Compute the integral $\int_{-1}^1 \frac{|x-y|^{\alpha}}{(1 - x^2)^{\frac{1+\alpha}{2}}}dx = \frac{\pi}{\cos(\pi \alpha/2)}$ $$
\mbox{How to prove that}\
\int_{-R}^{R}\frac{\left\vert x - y\right\vert^{\alpha}}
{\left(R^{2} - x^{2}\right)^{\large\left(1+\alpha\right)/2}}
\,\mathrm{d}x = \frac{\pi}{\cos\left(\pi \alpha/2\right)}\
{\Large ?},
$$
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.

May be this is a more elementary solution.
Change the variable $x$ to $Rx$ and put $c = \dfrac{y}{R}$.
We are ready if we can prove that
\begin{equation*}
f(c) = \int_{-1}^{1}\dfrac{|x-c|^{\alpha}}{(1-x^2)^{\frac{1+\alpha}{2}}}\, dx = \underbrace{ \int_{-1}^{c}\dfrac{(c-x)^{\alpha}}{(1-x)^{\frac{1+\alpha}{2}}(1+x)^{\frac{1+\alpha}{2}}}\, dx}_{= I_1} + \underbrace{\int_{c}^{1}\dfrac{(x-c)^{\alpha}}{(1-x^2)^{\frac{1+\alpha}{2}}}\, dx}_{= I_2}
\end{equation*}
does not depend on $c\in (-1,1)$. Before we prove that $f'(c) = 0$ we make some variable substitutions.
In $I_1$ we put
\begin{equation*}
x= \dfrac{1+c}{t+1}-1.
\end{equation*}
Then
\begin{equation*}
I_1 = (1+c)^{\frac{\alpha+1}{2}}\int_{0}^{\infty}\dfrac{t^{\alpha}}{(2t+1-c)^{\frac{\alpha+1}{2}}(t+1)}\, dt.
\end{equation*}
In the integral $I_2$ we use the substitution $x=-t$.
Then
\begin{equation*}
I_2 = \int_{-1}^{-c}\dfrac{((-c)-t)^{\alpha}}{(1-t^2)^{\frac{1+\alpha}{2}}}\, dt = I_1(-c).
\end{equation*}
Now
\begin{gather*}
\dfrac{dI_1}{dc} = \dfrac{\alpha+1}{2}(1+c)^{\frac{\alpha-1}{2}}\int_{0}^{\infty}\dfrac{t^{\alpha}}{(2t+1-c)^{\frac{\alpha+1}{2}}(t+1)}\, dt +\\[2ex] \dfrac{\alpha+1}{2}(1+c)^{\frac{\alpha+1}{2}}\int_{0}^{\infty}\dfrac{t^{\alpha}}{(2t+1-c)^{\frac{\alpha+3}{2}}(t+1)}\, dt =\\[2ex]
\dfrac{\alpha+1}{2}(1+c)^{\frac{\alpha-1}{2}}\int_{0}^{\infty}\dfrac{t^{\alpha}}{(2t+1-c)^{\frac{\alpha+1}{2}}(t+1)}\left(1+\dfrac{1+c}{2t+1-c}\right)\, dt =\\[2ex]
(\alpha+1)(1+c)^{\frac{\alpha-1}{2}}\int_{0}^{\infty}\dfrac{t^{\alpha}}{(2t+1-c)^{\frac{\alpha+3}{2}}}\, dt 
\end{gather*}
After a final scaling $t=\dfrac{1-c}{2}s$ we get
\begin{equation*}
\dfrac{dI_1}{dc} = (\alpha+1)(1-c^2)^{\frac{\alpha-1}{2}}\cdot 2^{-\alpha-1}\int_{0}^{\infty}\dfrac{s^{\alpha}}{(1+s)^{\frac{\alpha +3}{2}}}\, ds.
\end{equation*}
Furthermore
\begin{equation*}
I'_{2}(c) = -I'_1(-c) = -I'_1(c).
\end{equation*}
Consequently $f'(c) = 0$ and $f$ is constant.
