Hello I want to get estimate (24) but I don´t have an idea how to get this estimate out of the proof for 23. In the book where it comes from, the author didn´t even say any word how to get from (23) to (24). Anyone has an idea how to get the $\log$ in?
Let $\alpha \in (0,1)$. Then,
$(23)~~~\vert\zeta(s)\vert\leq\frac{3\vert\tau\vert^{1-\alpha}}{2\alpha(1-\alpha)}~~~(\sigma\ge\alpha,~\vert\tau\vert\ge 1).$
For every constant c
$(24)~~~\zeta(s)\ll \log(\vert\tau\vert)~~~(\vert\tau\vert\ge 2,~\sigma\ge 1-\frac{c}{\log(\vert\tau\vert)})$
Already shown:
$\vert\zeta(s)\vert=\vert\sum\limits_{n\leq N}n^{-s}-\frac{N^{1-s}}{1-s}-s\int\limits_{N}^{\infty}\{t\}t^{-s-1}dt\vert\leq\vert\sum\limits_{n\leq N}n^{-s}\vert+\vert\frac{N^{1-s}}{1-s}\vert+\vert s\vert\int\limits_{N}^{\infty}\vert\{t\}t^{-s-1}\vert dt\leq \vert\sum\limits_{n\leq N}n^{-s}\vert+\vert\frac{e^{(1-\sigma)\log(N)}e^{-i\tau\log(N)}}{\tau}\vert+\vert s\vert\int\limits_{N}^{\infty}\vert t^{-s-1}\vert dt\leq\sum\limits_{n\leq N}n^{-\sigma}+\frac{N^{1-\sigma}}{\vert\tau\vert}+\vert s\vert\biggl[\vert\frac{t^{-s}}{s}\vert\biggl]_{N}^{\infty}\leq\sum\limits_{n\leq N}n^{-\sigma}+\frac{N^{1-\sigma}}{\vert\tau\vert}+\vert s\vert\frac{N^{-\sigma}}{\vert s\vert}=\sum\limits_{n\leq N}n^{-\sigma}+\frac{N^{1-\sigma}}{\vert\tau\vert}+\frac{1}{2}\vert s\vert\frac{N^{-\sigma}}{\vert s\vert}+\frac{1}{2}\vert s\vert\frac{N^{-\sigma}}{\vert s\vert}\leq \sum\limits_{n\leq N}n^{-\sigma}+\frac{N^{1-\sigma}}{\vert\tau\vert}+\frac{1}{2}N^{-\sigma}+\frac{1}{2}\vert s\vert\frac{N^{-\sigma}}{\vert \sigma\vert}$
$=\sum\limits_{n\leq N}n^{-\sigma}+N^{1-\sigma}\vert\tau\vert^{-1}+\frac{1}{2}N^{-\sigma}+\frac{1}{2}\vert s\vert\sigma^{-1}N^{-\sigma}$
Further,
$\sum\limits_{n\leq N}n^{-\sigma}=\int\limits_{1}^{N}t^{-\sigma}dt-\int\limits_{1}^{N}\{t\}dt \leq 1+\int\limits_{1}^{N}t^{-\sigma}dt<\frac{N^{1-\sigma}}{1-\sigma}~~~(0<\sigma<1)$
If $\sigma\ge\alpha$ ,then ,
$\vert\zeta(s)\vert\leq\frac{N^{1-\sigma}}{1-\sigma}+ N^{1-\sigma}\vert\tau\vert^{-1}+\frac{1}{2}N^{-\sigma}+\frac{1}{2}\vert s\vert\sigma^{-1}N^{-\sigma}=N^{1-\sigma}\biggl(\frac{1}{1-\sigma}+\frac{1}{\vert\tau\vert}+\frac{\sigma+\vert \sigma+i\tau\vert}{2\sigma N}\biggl) \leq N^{1-\alpha}\biggl(\frac{1}{1-\alpha}+\frac{1}{\vert\tau\vert}+\frac{2\alpha+\vert\tau\vert}{2\alpha N}\biggl)$
Choose $N=[\vert\tau\vert]$, follows, $\vert\zeta(s)\vert\leq N^{1-\alpha}\biggl[\frac{1}{1-\alpha}+\frac{1}{\vert\tau\vert}+\frac{2\alpha+\vert\tau\vert}{2\alpha N}\biggl]\leq\frac{N^{1-\alpha}}{\alpha(1-\alpha)}\biggl[\alpha+\frac{\alpha(1-\alpha)}{\vert\tau\vert}+\alpha(1-\alpha)\frac{2\alpha-\vert\tau\vert}{2\alpha N}\biggl]= \frac{N^{1-\alpha}}{\alpha(1-\alpha)}\biggl[\alpha+\underbrace{\frac{\alpha(1-\alpha)}{\vert\tau\vert}}_{\leq\alpha(1-\alpha)}+(1-\alpha)(\underbrace{\frac{\alpha}{N}}_{\leq\alpha}+\underbrace{\frac{\vert\tau\vert}{2N}}_{\leq 1})\biggl] \leq\frac{N^{1-\alpha}}{\alpha(1-\alpha)}[\alpha+\alpha(1-\alpha)+(1+\alpha)(1-\alpha)]$
Max of [ ] is at$\frac{1}{2}$ and has value $\frac{3}{2}$. So we get $(23)$
See here p.10 for $c > 0, s= \sigma+it, \sigma \ge 1-\frac{c}{\log t} > 0,t \ge 3,x \le t$ then $|x^{1-s}| \le e^c$ so that $$\zeta(s)=\sum_{n\le t}n^{-s}-\frac{t^{1-s}}{1-s}-s\int_{t}^\infty\{x\}x^{-s-1}dx$$ $$ = O(\sum_{n \le t} n^{-1}) + O(1/t) + O(t \int_{t}^\infty x^{-\sigma-1}dx)$$ $$ = O(\log t) + O(1)+O(\frac{t^{1-\sigma}}{\sigma})$$ $$ = O(\log t)$$