# What's the Fourier transform of these functions?

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)

• What is $n$? (need more characters) Commented Aug 16, 2012 at 11:51
• n is the dimension of the space Commented Aug 16, 2012 at 13:17
• For $|x|^\alpha$, see THIS ANSWER. Commented May 8, 2021 at 17:16

${1.}$ 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 $$({\cal P}\frac1{|x|},\varphi)= \int_{|x|\le 1}\frac{\varphi(x)-\varphi(0)}{|x|}\,dx+ \int_{|x|> 1}\frac{\varphi(x)}{|x|}\,dx.$$