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What is the Fourier transform of $f(x)=\frac{1}{|x|^{\alpha}}$ ? s.t $0<\alpha<n$ and $|x|=||x||_2$

I tried but with no result.. can you give me a hint?

sara
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  • Please attach your progress on the question. It allows us to help you better. Pardon my grammar :) – user79161 Jan 01 '21 at 12:36
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    Using rotation invariance and homogeneity could help (but it is not sufficient) – Romulus Augustulus Jan 01 '21 at 12:38
  • @RomulusAugustulus sry but i didnt understand...some help pls or some link pls – sara Jan 01 '21 at 12:41
  • @user79161 i tried but i had no result... so my progress does not benefit – sara Jan 01 '21 at 12:42
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    You can prove that the Fourier transform of a distribution invariant by rotation is rotation invariant, and that the Fourier transform of a homogeneous distribuzione is homogeneous (with a different degree of homogeneity). So your Fourier transform must be of the form $\frac{C}{|x|^{\beta}}$ with $\beta$ easily computable. However to compute $C$ you can't no more use symmetries – Romulus Augustulus Jan 01 '21 at 12:51
  • @RomulusAugustulus okay thank i'll try. .. i have another question if I have $u \in H^s(\mathbb{R^n})$ and here we are taking the theorem of schwartz space... so my dr. put the fourier of the derivative of u as $FD^su=|2\pi\xi|^sFu$ , so how! Coz I know it should be as if $FD^2u=(2i\pi\xi)^2Fu$ and not as a norm or absolute value! – sara Jan 01 '21 at 12:56
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    Does this answer your question? Computing Fourier transform of power law – Infiniticism Jan 02 '21 at 04:33
  • @Iridescent Yes it might I'll see it. But how did $e^{-2\pi i k \cdot x}$ become $e^{-2\pi i \lvert k \rvert r \cos{\theta}}$!! – sara Jan 02 '21 at 04:50

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