TITLE: Bounds for $\zeta$ function on the $1$-line
QUESTION [9 upvotes]: I was going over my notes from a class on analytical number theory and we use a bound for the $\zeta$ function on the $1$ line as $\vert \zeta(1+it) \vert \leq  \log(\vert t \vert) + \mathcal{O}(1)$ for $t$ bounded away from $0$, say $\vert t \vert \geq 1$. I don't seem to have a proof for this in my notes.
How does one prove the following bound for $\zeta(s)$ on the one line?

$$\zeta(1+it) \leq \log(|t|) + \mathcal{O}(1) \, \, \forall t \geq 1$$

REPLY [10 votes]: Analytic Continuation of $\boldsymbol{\zeta(z)}$ with Euler-Maclaurin
The Euler-Maclaurin Sum formula says that, as a function of $n$,
$$
\sum_{k=1}^n\frac{1}{k^z}=\zeta^\ast(z)+\frac{1}{1-z}n^{1-z}+\frac12n^{-z}+O\left(zn^{-1-z}\right)\tag{1}
$$
for some $\zeta^\ast(z)$. Note that for $\mathrm{Re}(z)>1$, $\zeta^\ast(z)=\zeta(z)$.
For all $z\in\mathbb{C}\setminus\{1\}$, define
$$
\zeta_n(z)=\sum_{k=1}^n\frac{1}{k^z}-\frac{1}{1-z}n^{1-z}\tag{2}
$$
Note that each $\zeta_n$ is analytic and equation $(1)$ says that 
$$
\zeta_n(z)=\zeta^\ast(z)+\frac12n^{-z}+O\left(zn^{-1-z}\right)\tag{3}
$$
which says that for $\mathrm{Re}(z)>0$,
$$
\lim_{n\to\infty}\zeta_n(z)=\zeta^\ast(z)\tag{4}
$$
and the convergence is uniform on compact subsets of $\mathbb{C}\setminus\{1\}$. Thus, the $\zeta^\ast(z)$ defined in $(4)$ is analytic and agrees with $\zeta(z)$ for $\mathrm{Re}(z)>1$. Thus, $\zeta^\ast(z)$ is the analytic continuation of $\zeta(z)$ for $\mathrm{Re}(z)>0$.

Application to $\boldsymbol{|\zeta(1+is)|}$
Therefore, using $(1)$, we get that
$$
\zeta(1+it)=\sum_{k=1}^n\frac{1}{k^{1+it}}+\frac{1}{it}n^{-it}-\frac{1}{2n}n^{-it}+O\left(\frac{t}{n^2}\right)\tag{5}
$$
So as long as $n\ge|t|$, we get that
$$
\zeta(1+it)=\sum_{k=1}^n\frac{1}{k^{1+it}}+O\left(\frac1t\right)\tag{6}
$$
Using the Euler-Maclaurin Sum formula again, we get that
$$
\sum_{k=1}^n\frac1k=\log(n)+\gamma+O\left(\frac1n\right)\tag{7}
$$
Letting $n=\lceil|t|\rceil$, and using $(6)$ and $(7)$, we get for large $|t|$,
$$
|\zeta(1+it)|\le\log(|t|)+C\tag{8}
$$