Is there an easy way to see that the $n$-dimensional Fourier transform of $1/|x|^a$ is equal to $1/|x|^{n-a}$ (up to multiplicative constant) where $x$ is an $n$ dimensional vector (assuming that $ 0 < a < n $) ?


1 Answer 1


Yes... you use the way the definition of Fourier transform changes under rotations and dilations. Namely, recall that $$\hat{f}(\xi) = \int_{R^n} f(x) e^{-2\pi i x \cdot \xi} \,dx$$ So that if $f(x) = |x|^{-a}$ this becomes $$\hat{f}(\xi) = \int_{R^n} |x|^{-a} e^{-2\pi i x \cdot \xi} \,dx$$ Note this is an improper integral, so you have to be careful here with how one interprets the above. Next, note that if $R$ is a rotation, then $Rx \cdot \xi = x \cdot R^{-1}{\xi}$. So changing variables in the above, turning $x$ into $Rx$, and observing that $|x| = |Rx|$, leads to $$\hat{f}(\xi) = \int_{R^n} |x|^{-a} e^{-2\pi i x \cdot R^{-1}\xi} \,dx$$ $$ = \hat{f}(R^{-1}\xi)$$ Since $R$ is any rotation this means that $\hat{f}$ is radial. Next, change variables $x \rightarrow tx$ for some $t > 0$ in the above definition of $\hat{f}$. This time you get $$\hat{f}(\xi) = \int_{R^n} |tx|^{-a} e^{-2\pi i tx \cdot \xi} t^n \,dx$$ $$ = |t|^{n-a}\int_{R^n} |x|^{-a} e^{-2\pi i x \cdot t\xi} \,dx$$ So $\hat{f}(\xi) = |t|^{n-a} \hat{f}(t\xi)$, or equivalently $\hat{f}(t\xi) = |t|^{a-n}\hat{f}(\xi)$. This means on each radial line, $\hat{f(t\xi)}$ is equal to $C|t|^{a-n}$. And since $\hat{f}(\xi)$ is a radial function, $C$ is independent of which radial line you're on. This gives what you're looking for.

  • $\begingroup$ thanks zaricuse! this type of scaling argument is exactly what I was looking for! (btw, I think you lost a factor of $t^n$ in the last equation) $\endgroup$
    – phil ansze
    Nov 21, 2010 at 21:39
  • $\begingroup$ Thanks. Ok I corrected that equation. $\endgroup$
    – Zarrax
    Nov 21, 2010 at 21:45

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