Riemann–Von Mangoldt Formula – Detailed Explanation

Question

Let $$N(T) = \#\{\rho \in \mathbb{C}: \zeta(\rho) = 0,\, \operatorname{Im} \rho \in (0,T]\}$$ denote the number of zeros of $\zeta(s)$, counting multiplicities, with imaginary part lying in the interval $(0,T]$, that is, with imaginary part greater than $0$ and less than or equal to $T$. For example, one has $N(50) = 10$, since there are exactly 10 zeros of $\zeta(s)$ with imaginary part lying in the interval $(0,50]$. The Riemann–von Mangoldt formula, conjectured by Riemann in 1859 and proved by von Mangoldt in 1905, states that
$$N(T)={\frac {T}{2\pi }}\log {{\frac {T}{2\pi }}}-{\frac {T}{2\pi }}+O(\log {T}) \qquad (T \to \infty),$$ or, equivalently,
$$N(2 \pi T)=T \log T-T+O(\log {T}) \qquad (T \to \infty).$$
I'm wondering if more is known. In particular, is there a known asymptotic expansion of $N(T)$ or $N(2 \pi T)$, or, perhaps even, an explicit formula? (Pardon if there is an obvious reference for this. I've been working in analytic number theory for only the last few years, and there are still some gaps in my knowledge that I'm trying to fill.)

Answer 1

As $T\to\infty$, we have $$ N(T) = \frac{T}{2\pi}\log\frac{T}{2\pi}-\frac{T}{2\pi}+\frac{7}{8}+\frac{1}{\pi}\int_{\frac{1}{2}}^{\infty}\mathrm{Im}\Big(-\frac{\zeta'}{\zeta}(\sigma+iT)\Big)d\sigma+O(T^{-1}). $$ This is proved in Chapter 15 of Davenport's Multiplicative Number Theory. The error term $O(T^{-1})$ is a truncation for the asymptotic expansions for the arctan and gamma functions. The contribution from the arctan function consists of lower order terms in a Taylor expansion, and the contribution from the gamma function consists of lower order terms in the Stirling expansion.

EDIT: Because of the apparent lack of clarity regarding "arg" in this result, I replaced "$\arg \zeta(1/2+iT)$" with the corresponding integral, which should not be ambiguous. I hope this helps.