Computing the Fourier transform of exponential decay in $\mathbb{R}^2$

Question

I am trying to compute this $\textbf{Fourier transform in}$ $\mathbb{R}^2$ $$ I(\mathbf{k})\equiv\mathcal{F}\big(e^{-a|\mathbf{x}|}\big)(\mathbf{k}) = \int_{\mathbb{R}^2} e^{-a|\mathbf{x}|}e^{i\ \mathbf{k}\cdot \mathbf{x}}\ \mathrm{d}x^2\quad,$$ where $\mathbb{R} \ni a > 0$.

Due to the rotational symmetry, I rewrite the above integral in polar coordinates and I find $$ I(\mathbf{k})= \int_{0}^{2\pi}\int_{0}^{\infty} e^{-ar\ +\ i|\mathbf{k}|r\cos{\theta}}\ r\ \mathrm{d}r\ \mathrm{d}\theta\quad,$$ where $r = |\mathbf{x}|$.

I'd like to get rid of the $\cos{\theta}$ term in the exponential with a change of variables, but setting $u=\cos{\theta}$ is not possible because I am missing a $\sin{\theta}$. This is because $\mathrm{d}u = -\sin{\theta}\ \mathrm{d} \theta$. I can make this $\sin{\theta}$ appear if I integrate by parts, but then a $\theta$ appears as well and the change of variables $u=\cos{\theta}$ still won't work.

However, this change of variables works in $\mathbb{R}^3$ because the differential in spherical coordinates contains a $\sin{\theta}$ term. This has allowed me to compute the above integral in $\mathbb{R}^3$.

Then I found that $$ \int_{0}^{2\pi} e^{-i\beta\cos{\theta}}\ \mathrm{d}\theta = 2\pi\mathcal{J}_0(\beta)\quad,$$ where $\mathcal{J}_0$ is the Bessel function of the first kind, and $\beta \in \mathbb{R}$. Thus I could possibly write $$ I(\mathbf{k}) = 2\pi \int_{0}^{\infty}e^{-ar}\mathcal{J}_0\big(i|\mathbf{k}|r\big)\ r\ \mathrm{d}r \quad,$$ but unfortunately I can't proceed further.

What I'd like is to compute the above integral. Is there any other way I could obtain a closed form solution? It seems like a closed form solution in $\mathbb{R}^1$ and $\mathbb{R}^3$ exists, what about $\mathbb{R}^2$ though?

Thanks in advance!

Answer 1

The general answer in dimension $d$ is $$ \boxed{\mathcal F(e^{-a\,|x|}) = \frac{2\,a \,(2\,\pi)^d}{\omega_{d+1}\,(a^2 + |x|^2)^{(d+1)/2}}} $$ where $\omega_{d} = \frac{2\,\pi^{d/2}}{\Gamma(d/2)}$ is the size of the unit sphere in $\mathbb R^d$. In particular in dimension $2$ and $3$ you get $$ \begin{align*} \mathcal F(e^{-a\,|x|}) &= \frac{2\,a \,\pi}{\left(a^2 + |x|^2\right)^{3/2}} &\text{if } d=2 \\ \mathcal F(e^{-a\,|x|}) &= \frac{8\,a \,\pi}{\left(a^2 + |x|^2\right)^{2}} &\text{if } d=3 \end{align*} $$

