Contour Integration: $\int_0^\infty\frac{1}{x^a(1-x)}\,dx$ for $0<a<1$.

You use an almost-keyhole contour, except that you indent both paths above and below the real axis with a small semicircle to avoid the pole at $z=1$:

In doing this, you end up with not $4$, but $8$ contour segments. I will avoid writing them all out by noting that the integrals over the outer circular arc and inner circular arc at the origin vanish in the limits of their radii going to $\infty$ and $0$, respectively. We are left with

$$\oint_C dz \frac{z^{-a}}{1-z} = \int_{\epsilon}^{1-\epsilon} dx \frac{x^{-a}}{1-x} + i \epsilon \int_{\pi}^0 d\phi\, e^{i \phi} \frac{(1+\epsilon e^{i \phi})^{-a}}{-\epsilon e^{i \phi}} + \int_{1+\epsilon}^{\infty} dx \frac{x^{-a}}{1-x} \\+e^{-i 2 \pi a} \int_{\infty}^{1+\epsilon} dx \frac{x^{-a}}{1-x} +e^{-i 2 \pi a} i \epsilon \int_{2 \pi}^{\pi} d\phi\, e^{i \phi} \frac{(1+\epsilon e^{i \phi})^{-a}}{-\epsilon e^{i \phi}} +e^{-i 2 \pi a} \int_{1-\epsilon}^{\epsilon} dx \frac{x^{-a}}{1-x} $$

Combining like terms, we get

$$\oint_C dz \frac{z^{-a}}{1-z} = \left ( 1-e^{-i 2 \pi a}\right ) PV\int_{0}^{\infty} dx \frac{x^{-a}}{1-x} + \left ( 1+e^{-i 2 \pi a}\right ) i \pi = 0$$

because of Cauchy's Theorem. $PV$ denotes the Cauchy principal value. After a little algebra, the result is

$$PV\int_{0}^{\infty} dx \frac{x^{-a}}{1-x} = -i \pi \frac{1+e^{-i 2 \pi a}}{1-e^{-i 2 \pi a}}=-\pi \cot{\pi a}$$

EXAMPLE

Let's check the result for $a=1/2$. This would imply that

$$PV \int _{0}^{\infty} dx \frac{1}{\sqrt{x} (1-x)} = 0$$

Consider

$$\begin{align}\underbrace{\int_0^{1-\epsilon} dx \frac{1}{\sqrt{x} (1-x)}}_{x=1/u} &= \int_{1/(1-\epsilon)}^{\infty} \frac{du}{u^2} \frac{\sqrt{u}}{1-(1/u)} \\ &= -\int_{1+\epsilon}^{\infty} du \frac{1}{\sqrt{u} (1-u)}\end{align}$$

Thus

$$\int_0^{1-\epsilon} dx \frac{1}{\sqrt{x} (1-x)} + \int_{1+\epsilon}^{\infty} du \frac{1}{\sqrt{u} (1-u)} = 0$$

or

$$PV \int _{0}^{\infty} dx \frac{1}{\sqrt{x} (1-x)} = 0$$

as was to be demonstrated.