# Explicit estimates for $N(T,\chi)$ (not $N(T,\chi)+N(T,\overline{\chi})$)

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 $$T=0$$?

• Important self-correction: I should have said $\arcsin$, not $\arctan$. You still get a tighter upper bound. Dec 6, 2021 at 7:48
• 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.) Dec 6, 2021 at 7:50

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 $$|\arg L(2+it,\chi)-\arg L(2,\chi)|\le 2\log\zeta(2).$$
• 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). Dec 4, 2021 at 17:48