Fourier transform of $1/|x|$ as a tempered distribution

We know that as tempered distributions $$S'$$ on $$\mathbb{R}^3$$, the Fourier transform of $$1/4\pi|x|$$ is $$1/k^2$$. There are many ways to argue that this is true. One particular way I have seen in physics is that you compute

$$\int_{|x| Then by taking $$R \rightarrow \infty$$, we see that the second term $$\rightarrow 0$$ as a tempered distribution, which does kind of make sense since you expect the cosine term to oscillate so rapidly that it annihilates any Schwartz function. However, could any one provide a rigorous argument of how the oscillations annihilates Schwartz functions?

EDIT: I just realized that $$\cos{|k|R}$$ acting on a Schwartz function $$\phi (k)$$ is basically the Fourier transform of the Schwartz function $$\phi (k)$$ at $$R$$ (maybe some linear combination or you may need a bound on $$|\phi|$$). Since the Fourier transform maps Schwartz functions to Schwartz functions, we see that it must $$\rightarrow 0$$ as $$R \rightarrow \infty$$.

• One common way you could make the oscillations canceling out rigorous is through "integration by parts" and taking the distributional limit that way. For example, in 1D, $\lim_{k\to \infty} \int e^{ikx}\phi(x)dx = \lim_{k\to \infty} \frac{i}{k}\int e^{ikx} \phi'(x) dx = 0$ by dominated convergence since $\phi$ is a smooth, compactly supported function. But there is another way to compute this Fourier transform which generalizes to higher dimensions which I will type up as an answer. – Ninad Munshi Sep 18 '19 at 7:02

Instead of doing that trick, we will use a clever property of the Fourier Transform involving homogeneity.

$$\mathbf{\text{Def.}}$$ A function $$f\in\mathcal{S}(\Bbb R^n)$$ is homogeneous of degree $$s$$ if $$\forall a \neq 0$$

$$f(ax) = a^sf(x)$$

$$\mathbf{\text{Lemma.}}$$ Let $$f$$ be homogeneous degree $$s$$. Then $$\hat{f}$$ is homogeneous degree $$-n-s$$.

Proof: $$\begin{split} \hat{f}(ak) &= \frac{1}{(2\pi)^{\frac{3}{2}}}\int_{\Bbb R^n}f(x)e^{-i(ak) \cdot x}dx \\ &= \frac{1}{(2\pi)^{\frac{3}{2}}}\int_{\Bbb R^n}f\left(\frac{u}{a}\right)e^{-ik \cdot u}\frac{du}{a^n}\\ &=\frac{1}{(2\pi)^{\frac{3}{2}}}\int_{\Bbb R^n}\left(\frac{1}{a}\right)^sf(u)e^{-ik \cdot u}\frac{du}{a^n} = a^{-n-s}\hat{f}(k) \end{split}$$

Even though we proved this in the Schwartz function case, the same property applies to tempered distributions, albeit with a lot more work.

Notice that $$|x|^{-1}$$ is radial and homogeneous degree $$-1$$. Using the above lemma, we have that its Fourier transform should be homogeneous with degree $$-3+1 = -2$$. The only possible radial homogeneous distribution with that degree is $$C|k|^{-2}$$ for some constant $$C$$.