My notes state that if $s=\sigma+it$ with $\sigma>1$ and $t\in\mathbb{R}$, then for any $x\in\mathbb{N}$ we have
$$\zeta(s)=\sum_{n\leq x} \frac{1}{n^s} + \frac{x^{1-s}}{s-1} + O\left(\frac{|s|}{x^{\sigma}}\right)$$
and in particular that if $\sigma>1$ and $|t| \geq 2$ then $\zeta(\sigma+it)=O(\log t)$.
The big O notation is not clarified here and so I don't understand what it means. Is the implicit constant independent of all three variables $x,\sigma,t$? Because we later say that
$$\int_{x}^\infty \frac{dw}{w^{\sigma+1}} = \frac{x^{-\sigma}}{\sigma} = O\left(x^{-\sigma}\right)$$ in which case would I be right in saying that this is just because $\sigma>1$ and $\frac{1}{\sigma}$ is not part of the (absolute) implicit constant?
For the second part the formula is used with $x=\lfloor |t| \rfloor$ and then we argue that $\frac{x^{1-s}}{s-1} = O\left(\frac{1}{|t|}\right)$ and that $\frac{|s|}{x^{\sigma}}=O(1)$ which I don't know how to justify either (edit: actually if I am using the right definition for the big O notation, I think the first one is true with an implicit constant $1$ and I suppose the second becomes clear with some algebra. The main confusion here was that there was no justification for these two except '$\sigma>1$,' implying that it was immediate). Help would be appreciated.