number of n.t. zeros in an interval in the limit

Question

Motivation:

We know that the density of non-trivial zeros of $\zeta(s)$ increases with $t$. But how dense does it get? Assuming RH.

Summary

1. Trying to solve for number of zeros in the length (measured in $t$) of one cycle of $\cos\nu(t)$.
2. $\frac{dN(t)}{dt}$ gives me number of zeros per interval $dt$. Now $\cos\nu(t)$ is periodic (locally) and so we need $d\nu(t) = 2\pi$ to find how much does $\nu(t)$ have to change to get length of one cycle of $\cos\nu(t)$. ($\cos\nu(t) = \cos(\nu(t) + 2\pi) = \cos(\nu(t) + d\nu(t))$).
3. So in $\frac{d\nu(t)}{dt}$ I plug in $d\nu(t) = 2\pi$ to give me the interval $dt$ needed to equal one cycle of $\cos\nu(t)$. Doing so, I would get $\begin{align*} dt = \frac{4\pi}{\log \frac{t}{2\pi}}\end{align*}$ which can be plugged into $\frac{dN(t)}{dt}$ giving me $dN(t)$ which is the number of zeros in an unit interval $dt$ and we also know that this $dt$ is equal to length of one cycle of $\cos\nu(t)$.

See pages 119-120 of H.M.Edwards Riemann's Zeta Function for reference.

\begin{align} N(t) &= \frac{t}{2\pi} \log\left(\frac{t}{2\pi}\right)- \frac{t}{2\pi} + O\left(\frac{1}{t}\right)\\\\ \implies \frac {d N(t)}{dt} &= \frac{1}{2\pi}\log{\frac{t}{2\pi}} \tag{1} \\\\ Z(t) &= e^{i\nu(t)}\zeta(\frac{1}{2} + it) \\\\ \nu(t) &= Im \log \prod\left(\frac{it}{2} - \frac{3}{4}\right) - \frac{t}{2} \log \pi \\\\ &= \frac{t}{2}\log \frac{t}{2\pi} - \frac{t}{2} -\frac{\pi}{8} + \frac{1}{48t} + \frac{7}{5760t^3} + \cdots \end{align}

differentiating $\nu(t)$ and ignoring higher order terms of $\frac{1}{t}$ gives us: \begin{align*} \frac{d\nu(t)}{dt} = \frac{1}{2}\log\left(\frac{t}{2\pi}\right) \end{align*}

since $\cos\nu(t) = \cos(\nu(t) + 2\pi)$, substituting $2\pi$ for $d\nu(t)$, we get \begin{align*} dt = \frac{4\pi}{\log \frac{t}{2\pi}} \end{align*}

Substituting for $dt$ in (1), we get \begin{align*} \lim_{t \to \infty} dN(t) = \frac{4\pi}{\log \frac{t}{2\pi}} \frac{\log\frac{t}{2\pi}}{2\pi} = 2 \end{align*}

Let us put this to test. In the interval $t_1 = 30610045718.27011 \le t \le t_2 = 30610045999.548215$ there are 1000 zeros, which implies: \begin{align*} \frac{dN(t)}{dt} = \frac{1000}{t_2-t_1} = 3.5552. \end{align*} One cycle of $\cos\nu(t)$ in this range of $t$ will have a length of \begin{align*} dt = \frac{4\pi}{\log \frac{t}{2\pi}} = 0.56334. \end{align*}

Hence number of zeros in one cycle of $\cos\nu(t)$ is: $dN(t) = 0.56334*3.5552 = 2.00278$. For large $t$, this should converge to $2$.

Similar analysis will show that even for Dirichlet L-series, $\lim\limits_{t \to \infty} dN(t,\chi) = 2$.

My Question: Since the Davenport-Hielbronn Zeta function has zeros on and off the CL, I am curious to know what is $\lim\limits_{t \to \infty} dN(t)$ for D-H zeta? I conjecture that this will be different from $2$. (obviously there is a $\cos\nu(t)$ equivalent for DH Zeta different from the one used above).