11
$\begingroup$

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

$\endgroup$
9
  • 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 '16 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 $\endgroup$ – theveryanonym Oct 17 '16 at 13:47
  • $\begingroup$ Have you tried to show that $I'(y)=0$? $\endgroup$ – tired Oct 17 '16 at 14:04
  • $\begingroup$ A change of variables, Euler's beta function and the reflection formula for the $\Gamma$ function give a way. $\endgroup$ – Jack D'Aurizio Oct 17 '16 at 19:15
  • 2
    $\begingroup$ @JackD'Aurizio if $y = 0$, indeed. Otherwise I cannot see it. $\endgroup$ – theveryanonym Oct 17 '16 at 21:23
5
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ very nice, thank you @Nemo! $\endgroup$ – theveryanonym Oct 18 '16 at 14:34
  • $\begingroup$ Could you elaborate on why you can assume $R=1$ with loss of generality? $\endgroup$ – Daniel R Oct 18 '16 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 $ $\endgroup$ – theveryanonym Oct 19 '16 at 9:54
  • $\begingroup$ @theveryanonym Ah, as simple as that. Thanks. $\endgroup$ – Daniel R Oct 19 '16 at 9:56
5
$\begingroup$

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,

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

$\endgroup$
0
4
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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. $\square$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.