Timeline for Effective, explicit bound on $L'(1,\chi)/L(1,\chi)$

5 events

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Sep 22, 2019 at 12:21	answer	added	H A Helfgott		timeline score: 5
Sep 16, 2019 at 21:49	comment	added	H A Helfgott		Interesting, but I want $\leq c \sqrt{q}$ with a nice, clean constant, not just $O(\sqrt{q})$.
Sep 16, 2019 at 21:27	comment	added	MyNinthAccount		Another way to go about this could be: consider $\sum_{p^v} {\chi(p^v)\Lambda(p^v)\over p^v}e^{-p^v/X}=\int X^s \Gamma(s) {L'\over L}(s+1){ds\over2\pi i}$ and move the contour left. You are willing to accept $O(\sqrt q)$ and so can grandiosely take $X\approx e^{\sqrt q}$ if desired to truncate the sum, while the poles contribute $(L'/L)(1)+\Gamma(\beta-1)X^{\beta-1}+O(...)$ where bounding the other zeros shouldn't be too hard. The $\Gamma(\beta-1)$ then gives $O(\sqrt q)$ again.
Sep 16, 2019 at 18:24	comment	added	MyNinthAccount		I think one does get something rather easily, as when $\chi$ is exceptional then $L(s,\chi)$ is rather much like $\zeta(2s)/\zeta(s)$. Looking at the imaginary quadratic case, if the class number is $1$ we thus have $L'(1,\chi)\sim\pi^2/6$, with in general there additionally being a product by a reciprocal sum over minima. One can work analogously in the real quadratic case as done by Goldfeld and Schinzel. Whether or not this can be done cleanly and explicitly is a different question. But one should get $(\pi^2/6)/(\pi/\sqrt q)$. archive.numdam.org/item/ASNSP_1975_4_2_4_571_0
Sep 16, 2019 at 17:10	history	asked	H A Helfgott	CC BY-SA 4.0