The integral of x^(alpha-1) / (1 + x) over (0, infinity) is pi / sin(pi alpha)

The factor $R(z)=1\mathrm{/}(1+z)$ has no pole on $[0,\mathrm{\infty })$. Since $0<\mathrm{Re}\alpha <1$, the absolute value of the real integrand is $O(x^{\mathrm{Re}\alpha -1})$ near $0$ and $O(x^{\mathrm{Re}\alpha -2})$ at infinity, so the improper integral converges. On the inner keyhole circle the arc integral is $O({\epsilon }^{\mathrm{Re}\alpha })$, and on the outer circle it is $O(R^{\mathrm{Re}\alpha -1})$; both tend to $0$. Thus every hypothesis of [L1] holds.

The only pole away from the positive real axis is the simple pole at $-1$. On the chosen branch, $(-1{)}^{\alpha -1}=e^{i\pi (\alpha -1)}=-e^{i\pi \alpha }$, so $\mathrm{Res}(f,-1)=-e^{i\pi \alpha }$.

Applying [L1] using step 1.1 and substituting the residue from step 1.2 gives $(1-e^{2\pi i\alpha }){\int }_0^{\mathrm{\infty }}\frac{x^{\alpha -1}}{1+x}dx=-2\pi i{e}^{i\pi \alpha }\mathrm{.}$ Since $1-e^{2\pi i\alpha }=-2i{e}^{i\pi \alpha }\sin (\pi \alpha )$, division yields ${\int }_0^{\mathrm{\infty }}\frac{x^{\alpha -1}}{1+x}dx=\frac{\pi }{\sin (\pi \alpha )}$.