I'm stuck proving the following:
for all $t$ such that $|t|$ sufficiently large and if $\sigma > 1 - \frac{100}{log |t|} $ $\zeta(\sigma + it) = O(log|t|)$
Questions:
- Do we only consider the real part and if so why?
- In what we should obtain (below) how do we obtain the |t| in the denominator of the first error term?
- In what we should obtain (below) how do we get the $|\sigma + it|$ in the error term, the other part clearly follows by computing the integral
- How are these error terms O(1)?
Attempt:
Letting $x=|t|$ then by definition of $\zeta(s) = \sum_{n \leq |t|} \frac{1}{n^{s}} + \frac{|t|^{1-s}}{s-1} + \frac{\{|t|\}}{|t|^{s}} - s \int_{|t|}^{\infty} \{w\} \frac{dw}{w^{s+1}}$
After substituting in for $\sigma$ (rewriting $s = \sigma + it$) I obtain
$$ \sum_{n \leq |t|} \frac{1}{n^{\sigma + it}} + \frac{|t|^{\frac{100}{log|t|} -it}}{{-\frac{100}{log|t|}-it}} + \frac{\{|t|\}}{|t|^{\sigma + it}} - s \int_{|t|}^{\infty} \{w\} \frac{dw}{w^{s+1}} $$
I should somehow obtain
$$\zeta(\sigma + it) = \sum_{n \leq |t|} \frac{1}{n^{\sigma + it}} + O(\frac{|t|^{\frac{100}{log|t|}}}{|t|}) + O(|\sigma + it||t|^{- \sigma})$$ $$ = \sum_{n \leq |t|} \frac{1}{n^{\sigma + it}} + O(1)$$
Thanks.
If you are referring to the exponents, then yes, we consider only the real part. The reason is that we are using the real-valued logarithm of $a > 0$ to define $a^s = \exp (s\cdot \log a)$, and $\lvert \exp w\rvert = \exp (\operatorname{Re} w)$ (since $\lvert \exp w\rvert^2 = (\exp w)(\overline{\exp w}) = (\exp w)(\exp \overline{w}) = \exp (w + \overline{w}) = \exp(2\operatorname{Re} w)$). So $\lvert a^s \rvert = a^{\sigma}$ for $a > 0$.
The denominator is $s-1$, and we have $\lvert s-1\rvert \geqslant \lvert \operatorname{Im} (s-1)\rvert = \lvert t\rvert$. This yields the estimate $$\biggl\lvert \frac{\lvert t\rvert^{1-s}}{s-1}\biggr\rvert = \frac{\lvert t\rvert^{1-\sigma}}{\lvert s-1\rvert} \leqslant \frac{\lvert t\rvert^{1-\sigma}}{\lvert t\rvert} = \lvert t\rvert^{-\sigma} < \lvert t\rvert^{\frac{100}{\log \lvert t\rvert} - 1} = \frac{e^{100}}{\lvert t\rvert}$$ for $\lvert t\rvert > 1$, since then $\lvert t\rvert^{-\sigma}$ is decreasing in $\sigma$ and $\lvert t\rvert^{b/\log \lvert t\rvert} = \exp \bigl(\frac{b}{\log \lvert t\rvert}\cdot \log \lvert t\rvert\bigr) = e^b$.
The next term is fairly direct, $$\biggl\lvert \frac{\lbrace \lvert t\rvert\rbrace}{\lvert t\rvert^s}\biggr\rvert = \frac{\lbrace\lvert t\rvert\rbrace}{\lvert t\rvert^{\sigma}} < \frac{1}{\lvert t\rvert^{\sigma}} < \frac{e^{100}}{\lvert t\rvert}$$ for $\lvert t\rvert \geqslant 1$ since $0 \leqslant \lbrace \lvert t\rvert\rbrace < 1$.
For the integral term, we cannot go quite so far. The integral only makes sense for $\operatorname{Re} s > 0$ (perhaps also for $\operatorname{Re} s = 0$ and $s\neq 0$, but not beyond that), so we must either restrict to $\lvert t\rvert \geqslant e^{100}$ or modify the lower bound for $\sigma$ to $$\sigma > \max \; \Biggl\{ 0, 1 - \frac{100}{\log \lvert t\rvert}\Biggr\}\,.$$ Well, since the straightforward estimate of the integral, $$\Biggl\lvert \int_{\lvert t\rvert}^{\infty} \frac{\lbrace w\rbrace}{w^{1+s}}\,dw\Biggr\rvert \leqslant \int_{\lvert t\rvert}^{\infty} \frac{\lbrace w\rbrace}{w^{1+\sigma}}\,dw \leqslant \int_{\lvert t\rvert}^{\infty} \frac{dw}{w^{1+\sigma}} = \frac{1}{\sigma\cdot \lvert t\rvert^{\sigma}}\,,$$ produces a $\sigma$ in the denominator, we must keep $\sigma$ away from zero, so we actually need a stronger restriction, $$\sigma \geqslant \max\; \Biggl\{ c, 1 - \frac{100}{\log \lvert t\rvert}\Biggr\}$$ for a fixed $c \in (0,1)$, or $\lvert t\rvert \geqslant t_0$ for a fixed $t_0 > e^{100}$.
With that, we can estimate the last term by $$\Biggl\lvert s\int_{\lvert t\rvert}^{\infty} \frac{\lbrace w\rbrace}{w^{1+s}}\,dw\Biggr\rvert \leqslant \frac{\lvert \sigma + it\rvert}{\sigma\lvert t\rvert^{\sigma}} \leqslant \frac{\sigma + \lvert t\rvert}{\sigma\lvert t\rvert^{\sigma}} = \lvert t\rvert^{-\sigma} + \frac{\lvert t\rvert^{1-\sigma}}{\sigma}\,.$$ We had estimated $\lvert t\rvert^{-\sigma}$ above, and with that we obtain $$\Biggl\lvert s\int_{\lvert t\rvert}^{\infty} \frac{\lbrace w\rbrace}{w^{1+s}}\,dw\Biggr\rvert \leqslant \frac{e^{100}}{\lvert t\rvert} + \frac{e^{100}}{\sigma} \leqslant e^{100}\biggl(\frac{1}{\lvert t\rvert} + \frac{1}{c}\biggr)\,.$$ Thus each of the terms except the partial sum is bounded and we have $$\zeta(\sigma + it) = \sum_{n \leqslant \lvert t\rvert} \frac{1}{n^{\sigma +it}} + O(1)$$ in the indicated region. The remaining sum is estimated using $$\lvert n^{-\sigma - it}\rvert = n^{-\sigma} = \frac{n^{1-\sigma}}{n} \leqslant \frac{\lvert t\rvert^{1-\sigma}}{n} \leqslant \frac{e^{100}}{n}$$ for $\sigma < 1$, and $$\lvert n^{-\sigma-it}\rvert = n^{-\sigma} \leqslant \frac{1}{n}$$ for $\sigma \geqslant 1$.