I'm trying to compute the Fourier transform of $$f(\mathbf{r}) = \frac{1}{r^\alpha}$$ where $\mathbf{r} \in \mathbb{R}^n$. For sufficiently large $\alpha$, the Fourier transform exists. One well-known example in physics is the case $\alpha = n-2$, which is the Coulomb potential; its Fourier transform is $1/k^2$.

For a general $\alpha$, we have $$\widetilde{f}(\mathbf{k}) \propto \frac{1}{k^{n-\alpha}}$$ by ‘physical arguments'. Specifically, since $f$ is rotationally invariant, $\widetilde{f}$ must be as well. Since $f$ is scale invariant, so must $\widetilde{f}$, so it must be a power law. The exponent of the power law can then be found by dimensional analysis.

I tried to do the calculation explicitly to get the proportionality constant, but it got very messy. How is that calculation done?


1 Answer 1


Provided that $\alpha$ is in the right range, it is quite easy to prove that the Fourier transform is radial and has the scaling you calculate; it then reduces to the calculation of an integral to find the constant.

For $f(r)$ integrable and radial, the Fourier transform is also radial, because we can write $$ \int_{\mathbb{R}^n} f(\lvert x \rvert) e^{-2\pi i k \cdot x} \, dx = \int_0^{\infty} f(r) r^{n-1} \left( \int_{S^{n-1}} e^{-2\pi i \lvert k \rvert r \cos{\theta}} \, dn \right) \, dr, $$ and the inside integral is a function of $r$. Which function? It turns out that it is close to a Bessel function; in fact, we have $$ \int_{S^{n-1}} e^{-2\pi i a \cos{\theta}} \, dn = 2\pi a^{1-n/2} J_{n/2-1}(2\pi a), $$ which we can show by expanding the exponential in a power series and integrating term-by-term. Thus the Fourier transform of $r^{-\alpha}$ is $$ \int_{0}^{\infty} r^{n-\alpha-1} 2\pi (\lvert k \rvert r)^{1-n/2} J_{n/2-1}(2\pi \lvert k \rvert r) \, dr, $$ and setting $u= \lvert k \rvert r$ gives the correct scaling. One would then evaluate $ 2\pi \int_{0}^{\infty} u^{n/2-\alpha} J_{n/2-1}(2\pi u) \, du $, but there is an easier way.

We have $$ \frac{1}{r^{\alpha}} = \frac{2\pi^{\alpha/2}}{\Gamma(\alpha/2)} \int_{0}^{\infty} \lambda^{\alpha-1} e^{-\pi \lambda^2 r^2} \, d\lambda, $$ and the latter is easy to Fourier transform: interchanging the order of integration, we have $$ \int_{\mathbb{R}^n} e^{-\pi \lambda^2 \lvert x \rvert^2} e^{-2\pi i k \cdot x} \, dx = \lambda^{-n}e^{-\pi \lvert k \rvert^2/\lambda^2} $$ Now, $$ \frac{2\pi^{\alpha/2}}{\Gamma(\alpha/2)} \int_{0}^{\infty} \lambda^{\alpha-n-1} e^{-\pi \lvert k \rvert^2/ \lambda^2 } \, d\lambda = \frac{2\pi^{\alpha/2}}{\Gamma(\alpha/2)} \int_{0}^{\infty} \mu^{(n-\alpha)-1} e^{-\pi \lvert k \rvert^2 \mu^2 } \, d\mu \\ = \frac{2\pi^{\alpha/2}}{\Gamma(\alpha/2)} \frac{\Gamma((n-\alpha)/2)}{2\pi^{n/2-\alpha/2}} \frac{1}{\lvert k \rvert^{n-\alpha}} \\ = \frac{\pi^{\alpha-n/2}\Gamma((n-\alpha)/2)}{\Gamma(\alpha/2)} \frac{1}{\lvert k \rvert^{n-\alpha}}, $$ setting $\mu = 1/\lambda$, and the result holds if $0<\alpha<n$.

  • $\begingroup$ To some extent this result holds for all $\alpha$, but you need to interpret the positive powers of $|k|$ as a distribution with suitable subtractions. Look up the distribution called $x^\alpha_+$ to see what this means. $\endgroup$
    – mike stone
    Commented Jul 18, 2017 at 16:25
  • $\begingroup$ I particularly enjoy the second way (+1) $\endgroup$
    – idm
    Commented Jan 1, 2018 at 13:43
  • $\begingroup$ btw, I think there is a small mistakes. Isn't it $$\frac{1}{r^\alpha }=\frac{2\pi^{\alpha /2}}{\Gamma(\alpha /2)}\int_0^\infty \lambda ^{\alpha -1}e^{-\pi\lambda ^2r^2}\mathrm d \lambda \ ?$$ $\endgroup$
    – idm
    Commented Jan 1, 2018 at 13:56
  • 2
    $\begingroup$ I learned $x_+^\lambda$ from Gerard Friedlander's lectures. These are published as "Introduction to the Theory of Distributions". There is a second edition with co-author Joshi, I only know the first edition. $\endgroup$
    – mike stone
    Commented Nov 27, 2019 at 20:42
  • 1
    $\begingroup$ No: the first line expresses the change of variables $x=rn$, where $n$ is a unit vector, and $\cos \theta = k \cdot n / |k|$. $\endgroup$
    – Chappers
    Commented Jun 16, 2020 at 9:14

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