I'm trying to rewrite an integral of the form $$\int_{-∞}^∞ (x^2+k^2)^{-s/2}e^{i(xc_1+kc_2)}dx$$ ($s>0,c_1,c_2,k\in\mathbb R$) in such a way that it is positive (for some $s$).
To do so I'd like use a contour integral like the following one and use that (because of Cauchy $\oint=0$ and exponential decay) we have $\int_{-iR+\varepsilon}^{-ik_2+\varepsilon}+\int_{-ik_2-\varepsilon}^{-iR-\varepsilon}=-\int_\mathbb{R}$ (in the limits $R\rightarrow\infty, \varepsilon\rightarrow 0$).
Choose the branch cut of the root to be such that it "goes down" (like in the picture).
I thought to get a branch cut like that I'd have to set $z^{-s/2}=|z|^{-s/2}(e^{i\text{Arg}(z)})^{-s/2}$ for $\text{Arg}z\in[-\pi/2,3\pi/2)$, but then the integrals along the branch cut would get factors of $e^{i\frac{-\pi}{2}\frac{-s}{2}}$ and $e^{i\frac{3\pi}{2}\frac{-s}{2}}$ (one of which gets multiplied by $-1$ in order to interchange the bounds of integration so that the two integrals can be added up). Now this doesn't look very nice, compared to choosing $\text{Arg}z\in[-\pi,\pi)$ which would give something like $e^{\pm iπs/2}$ which then adds up to $2\sin(πs/2)$.
(The nice thing about choosing the branch cut like in the picture is that the imaginary part in the integrand vanishes which yields the desired positivity for appropriate $s$.)
Did I misunderstand something about how to choose branch cuts?