TITLE: Compute the integral $\int_{-1}^1 \frac{|x-y|^{\alpha}}{(1 - x^2)^{\frac{1+\alpha}{2}}}dx = \frac{\pi}{\cos(\pi \alpha/2)}$
QUESTION [11 upvotes]: $$
\mbox{How to prove that}\
\int_{-R}^{R}\frac{\left\vert x - y\right\vert^{\alpha}}
{\left(R^{2} - x^{2}\right)^{\large\left(1+\alpha\right)/2}}
\,\mathrm{d}x = \frac{\pi}{\cos\left(\pi \alpha/2\right)}\
{\Large ?},
$$
where $-1 < \alpha < 1$, $-R \le y \le R$.
Since the right hand side does not depend on $y$, I suppose, there must be some physical interpretation. I'll be grateful for any hints.

REPLY [2 votes]: Here is a geometric proof of the identity.

Let $I$ denote the integral. Substituting $x=R\sin\theta$ and $k=-y/R \in [-1, 1]$, we get
$$ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left| \frac{\sin\theta-k}{\cos\theta} \right|^{\alpha} \, \mathrm{d}\theta. $$
In order to simplify $I$, we introduce the function
$$ F_k(t) := \operatorname{length}\left(\left\{ \theta \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right) : \left| \frac{\sin\theta-k}{\cos\theta} \right| \leq t\right\}\right) $$
and note that $I$ is recast as
$$ I = \int_{0}^{\infty} t^{\alpha} \, \mathrm{d}F_k(t) $$
by the "change of variables". Now we will make use of the following claim:

Claim. $F_k$ does not depend on $k \in [-1, 1]$.

Once this claim is proved, we may conveniently use $F_0(t) = 2\arctan t$ to find that
$$ I
= \int_{0}^{\infty} t^{\alpha} \, \mathrm{d}F_0(t)
= 2\int_{0}^{\infty} \frac{t^{\alpha}}{t^2+1} \, \mathrm{d}t
= \frac{\pi}{\cos(\pi\alpha/2)}, $$
where the last step can be verified in a routine way. $\square$

Proof of Claim. Note that $\frac{\sin\theta-k}{\cos\theta}$ is the slope of the line joining $(0, k)$ to $(\cos\theta, \sin\theta)$. Under this geometric interpretation, $F(t)$ corresponds to the length of the circular arc $AB$ shown below:

Let $l \in [-1, 1]$ and assume WLOG that $l < k$. Then $F_k(t) - F_{l}(t)$ is equal to the length of the arc $AC$ minus the length of the arc $BD$:

Now we reflect the arc $BD$ about the $y$-axis to obtain the arc $B'D'$. Then the two arcs $AC$ and $B'D'$ are congruent by the symmetry, and hence, they have the same length. This proves that $F_k(t) - F_l(t) = 0$ and therefore the claim follows. $\square$