The problem is to compute the following integral:

$$\int_{-1}^1 dx \frac{|x-y|^{\alpha}}{(1-x^2)^{(1+\alpha)/2}}$$

I will show how to compute this integral using Cauchy’s theorem. It was remarked that it should not be possible to use Cauchy’s theorem, as Cauchy’s theorem only applies to analytic functions, and an absolute value certainly does not qualify. True. Nevertheless, for the special case of the integral in question, things work out quite nicely as you will see.

Consider the contour integral

$$\oint_C dz \frac{(z-y)^{\alpha}}{(z^2-1)^{(1+\alpha)/2}} $$

where $C$ is the following contour:

The radii of the small arcs is $\epsilon$ and the large arc is $R$. The left arc is centered at $z=-1$, the right arc is centered at $z=1$, and the center arc is centered at $z=y$.

Note that the integrals about the small arcs vanish in the limit as $\epsilon \to 0$. The integrals along the segment $AB$ cancels with that along $KL$. Thus, the contour integral is equal to, in this limit,

$$e^{-i \left ( \frac{1+\alpha}{2}\right ) \pi} \left [e^{i \pi \alpha} \int_{-1}^y dx \frac{(y-x)^{\alpha}}{(1-x^2)^{(1+\alpha)/2}} + \int_y^1 dx \frac{(x-y)^{\alpha}}{(1-x^2)^{(1+\alpha)/2}}\right ]\\ – e^{i \left ( \frac{1+\alpha}{2}\right ) \pi} \left [e^{-i \pi \alpha} \int_{-1}^y dx \frac{(y-x)^{\alpha}}{(1-x^2)^{(1+\alpha)/2}} + \int_y^1 dx \frac{(x-y)^{\alpha}}{(1-x^2)^{(1+\alpha)/2}}\right ]\\ + i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \frac{(R e^{i \theta}-y)^{\alpha}}{(R^2 e^{i 2 \theta}-1)^{(1+\alpha)/2}}$$

In the limit as $R \to \infty$, the last integral about the big arc approaches $i 2 \pi$. Meanwhile, the first four integrals may be greatly simplified to be

$$-i 2 \sin{\left [\left ( \frac{1-\alpha}{2} \right ) \pi \right ]} \int_{-1}^y dx \frac{(y-x)^{\alpha}}{(1-x^2)^{(1+\alpha)/2}} – i 2 \sin{\left [\left ( \frac{1+\alpha}{2} \right ) \pi \right ]} \int_y^1 dx \frac{(x-y)^{\alpha}}{(1-x^2)^{(1+\alpha)/2}}$$

which may be written in more compact form such that the contour integral is equal to

$$-i 2 \cos{\left ( \frac{\pi \alpha}{2} \right )} \int_{-1}^1 dx \frac{|x-y|^{\alpha}}{(1-x^2)^{(1+\alpha)/2}} + i 2 \pi$$

By Cauchy’s theorem, the contour integral is zero. Therefore,

$$\int_{-1}^1 dx \frac{|x-y|^{\alpha}}{(1-x^2)^{(1+\alpha)/2}} = \frac{\pi}{\cos{\left ( \frac{\pi \alpha}{2} \right )}} $$

Note how the absolute value was introduced naturally because we got to the point where we could simply add the pieces of the integral back together. In general, the above methodology will not work. But it is nice to be able to recognize when it will work.