Choosing branch cuts for complex integration When calculating integrals like $$\int_0^\infty \frac{x^\alpha}{1 + x^2}dx$$ for $\alpha \in (-1,1 )$, it is convenient to take the branch cut of the integrand along the positive real axis and then use the keyhole contour.
I was wondering is there a way to use the principal branch cut, which runs along the negative real axis, to calculate this kind of integrals? I'm assuming some trivial manipulation of the integrand for $x>0$ would do the trick, but I fail to see it.

Using the principal branch, we can write the integral $\oint_C \frac{z^a}{z^2+1}\,dz$, where $C$ is comprised of (i) the real line segment from $-R$ to $R$ and (ii) the semi-circle in the upper-half plane, centered at the origin and with radius $R$, as
$$\begin{align}
\oint_C \frac{z^a}{z^2+1}\,dz&=\int_{-R}^0 \frac{x^a}{x^2+1}\,dx+\int_0^R \frac{x^a}{x^2+1}\,dx+\int_0^\pi \frac{(Re^{i\phi})^a}{(Re^{i\phi}))^2+1}\,(iRe^{i\phi}))\,d\phi\\\\
&=(1+e^{i\pi a})\int_0^R \frac{x^a}{x^2+1}\,dx+\int_0^\pi \frac{(Re^{i\phi})^a}{(Re^{i\phi}))^2+1}\,(iRe^{i\phi}))\,d\phi\tag1
\end{align}$$
As $R\to \infty$ the second integral on the right-hand side of $(1)$ approaches $0$.  Hence, taking this limit and invoking the reside theorem we find that 
$$\begin{align}
\int_0^R \frac{x^a}{x^2+1}\,dx&=\frac1{1+e^{i\pi a}}\,(2\pi i) \text{Res}\left(\frac{z^a}{z^2+1}, z=i\right)\\\\
&=\frac{\pi e^{i\pi a/2}}{1+e^{i\pi a}}\\\\
&=\frac{\pi}{2\cos(\pi a/2)}
\end{align}$$
