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 ?
The Fourier transform of $e^{it|x|}$ ?
(the Fourier transforms I have mentioned here are in the sense of temperate distributions)
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Davide Giraudo
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Tomas
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What is $n$? (need more characters) – Fabian Aug 16 '12 at 11:51
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n is the dimension of the space – Tomas Aug 16 '12 at 13:17
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For $|x|^\alpha$, see THIS ANSWER. – Mark Viola May 08 '21 at 17:16
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${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 $$ ({\cal P}\frac1{|x|},\varphi)= \int_{|x|\le 1}\frac{\varphi(x)-\varphi(0)}{|x|}\,dx+ \int_{|x|> 1}\frac{\varphi(x)}{|x|}\,dx. $$
Andrew
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