exp(-c\sqrt{\log x})$$$$
since we may suppose that $$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}).

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 $$$$\label{4} \psi(x)=\sum_{n\leq x}\Lambda(n)=x+O(x(\log x)^2).$$$$

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.

• When you plug in $T=\exp(\sqrt{c\log t})$, then the error term will be significantly smaller than $x(\log x)^2$. Nov 23, 2022 at 20:58
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 $$x$$.