Prove that $|\zeta(\sigma+it)|=O(\log t)$ uniformly for $1\le\sigma\le 2$ and $t\ge 10$.
The solution given by my lecturer is as follows. Recall the approximate formula for zeta, given by $$\zeta(s)=\sum_{n=1}^N\frac{1}{n^s}+\frac{N^{1-s}}{s-1}-s\int_N^{\infty}\frac{\{t\}}{t^{s+1}}dt$$ for any integer $N>0$ and $\textrm{Re}(s)>0$. Substituting $s=\sigma+it$ into this formula, we have three terms which are (respectively) bounded above by $O(\log N), O(1)$ and $O(t)\int_N^{\infty}\frac{\{t\}}{t^{\sigma+1}}dt\ll t\int_N^{\infty}\frac{1}{t^2}dt=O(t/N)$.
I find this confusing - it makes sense to me that we should use the approximate formula here, but I don't understand the bounds we have here. For instance, to bound the first term, we could do an integral comparison, such as $$\sum_{n=1}^N\frac{1}{n^s}\sim\int_1^N\frac{1}{x^s}dx=\left[\frac{x^{1-s}}{1-s}\right]^{N}_1=\frac{N^{1-s}-1}{1-s},$$ and I don't see how this is $O(\log N)$.
Are we supposed to write $s=\sigma+it$ here or does it not matter? If we write $s=\sigma+it$ we would have this bound as $$\frac{N^{1-\sigma-it}-1}{1-\sigma-it}$$ but it isn't clear to me how we would get the required bound here. As for the middle term, I accept this is $O(1)$. For the last term, the presence of $t$ in the bound confuses me.
