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

  • 2
    $\begingroup$ I'm curious where you found the integral. You mention some physical interpretation and also gave the mathematical physics tag. Could you clarify? $\endgroup$
    – mickep
    Oct 17, 2016 at 13:40
  • 2
    $\begingroup$ I found it in Matheron, 1974, page 10. Here is the link cg.ensmp.fr/bibliotheque/public/MATHERON_Rapport_00186.pdf. Physical interpretation is just a guess, it looks a bit like Riesz potential, for example. I thought they might be a simple physical explanation why the integral does not depend on $y$, but have no idea what it could be $\endgroup$ Oct 17, 2016 at 13:47
  • $\begingroup$ Have you tried to show that $I'(y)=0$? $\endgroup$
    – tired
    Oct 17, 2016 at 14:04
  • $\begingroup$ A change of variables, Euler's beta function and the reflection formula for the $\Gamma$ function give a way. $\endgroup$ Oct 17, 2016 at 19:15
  • 2
    $\begingroup$ @JackD'Aurizio if $y = 0$, indeed. Otherwise I cannot see it. $\endgroup$ Oct 17, 2016 at 21:23

4 Answers 4


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.

  • $\begingroup$ very nice, thank you @Nemo! $\endgroup$ Oct 18, 2016 at 14:34
  • $\begingroup$ Could you elaborate on why you can assume $R=1$ with loss of generality? $\endgroup$
    – Daniel R
    Oct 18, 2016 at 17:04
  • $\begingroup$ @DanielR change of variable $x = tR$, $\int_{-R}^R \frac{|x - y|^{\alpha}}{(R^2 - x^2)^{\frac{1+\alpha}{2}} } = \int_{-R}^R \frac{|x - y|^{\alpha}}{R^{1 + \alpha}(1- x^2/R^2)^{\frac{1+\alpha}{2}} } = \int_{-1}^1 \frac{|tR - y|^{\alpha}}{R^{1 + \alpha}(1- t^2)^{\frac{1+\alpha}{2}} } R dt = \int_{-1}^1 \frac{R^{\alpha}|t - y/R|^{\alpha}}{R^{\alpha}(1- t^2)^{\frac{1+\alpha}{2}} } dt = \int_{-1}^1 \frac{|t - y'|^{\alpha}}{(1- t^2)^{\frac{1+\alpha}{2}} } dt $ $\endgroup$ Oct 19, 2016 at 9:54
  • $\begingroup$ @theveryanonym Ah, as simple as that. Thanks. $\endgroup$
    – Daniel R
    Oct 19, 2016 at 9:56

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.


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:

enter image description here

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,

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


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:

Figure 1: Geometric Interpretation of F(t)

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 $BD$:

Figure 2

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. $\square$


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