I'm stuck proving the following:

for all $t$ such that $|t|$ sufficiently large and if $\sigma > 1 - \frac{100}{log |t|} $ $\zeta(\sigma + it) = O(log|t|)$


  • 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)?


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

$$\zeta(\sigma + it) = \sum_{n \leq |t|} \frac{1}{n^{\sigma + it}} + O(\frac{|t|^{\frac{100}{log|t|}}}{|t|}) + O(|\sigma + it||t|^{- \sigma})$$ $$ = \sum_{n \leq |t|} \frac{1}{n^{\sigma + it}} + O(1)$$



1 Answer 1


Do we only consider the real part and if so why?

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 $a > 0$.

In what we should obtain (below) how do we obtain the $\lvert t\rvert$ in the denominator of the first error term?

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 $\sigma \geqslant 1$.


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