Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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.
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 $\log(x)$.
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 $t > 1$ : $$|\zeta(s)| < 2e^A+2+\log |t|, \qquad {\scriptstyle(\text{and differentiating everything) }}\quad |\zeta'(s)| < 2e^A+2+\log^2 |t|$$
