TITLE: Explicit estimates for $N(T,\chi)$ (not $N(T,\chi)+N(T,\overline{\chi})$)
QUESTION [7 upvotes]: Let $N(T,\chi)$ denote the number of zeros of $L(s,\chi)$ with imaginary part between $0$ and $T$, with any zero with imaginary part equal to $T$ or to $0$ (not that the latter kind really exists) counting as half a zero. Here I am following the convention in Montgomery-Vaughan, rather than that in part of the literature, where $N(T,\chi)$ means what I would call $N(T,\chi) + N(T,\overline{\chi})$.
The explicit literature generally (McCurley, Trudgian, Bennett-Martin-O'Bryant-Rechnitzer...) generally bounds $N(T,\chi) + N(T,\overline{\chi})$. The question is: what kind of explicit bounds we can extract from their proofs for $N(T,\chi)$?

The first step is easy: we can express $N(T,\chi)$ as $\text{main term} + S(T,\chi)-S(0,\chi)$, as in Montgomery-Vaughan, Thm. 14.5, where $S(T,\chi) = \frac{1}{\pi} \arg L(1/2+iT,\chi)$. One would then decompose
$$S(T,\chi)-S(0,\chi) = \frac{1}{\pi} \left(\arg L(\sigma+i T,\chi)|_{\sigma=\sigma_0}^{1/2} +  \arg L(\sigma_0+i t,\chi)|_{t=0}^T + 
\arg L(\sigma,\chi)|_{\sigma=1/2}^{\sigma_0}\right)$$
for some $\sigma_0>1$ of our choice. The literature gives the bound $2 \log \zeta(\sigma_0)$ on $\left|\arg L(\sigma_0+it)|_{t=-T}^T\right|$. The reason is a mystery to me -- it is obvious that $2 \sum_p \arcsin p^{-\sigma}$ is a tighter upper bound on $\left|\arg L(\sigma_0+it)|_{t=-T}^T\right|$ (and it is easy to compute). I do not know how to do better than $2 \sum_p \arcsin p^{-\sigma}$ as an upper bound on
$\left|\arg L(\sigma_0+it)|_{t=0}^T\right|$, and suspect one cannot, in general, as $t$
and $\chi$ could conspire.
The bulk of the explicit literature deals with bounding $\arg L(\sigma+i T,\chi)|_{\sigma=\sigma_0}^{1/2}$. Is there a better bound on $\arg L(\sigma,\chi)|_{\sigma=1/2}^{\sigma_0}$ than what one would get just by setting $T=0$?

REPLY [2 votes]: For the first question about $2\log\zeta(\sigma_0)|$, I think the reasoning is this:
For $\sigma>2$ we have
$$|L(s,\chi)-1|\le \sum_{n=2}^\infty\frac{1}{n^\sigma}=\zeta(\sigma)-1<1.$$
Hence $\log L(s,\chi)$ can be defined by the power series
$$\log L(s,\chi)=-\sum_{k=1}^\infty \frac{1}{k}(1-L(s,\chi))^k$$
In particular  $|\Im \log L(2+it,\chi)|<\pi/2$ and coincide with $\arg(L(2+it,\chi))$. Also this is equal to
$$\log L(2+it)=\sum_{n=2}^\infty \frac{\Lambda(n)}{\log n}\frac{\chi(n)}{n^{2+it}}$$
It follows that
$$|\arg L(2+it)|\le |\log L(2+it)|\le \sum_{n=2}^\infty \frac{\Lambda(n)}{\log n}\frac{1}{n^{2}}=\log\zeta(2).$$
Now it follows that
$$|\arg L(2+it,\chi)-\arg L(2,\chi)|\le 2\log\zeta(2).$$