Fourier transform of $1/r$ in two dimensions

Question

It it well known that in 3d, the Fourier transform of $1/r$ is $4\pi/k^2$.

How about in 2d? As it is homogeneous, we know the result should be proportional to $1/k$. But what is the pre-factor?

For that we have to finish the integral.

Answer 1

We shall adopt the convention that for an $L^1$ function, $f\in \mathbb{R^2}$, the Fourier Transform is defined as

$$\mathscr{F}\{f\}(\vec k)=\int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y)e^{i(k_xx+k_yy)}\,dx\,dy\tag1$$

Then, for $\phi\in \mathbb{S}$, and $\psi(x,y)=\frac1{\sqrt{x^2+y^2}}$, using $(1)$, we have in distribution

$$\begin{align} \langle \mathscr{F}\{\psi\},\phi\rangle&=\langle \psi,\mathscr{F}\{\phi\}\rangle\\\\ &=\int_{-\infty}^\infty \int_{-\infty}^\infty \frac1{\sqrt{x^2+y^2}}\left(\int_{-\infty}^\infty \int_{-\infty}^\infty \phi(k_x,k_y)e^{ik_xx+k_yy}\,dk_x\,dk_y\right)\,dx\,dy\tag2 \end{align}$$

Transforming $(2)$ to polar coordinates $(x,y)\mapsto(r,\theta)$ and $(k_x,k_y)\mapsto (k,\alpha)$, and applying Fubini-Tonelli (FT), we find that

$$\begin{align} \langle \mathscr{F}\{\psi\},\phi\rangle&=\int_0^{2\pi}\int_0^\infty \frac1r\left(\int_0^{2\pi}\int_0^\infty \phi(k,\alpha)e^{i\vec k\cdot\vec r}\,k\,dk\,d\alpha\right)\,r\,dr\,d\theta\\\\ &\overbrace{=}^{\text{FT}}\lim_{L\to \infty}\int_0^{2\pi}\int_0^\infty \phi(k,\alpha)\left(\int_0^{2\pi}\int_0^L e^{i\vec k\cdot\vec r}\,dr\,d\theta\right)\,k\,dk\,d\alpha\\\\ &=\lim_{L\to \infty}\int_0^{2\pi}\int_0^\infty \phi(k,\alpha)\left(\int_0^{2\pi}\int_0^L e^{i kr\cos(\theta)}\,dr\,d\theta\right)\,k\,dk\,d\alpha\\\\ &=\lim_{L\to \infty}\int_0^{2\pi}\int_0^\infty \phi(k,\alpha)\left(2\pi\int_0^L J_0(kr)\,dr\right)\,k\,dk\,d\alpha\\\\ &= \int_0^{2\pi}\int_0^\infty \left(\frac{2\pi}k\right)\phi(k,\alpha)\,k\,dk\,d\alpha\tag3 \end{align}$$

where we used the fact that $\int_0^\infty J_0(x)\,dx=1$.

Hence, we conclude from $(3)$ that in distribution,

$$\mathscr{F}\{\psi\}(k_x,k_y)=\frac{2\pi}{\sqrt{k_x^2+k_y^2}}$$

Answer 2

One can avoid doing the full computation thanks to the dual definition of the Fourier transform and using the fact that the Fourier transform of a Gaussian is a Gaussian. For example, taking the Fourier transform convention $\widehat f(y) =\int e^{-2i\pi x\cdot y} f(x)\,\mathrm d x$ then $\varphi(x) = e^{-\pi |x|^2}$ is its own Fourier transform so if you know that $\widehat f = c\,f$ then $$ c\, \langle f,\varphi\rangle = \langle \widehat f,\varphi\rangle = \langle f,\widehat\varphi\rangle = \langle f,\varphi\rangle $$ so $c=1$.