I´m working through a book for my seminar and I need a clue to get further. So in the Theorem I am stucked at, it was proofen:
For $\alpha \in (0,1)$:
$\vert\zeta(s)\vert\leq\frac{3\vert\tau\vert^{1-\alpha}}{2\alpha(1-\alpha)}~~~(\sigma\ge\alpha,~\vert\tau\vert\ge 1)$
But there was a second claim, which is not further mentioned in the proof:
$\zeta(s)\ll \log(\vert\tau\vert)~~~(\vert\tau\vert\ge 2,~\sigma\ge 1-\frac{c}{\log(\vert\tau\vert)})$ For any constant c
Anyone can give me a clue hwo to get from the first, prooven, claim to the second one?
It is proven in every text on the prime number theorem. For $a \in [0,1]$ taking the geometric mean of two different bounds $$\begin{eqnarray}n^{-s}-\int_n^{n+1} x^{-s}dx &=& O(n^{-s}) \\& =&\int_n^{n+1}\int_n^x st^{-s-1}dtdx \\ &=& O(s n^{-s-1}) \\ &=& O( (sn^{-s-1})^a (n^{-s})^{(1-a)})\\&= &O(s^a n^{-s-a})\end{eqnarray}$$
For $\sigma > 1-\frac{c}{\log |t|}$, $|t| > 1, a = \frac{2c}{\log |t|}$ we obtain $$\begin{eqnarray}\zeta(s) &=&\frac{1}{s-1}+ \sum_{n\ge 1} (n^{-s}-\int_n^{n+1} x^{-s}dx)\\ &=& \sum_{n \le |t|} n^{-s}- \frac{\lfloor |t|\rfloor^{1-s}}{s-1}+\sum_{n > |t|} (n^{-s}-\int_n^{n+1} x^{-s}dx) \\ &=& \sum_{n \le |t|}O(n^{-1})+O(1) + \sum_{n > |t|} O(s^a n^{-s-a})\\ &=& O(\log |t|) + O(1) + O(|t|^a \frac{1}{\sigma+a-1})\\ &=& O(\log |t|) + O( \frac{1}{1-\frac{c}{\log |t|}+\frac{2c}{\log |t|}-1})\\&=& O(\log |t|)+O(\log|t|)\end{eqnarray}$$