My book says that
$$\int_{r-i\infty}^{r+i\infty} O(\vert s \vert^{-r}) \vert ds \vert = K\int_{0}^{\infty} \frac{1}{(1+t)^r}dt $$
( where K, $r \space(r> 0)$ constants, s complex )
But I can't understand it no matter how I tried.
In my calculation, I set $O(\vert s \vert^{-r}) = \vert s \vert^{-r},$ $\vert ds \vert = \sqrt{dx^2+dy^2} = \frac{dy}{sin\theta}, \space \space \vert s \vert^{-r} = (\sqrt{r^2+y^2})^{-r}, \space\space y=rtan\theta, \space dy=rsec^2\theta d\theta $
And then, the left side goes to $$\int_{-\pi/2}^{+\pi/2}sin^{-1}\theta \space cos^{r-2}\theta d\theta$$
Now that Beta function has a few of identities,
$$B(z_1, z_2) = 2\int_{0}^{+\pi/2}sin^{2z_1-1}\theta \space cos^{2z_2-1}\theta d\theta = \int_{0}^{\infty} \frac{t^{z_1-1}}{(1+t)^{z_1+z_2}}dt $$
So, for the right side, $z_1=1,\space z_2=r-1 \space \Rightarrow \space sin\theta cos^{2r-3}\theta$
the right and left intergrand don't match.
Where did I make a mistake?
I do not understand your book. If $s=r+it$ then with the change of variable $u=t^2/r^2$ you get $$\int_{-\infty}^{\infty}\frac{dt}{|r+it|^r}=\frac{1}{r^{r-1}}\int_0^{\infty}\frac{u^{-1/2}}{(1+u)^{r/2}}du=\frac{1}{r^{r-1}}B(\frac{1}{2},\frac{r-1}{2})\neq K B(1,r-1)=\frac{K}{r-1}.$$