Fourier Transform of $\frac{\sin(\|x\|)}{\|x\|}$

Question

I am having a problem finding the Fourier Transform of $$f(x) = \frac{\sin(\|x\|)}{\|x\|}$$

I am interested in the cases where $x \in \mathbb{R}^n, n= 2, 3$.

Any help much appreciated !

Answer 1

Formally $$\int_{\mathbb{R}}\int_{\mathbb{R}}\frac{\sin\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}e^{-isx}e^{-ity}\,dx\,dy =\int_{-\pi}^{\pi}\int_{0}^{+\infty}\sin(\rho)e^{-is\rho\cos\theta}e^{-it\rho\sin\theta}\,d\rho\,d\theta $$ equals $$ \int_{-\pi}^{\pi}\frac{d\theta}{(s\cos\theta+t\sin\theta)^2-1}=\int_{-\pi}^{\pi}\frac{d\phi}{(s^2+t^2)\sin^2\phi-1}=\frac{-2\pi}{\sqrt{1-(s^2+t^2)}} $$ and the Fourier transform of a radial function is a radial function, but there are issues in the application of Fubini's theorem since $\frac{\sin\|x\|}{\|x\|}$ does not belong to $L^1(\mathbb{R}^2)$ or $L^1(\mathbb{R}^3)$.

Answer 2

From Bracewell, the n-dimensional Fourier Transform

$$F(s_1, \dots, s_n) = \int_{-\infty}^{\infty} \dots \int_{-\infty}^{\infty} f(x_1,\dots, x_n) e^{-2\pi i (x_1 s_1 + \dots + x_n s_n)} \space dx_1 \dots dx_n $$

when there is n-dimensional symmetry, can be manipulated into

$$F(q) = \dfrac{2\pi}{q^{\frac{1}{2}n-1}} \int_{0}^{\infty} f(r) J_{\frac{1}{2}n-1}(2\pi qr)\space r^{\frac{1}{2}n} \space dr$$

Where $r$ and $q$ are the radii from the origin in the original domain and transform domain respectively.

For the case of $n=2$, you need to compute

$$F(q) = 2\pi \int_0^\infty \dfrac{\sin(r)}{r}J_0(2\pi qr)\,r\, dr = 2\pi \int_0^\infty \sin(r)J_0(2\pi qr) \,dr$$

For the case of $n=3$ note that

$$J_{\frac{1}{2}}(x) = \sqrt{\dfrac{2}{\pi x}}\sin(x)$$

so you need to compute

$$F(q) = \dfrac{2\pi}{\sqrt{q}} \int_0^\infty \dfrac{\sin(r)}{r}\sqrt{\dfrac{2}{\pi (2\pi qr)}}\sin(2\pi qr)\,r\sqrt{r}\, dr = \dfrac{2}{q}\int_0^\infty \sin(r)\sin(2\pi qr)\, dr$$

Update

It is worth noting that the n-spherically symmetric Fourier transform (aka Hankel Transform) is its own inverse. Also for the $n=3$ case

$$\begin{align*} f(r) &= \dfrac{2\pi}{\sqrt{r}} \int_0^\infty \dfrac{\delta\left(q-\frac{c}{2\pi}\right)}{2q}\sqrt{\dfrac{2}{\pi(2\pi rq)}}\sin(2\pi rq)q\sqrt{q} \, dq\\ \\ &= \dfrac{1}{r}\int_0^\infty \delta\left(q-\frac{c}{2\pi}\right)\sin(2\pi rq) \, dq\\ \\ &= \dfrac{\sin(cr)}{r}\\ \end{align*}$$