黎曼 zeta 函数辐角的大偏差
数论
2024-03-27 v2
摘要
令S ( t ) = 1 π ℑ log ζ ( 1 2 + i t ) S(t) = \frac{1}{\pi}\Im \log\zeta\left(\frac{1}{2}+it\right) S ( t ) = π 1 ℑ log ζ ( 2 1 + i t ) 。我们对集合{ t ∈ [ T , 2 T ] : S ( t ) ≥ V } \{t\in [T,2T] \colon S(t) \geq V\} { t ∈ [ T , 2 T ] : S ( t ) ≥ V } 的测度证明了无条件下界,其中log log T ≤ V ≪ ( log T log log T ) 1 / 3 \sqrt{\log\log T} \leq V \ll \left(\frac{\log T}{\log \log T}\right)^{1/3} log log T ≤ V ≪ ( l o g l o g T l o g T ) 1/3 。当V ≤ ( log T ) 1 / 3 − ε V \leq (\log T)^{1/3-\varepsilon} V ≤ ( log T ) 1/3 − ε 时,我们的界呈高斯形状,方差正比于log log T \log\log T log log T 。在端点V ≍ ( log T log log T ) 1 / 3 V \asymp \left(\frac{\log T}{\log \log T}\right)^{1/3} V ≍ ( l o g l o g T l o g T ) 1/3 处,我们的结果蕴含了Tsang所给出的关于S ( t ) S(t) S ( t ) 的最佳已知Ω \Omega Ω 定理。我们还解释了基于当前对zeta函数零点的认知,当V ≫ ( log T log log T ) 1 / 3 V \gg \left(\frac{\log T}{\log \log T}\right)^{1/3} V ≫ ( l o g l o g T l o g T ) 1/3 时该方法为何失效。在黎曼假设条件下,我们将结果推广至log log T ≤ V ≪ ( log T log log T ) 1 / 2 \sqrt{\log\log T} \leq V \ll \left(\frac{\log T}{\log \log T}\right)^{1/2} log log T ≤ V ≪ ( l o g l o g T l o g T ) 1/2 的范围。
引用
@article{arxiv.2101.01747,
title = {Large deviations of the argument of the Riemann zeta function},
author = {Alexander Dobner},
journal= {arXiv preprint arXiv:2101.01747},
year = {2024}
}
备注
21 pages. The results in this version are stronger than in v1. To appear in Mathematika