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.
 A: Without loss of generality one can assume $R=1$.
We will calculate Fourier transform of the function
$$
f(\beta)=\int_{-1}^1 \frac{|x-y|^{\alpha+i\beta}}{(1 - x^2)^{\frac{1+\alpha+i\beta}{2}}}dx
$$
and show that it is independent of $-1\le y\le 1$.
We have 
$$
F(t)=\int_{-\infty}^\infty f(\beta)e^{-i\beta t}dt=\int_{-1}^1 \frac{|x-y|^{\alpha}}{(1 - x^2)^{\frac{1+\alpha}{2}}}\cdot 2\pi\delta\left(t-\ln|x-y|+\frac{1}{2}\ln(1-x^2)\right) dx
$$
where $\delta$ is delta function. The roots of $t-\ln|x-y|+\frac{1}{2}\ln(1-x^2)=0$ can be found by solving quadratic equation
$$
x^2(1+e^{2t})-2xy+y^2-e^{2t}=0.
$$
They are $x_{1,2}=\frac{y\pm e^{t}\sqrt{1+e^{2t}-y^2}}{1+e^{2t}}$. Also $\frac{|x_{1,2}-y|}{(1 - x_{1,2}^2)^{\frac{1}{2}}}=e^t$. 
We now prove that $|x_{1,2}|\le 1$. It is enough to consider $0\le y\le 1$ and prove that $\frac{y+ e^{t}\sqrt{1+e^{2t}-y^2}}{1+e^{2t}}<1$. It can be easily checked that this is indeed true.
So we have using $\delta(g(t))=\sum_{t_i}\frac{\delta(t-t_i)}{|g'(t_i)|}$, where the sum is over the real roots of the function $g(t)$, that
$$
\delta\left(t-\ln|x-y|+\frac{1}{2}\ln(1-x^2)\right)=\sum_{i=1,2}\frac{(1-x_i^2)|x_i-y|}{1-x_iy}\delta(x-x_i).
$$
As a result
$$
F(t)=2\pi e^{\alpha t} \sum_{i=1,2}\frac{(1-x_i^2)|x_i-y|}{(1-x_iy)\sqrt{1-x_i^2}}=e^{(\alpha+1) t} \sum_{i=1,2}\frac{1-x_i^2}{1-x_iy}.
$$
Now using Viete's theorem one can show by trivial algebra that 
$$
\sum_{i=1,2}\frac{1-x_i^2}{1-x_iy}=\frac{2}{1+e^{2t}}.
$$
So $F(t)=2\pi \frac{e^{\alpha t}}{\cosh t}$ is independent of $y$. Calculating inverse Fourier transform we have for $-1 < \alpha < 1$
$$
f(\beta)=\int_{-\infty}^\infty  \frac{e^{(\alpha+i\beta) t}}{\cosh t}dt=\frac{\pi}{\cos\frac{\pi(\alpha+i\beta)}{2}}.
$$
Now put $\beta=0$ to complete the proof.
A: 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.
A: Contour integration works here.  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,

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

A: 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. $\square$
