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I am currently preparing for a course in analytic number theory and I wanted to get a heads start. In my preparation, I came across the following problem:

Show that for $|y|\geq 2$, $|\zeta(1+iy)| \leq C\log|y|$ for some constant $C.$

I am very weak when it comes to determining bounds such as this, and honestly don't know where to start. This is all very new to me. Any help is appreciated.

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  • $\begingroup$ For this bound, you must have more than the Dirichlet series definition, since the series only converges for real part greater than 1. Perhaps you could say where you are getting this problem. $\endgroup$ Commented Nov 21, 2013 at 1:28
  • $\begingroup$ It is from Stein's Complex Analysis and the chapter on the zeta function. $\endgroup$
    – Mike M.
    Commented Nov 21, 2013 at 23:01

2 Answers 2

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The proof I know requires more than the definition of the zeta function as a Dirichlet series. Here is an outline of the proof, along with the important estimates:

Firstly, you want to use the following defintion: $\displaystyle \zeta(s)=\frac{s}{s-1}-s\int_{1}^{\infty}\frac{\{t\}}{t^{s+1}}dt$.

Then, using Abel's summation you can show that $\displaystyle\sum_{n\leq x}\frac{1}{n^s}=\frac{s}{s-1}-\frac{1}{(s-1)x^{s-1}}-\frac{\{x\}}{x^{s}}-\int_{1}^{x}\frac{\{t\}}{t^{s+1}}dt$.

Combine the two expressions above to get $(*)$: $\displaystyle\zeta(s)= \sum_{n\leq x}\frac{1}{n^s}-s\int_{x}^{\infty}\frac{\{t\}}{t^{s+1}}dt+\frac{1}{(s-1)x^{s-1}}+\frac{\{x\}}{x^s}$.

The third fact, which is also proved using Abel's summation, is that $\displaystyle\sum_{n=1}^{x}{\frac{1}{n^s}} = O(\log(x))$ when $s=1+yi$. Indeed, show that $\displaystyle\sum_{n=1}^{x}{\frac{1}{n}} = O(\log(x))$ and then compare it to $\displaystyle\sum_{n=1}^{x}{\frac{1}{n^s}}$. If you aren't familiar with the Big O notation, this is equivalent to saying there is a $N\in\mathbb{N}$ and $c>0$ s.t. $\displaystyle \left|\sum_{n=1}^{x}{\frac{1}{n^s}}\right|\leq c\log(x)$ for all $x\geq N$.

To finish, you would need to show that the other three terms in $(*)$ don't grow faster than $\log(x)$.

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  • $\begingroup$ When you write $\{t\}$, is this the fractional part of $t$? $\endgroup$
    – Mike M.
    Commented Nov 24, 2013 at 4:16
  • $\begingroup$ Yes, that is what I mean. $\endgroup$ Commented Nov 24, 2013 at 4:35
  • $\begingroup$ your answer is impossible to understand. The target is $\mathcal{O}(\log t) =\mathcal{O}(\log |s|)$ $\endgroup$
    – reuns
    Commented Oct 20, 2016 at 6:37
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Let $s = \sigma+ it$. For $\sigma > 1$ and by analytic continuation for $Re(s) > 0$ : $$\zeta(s) - \sum_{n < N} n^{-s} = s \int_N^\infty \lfloor x-N \rfloor x^{-s-1}dx = \frac{N^{1-s}}{s-1} -s \underbrace{\int_N^\infty \{x\} x^{-s-1}dx}_{\textstyle< \int_N^\infty x^{-1-\sigma}dx < N^{-\sigma}}$$

Taking $N = \lceil t \rceil$, for $\sigma > 1-\frac{A}{\log |t|}$ we have for $n \le N$ : $$n^{-\sigma} = e^{-\sigma \ln n} < e^{-(1-\frac{A}{\log |t|}) \ln n} < e^{A} n^{-1} $$ and hence $|\sum_{n < N} n^{-s}| < \sum_{n=1}^{N-1} e^An ^{-1} < e^{A} \ln N$

i.e. for $\sigma > 1-\frac{A}{\log |t|}$ and $t > 1$ : $$|\zeta(s)| < 2e^A+2+\log |t|, \qquad {\scriptstyle(\text{and differentiating everything) }}\quad |\zeta'(s)| < 2e^A+2+\log^2 |t|$$

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  • $\begingroup$ Hello, I have some questions about your answer. First, I think the first equalities come from $$\zeta(s) = \sum_{1\leq n\leq x}{1\over n^s}-s\int_x^\infty{\{t\}\over t^{s+1}}\ dt+{1\over (s-1)x^{s-1}}+{\{x\}\over x^s}.$$ But if we plug in $x = N$, then $\{x\} = 0$ so it should be ${1\over s-1}N^{1-s}$. Next, I think you want $N = \lfloor |t|\rfloor$ so that $\log |t|>\log n$ for $n\leq N$. Finally, I can't understand why suddenly $|\zeta(s)|<2e^A+2+\log|t|$. I think $2$ is from ${1\over (s-1)}N^{1-s}$ but I can't see where the last integral of the first equation goes. Could you explain? $\endgroup$ Commented Nov 29, 2022 at 5:35
  • $\begingroup$ @onepotatotwopotato there was a mistake in the 1st line edited $\endgroup$
    – reuns
    Commented Nov 29, 2022 at 10:15

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