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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 $T=0$?

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  • $\begingroup$ Important self-correction: I should have said $\arcsin$, not $\arctan$. You still get a tighter upper bound. $\endgroup$ Dec 6, 2021 at 7:48
  • $\begingroup$ And yes, by the linear independence of $\pi$ and $\log 2, \log 3,\dotsc,\log p$ over $\mathbb{Q}$, $t$ and $\chi$ can conspire, and so the bound is tight, for $t$ and $\chi$ unbounded. (As @juan points out below - use Kronecker's theorem.) $\endgroup$ Dec 6, 2021 at 7:50

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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 $$|\arg L(2+it,\chi)-\arg L(2,\chi)|\le 2\log\zeta(2).$$

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  • $\begingroup$ Oh, I know how to derive that - I was just pointing out that (a) one can easily do better, (b) I don't know how to get rid of the factor of $2$ (and one most likely can't). $\endgroup$ Dec 4, 2021 at 17:48
  • $\begingroup$ The mystery is why people don't give a better bound :). $\endgroup$ Dec 4, 2021 at 17:49
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    $\begingroup$ @Helfgott I misunderstood you, I was surprised that you were unaware of this. I think the factor 2 can not be eliminated, perhaps using Kronecker's theorem we can prove this. $\endgroup$
    – juan
    Dec 4, 2021 at 18:09
  • $\begingroup$ That's also my impression. $\endgroup$ Dec 5, 2021 at 1:07

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