Can anyone help me compute the Fourier transform of $ 1/|x|^{n-\alpha} $ in $\mathbb{R}^n $ where $ 0 < \alpha < n $ ? Somehow it becomes the principal value of $ 1/|x|^\alpha $ which I can't figure out.

  • $\begingroup$ Hint: Isn't $1/|x|^{n-\alpha}$ regarded as a frequency versus $|x|$? $\endgroup$ Jul 12, 2013 at 11:36

1 Answer 1


This is not all that simple a calculation. A standard technique (see, e.g., Stein's Singular Integrals, III.3) is to remember that the Fourier transform of the Gaussian is another Gaussian, and to use that as an auxiliary function. That is to say, if

$$f(x) = e^{-\pi \delta |x|^2},$$ then $$\hat{f}(\xi) = \delta^{-\frac{n}{2}} e^{-\frac{\pi |\xi|^2}{\delta}}.$$

This is useful because if one invokes a change of variable in the definition of the Gamma function one retrieves the following identity: $$ \int_0^\infty e^{-\pi \delta |x|^2} \delta^{\beta -1} d\delta = \frac{\Gamma(\beta)}{(\pi|x|^2)^\beta}$$ Taking the Fourier transform of the above expression with respect to $x$ yields $$\int_0^\infty \delta^{\beta - \frac{n}{2} -1} e^{- \pi |\xi|^2 / \delta} d\delta = \int_0^\infty s^{-1 + \frac{n}{2} -\beta} e^{-\pi s |\xi|^2} ds = \frac{\Gamma(\frac{n}{2} -\beta)}{(\pi |\xi|^2)^{\frac{n}{2} -\beta}} $$ Setting $\beta = \frac{n-\alpha}{2}$, we get the desired result.

  • 3
    $\begingroup$ And the inevitability of the result follows from nice general properties of Fourier transform: $1/|x|^{n-\alpha}$ is a tempered distribution, rotation-invariant, and homogeneous under dilations. Fourier transform preserves rotation-invariance, and converts homogeneity of degree $n-\alpha$ to $\alpha$. Then the constant can be determined by the computation at the end of @RayYang's answer. $\endgroup$ Jul 12, 2013 at 14:51
  • $\begingroup$ @RayYang How did you identify your expression with the fourier transform of $ 1/|x|^{n-\alpha} $ ? $\endgroup$
    – smiley06
    Jul 13, 2013 at 11:53
  • $\begingroup$ Um, which specific part of the above is unclear, smiley? $\endgroup$
    – Ray Yang
    Jul 13, 2013 at 22:29
  • $\begingroup$ Got it now, thanks. $\endgroup$
    – smiley06
    Jul 16, 2013 at 14:33

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