---

For the proof I will use another definition of the Fourier transform which I am more familiar with $$ \mathfrak F(u)(x) = \int_{\mathbb R^d} e^{-2i\pi\,x\cdot y}\,u(y)\,\mathrm d y = \mathcal{F}(u)(-2\pi\,x) $$ Then the results reads $$ \mathfrak F(e^{-a\,|x|}) = \frac{2\,a \,(2\,\pi)^d}{\omega_{d+1}\,(a^2 + |2\pi\,x|^2)^{(d+1)/2}}, $$ By the scaling properties of the Fourier transform, it is sufficient to look for example at $a=2\pi$ and to prove $$ \mathfrak F(e^{-2π\,|x|}) = \frac{2}{\omega_{d+1}\,\langle x\rangle^{d+1}}, $$ where $\langle x\rangle = \sqrt{1+|x|^2}$, or equivalently, by the inversion theorem, $$ \mathfrak F\!\left(\frac{2}{\omega_{d+1}\,\langle x\rangle^{d+1}}\right) = e^{-2π\,|x|}, $$ A classic strategy is to make appear Gaussians. To do that, one can use the fact that $$ \frac{2}{\omega_c\,r^{c/2}} = \int_0^\infty t^{\frac{c}{2}-1} e^{-\pi r t}\, \mathrm d t $$ which follows by the simple change of variable $s = \pi\,r\,t$ in the left-hand side integral and the definition of the Gamma function. Hence, taking $r = 1+|x|^2$ and $c=d+1$ yields $$ \frac{2}{\omega_{d+1}\,\langle x\rangle^{d+1}} = \int_0^\infty t^{\frac{d-1}{2}} e^{-\pi t}\,e^{-\pi |x|^2 t}\, \mathrm d t $$ Now, just use the formula for the Fourier transform of Gaussians to get $\mathfrak F\!\left(\frac{2}{\omega_{d+1}\,\langle x\rangle^{d+1}}\right) = G(|x|)$ with $$ G(r) = \int_0^\infty t^{-\frac{1}{2}} e^{-\pi t}\,e^{-\pi r^2/ t}\, \mathrm d t $$ To verify that this is the exponential we were looking for, take the derivative to get $$ G'(r) = -2\pi\,r\,\int_0^\infty t^{-\frac{3}{2}} e^{-\pi t}\,e^{-\pi r^2/ t}\, \mathrm d t $$ and so by doing the change of variable $s = r^2/t$, $$ G'(r) = -2\pi\,\,\int_0^\infty s^{\frac{3}{2}-2} e^{-\pi r^2/s}\,e^{-\pi s}\, \mathrm d s = -2π\, G(r) $$ so $G(r) = C\,e^{-2π\,r}$ with $C = G(0) = ∫\frac{2}{\omega_{d+1}\,\langle x\rangle^{d+1}}\,\mathrm d x = 1 $, so $$ G(|x|) = e^{-2\pi\,|x|} $$

Answer 2

$$I(k,a)= \int_{0}^{2\pi}\int_{0}^{\infty} e^{-ar\ +\ ikr\cos{\theta}}\ r\ \mathrm{d}r\ \mathrm{d}\theta\quad$$ Where $k$ denotes $|\vec k| ; \,\,k,a>0$ $$I(k,a)=\int_{0}^{2\pi}d\theta\int_{0}^{\infty}e^{-r(a-ik\cos\theta)}rdr$$ $$=(z=r(a-ik\cos\theta))\int_{0}^{2\pi}\frac{d\theta}{(a-ik\cos\theta)^2}\int_{0}^{\infty}e^{-z}zdz=-\frac{4}{k^2}\int_{0}^{2\pi}\frac{e^{2i\theta}d\theta}{(e^{2i\theta}+\frac{2ia}{k}e^{i\theta}+1)^2}\,\,\,(z=e^{i\theta})$$ $$= -\frac{4}{ik^2}\oint_{|z|=1}\frac{zdz}{(z^2+\frac{2ia}{k}z+1)^2}$$ The roots of the denominator are $$z_{\pm}=-i\Big(\frac{a}{k}\pm\sqrt{\frac{a^2}{k^2}+1}\Big)$$ Only one root (pole of order 2) $z_{-}=-i\Big(\frac{a}{k}-\sqrt{\frac{a^2}{k^2}+1}\Big)$ lies inside the contour, therefore, taking the residue at $z=z_{-}$ $$I(k,a)=\frac{4i}{k^2}2\pi i\frac{d}{dz}\frac{z}{(z-z_{+})^2}\Big|_{z=z_{-}}=\frac{8\pi}{k^2}\frac{z_{-}+z_{+}}{(z_{-}-z_{+})^3}=\frac{8\pi}{k^2}\frac{\frac{2a}{k}}{8(\sqrt{\frac{a^2}{k^2}+1})^3}$$ $$I(k,a)=\frac{2\pi a}{(a^2+k^2)^{3/2}}$$ Quick check: at $k=0 \,\,\,I(0,a)=\frac{2\pi}{a^2}$, as it has to be.