Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
  1. The Fourier transform of $|x|^{\alpha}$.

    This is the Fourier transform of a homogeneous function, and there are several cases of various $\alpha$: when $a\leq -n$, it's not a temperate distribution; when $-n<\alpha<0$,then the Fourier transform is $c_{n}|\xi|^{-n+\alpha}$, where $c_{n}$ is some constant; when $\alpha=2k$,a positive even number, then it's Fourier transformation is $(-\Delta)^{k}\delta_{0}$.

    My question is when $\alpha$ is any positive number (not the even case), then what's the Fourier transform of it ?

  2. The Fourier transform of $e^{it|x|}$ ?

    (the Fourier transforms I have mentioned here are in the sense of temperate distributions)

share|improve this question
    
What is $n$? (need more characters) –  Fabian Aug 16 '12 at 11:51
    
n is the dimension of the space –  sun Aug 16 '12 at 13:17
add comment

1 Answer

${1.} $ The corresponding operator is denoted $(-\Delta)^{\alpha/2}$ and is called fractional Laplacian. Basically the Fourier transform of $|x|^\alpha$ "should be" a homogeneous function $|\xi|^{-n-\alpha}$. Since it is not locally summable it is not a kernel of an integral operator. But it defines an elliptic linear integro-differential operator of order $\alpha$. For $0<\alpha<2$ $$ (-\Delta)^{\alpha/2} u(x) = -c_{n,\alpha} \int_{\mathbb{R}^n}\frac{u(x-y)-2u(x)+u(x+y) }{|y|^{n+\alpha}}dy. $$

One-dimensional analogue can be this. Function $|x|^{-1}$ is not locally summable and thus doesn't define a regular distribution. But it can be used to define a distribution as $$ ({\cal P}\frac1{|x|},\varphi)= \int_{|x|\le 1}\frac{\varphi(x)-\varphi(0)}{|x|}\,dx+ \int_{|x|> 1}\frac{\varphi(x)}{|x|}\,dx. $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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