Computing Fourier transform of power law

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?

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.

• 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. Commented Jul 18, 2017 at 16:25
• I particularly enjoy the second way (+1)
– idm
Commented Jan 1, 2018 at 13:43
• btw, I think there is a small mistakes. Isn't it $$\frac{1}{r^\alpha }=\frac{2\pi^{\alpha /2}}{\Gamma(\alpha /2)}\int_0^\infty \lambda ^{\alpha -1}e^{-\pi\lambda ^2r^2}\mathrm d \lambda \ ?$$
– idm
Commented Jan 1, 2018 at 13:56
• 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. Commented Nov 27, 2019 at 20:42
• No: the first line expresses the change of variables $x=rn$, where $n$ is a unit vector, and $\cos \theta = k \cdot n / |k|$. Commented Jun 16, 2020 at 9:14