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 $$N(T)={\frac {T}{2\pi }}\log {{\frac {T}{2\pi }}}-{\frac {T}{2\pi }}+O(\log {T}) \ (T \to \infty),$$ or, equivalently, $$N(2 \pi T)=T \log T-T+O(\log {T}) \ (T \to \infty).$$ 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.)

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.)