# Fourier transform of $\frac{1}{r}$ (Coulomb potential)

When calculating the Fourier transform of a function of the form $f(\vec{r}) = \frac{1}{4 \pi \left|\vec{r}\right|}$, one encounters the problem that the resulting integral does not converge, i.e. $$\mathcal{F}\left[\frac{1}{4 \pi |\vec{r}|}\right]\!\!\left(\vec{p}\right) = \int_{\mathbb{R}^n} \frac{1}{4\pi |\vec{r}|} e^{-i \vec{p} \cdot \vec{r}}\mathrm{d}^3\! r = \frac{1}{\left|\vec{p}\right|} \int_0^\infty \sin\!\left(\left|\vec{p}\right| \cdot r\right) \mathrm{d}r$$ (see e.g. this arXiv paper, appendix A). This problem comes up when trying to express the Coulomb potential in momentum space.

In the link above, this is circumvented by introducing an attenuating factor $e^{- \lambda \left|\vec{r}\right|}$ and subsequently taking the limit $\lambda \to 0$, i.e. $$\mathcal{F}\left[\frac{1}{4 \pi |\vec{r}|}e^{- \lambda \left|\vec{r}\right|}\right]\!\!\left(\vec{p}\right) = \int_{\mathbb{R}^n} \frac{1}{4\pi |\vec{r}|} e^{- \left(\lambda\left|\vec{r}\right| + i \vec{p} \cdot \vec{r}\right)}\mathrm{d}^3\! r = \frac{1}{\left|\vec{p}\right|^2 + \lambda^2}$$ $$\stackrel{\lambda\to 0}{\Rightarrow} \mathcal{F}\left[\frac{1}{4 \pi |\vec{r}|}\right] = \frac{1}{\left|\vec{p}\right|^2}$$

My question is whether this is mathematically justified. Essentially, as I see it, this operation comes down to exchanging the order of taking the limit and the integration: $$\mathcal{F}\left[\lim_{\lambda\to 0}\frac{1}{4 \pi |\vec{r}|}e^{- \lambda \left|\vec{r}\right|}\right]\!\!\left(\vec{p}\right) \stackrel{?}{=} \lim_{\lambda\to 0}\mathcal{F}\left[\frac{1}{4 \pi |\vec{r}|}e^{- \lambda \left|\vec{r}\right|}\right]\!\!\left(\vec{p}\right)$$

I'm guessing the question can somehow be answered with the concept of uniform convergence, but mathematical details like these are not often part of my physics courses, so I'm not very familiar with its application.

• Another idea: $\frac{1}{r}$ is a locally integrable function in $\mathbb{R}^{n}/0$ which can be identified with a tempered distribution. Since $\frac{1}{r}$ is homogeneous of degree $1,$ its transform must be of the form $c \frac{1}{\vert \xi \vert ^{2}}$ (because your function is radial, the FT is rotationally invariant and it takes homogeneous distributrions of degree $k$ to homogeneous distributions of degree $-n-k$, with $n$ the dimension). – Qwertuy Nov 11 '14 at 0:34
• The constant $c$ you can compute using the following two facts: -The gaussian is an eigenfunction of the Fourier transform -Plancherel identity – Qwertuy Nov 11 '14 at 0:35
• If we interpret $f(\vec{r})\propto r^{-1}$ as a Coulomb potential, then the attenuated version is known as a Yukawa potential. So that would be a word worth searching under. – Semiclassical Nov 11 '14 at 0:47
• This Physics.SE question asks much the same question, though of course the audience in that case is different: physics.stackexchange.com/q/7462. – Semiclassical Nov 11 '14 at 1:02
• @Qwertuy Unfortunately, I don't really know the rigorous mathematical theory behind distributions (yet), so understanding that would be difficult to me. – Socob Nov 11 '14 at 1:18