This is about complex integral.
(The following is in the middle of proof relating to the poles left to the fundermental strip in Mellin Transform.)
Assume $f(s) = O({\vert s\vert}^{-r})\space$ with $\space r\gt 1 \space$, when $\vert s\vert \to \infty \space$ at $\space Re(s) = r$
Then, $$\lim_{T\to\infty} \vert \frac{1}{2\pi i}\int_{r-iT}^{r+iT} f(s)x^{-s}ds \vert \space\space \le \space\space \frac{1}{2\pi}\int_{r-i\infty}^{r+i\infty} \vert f(s) \vert x^{-Re(s)} \vert ds \vert \space\space = \space O(1)\int_{0}^{\infty}\frac{x^{-r}}{(1+t)^r}dt \space = \space O(x^{-r}) $$
I really can't understand the first equality.
I might not have enough knowledge of complex analysis,
I guess it reduces to $\int_{-\pi/2}^{+\pi/2}(sec\theta)^{-r} d\theta \space$ thing, and this is a dead end to me.
Can anyone explain the first equality?