inverse Fourier transform of $\frac{x^3y}{(x^2+y^2)^2}$

Question

Let $f(x,y)=\frac{x^3y}{(x^2+y^2)^2}$. What is the inverse Fourier transform of $f$?

Since $f$ is neither a $L^1$-function nor a Schwartz-function I am looking for an inverse Fourier transform in the sense of tempered distributions. I’ve tried several distributions but every single one failed. Does anyone have a hint?

Answer 1

We have $$\frac 1 {2 \pi} \int_{-\infty}^\infty \int_{-\infty}^\infty f(r) e^{i (x, y) \cdot (\xi, \zeta)} dx dy = \int_0^\infty r f(r) J_0( \rho r) dr.$$ If the one-dimensional functional $r_+^{-3}$ is understood as $$(r_+^{-3}, f) = \int_0^\infty r^{-3} \left( f(r) - f(0) - f'(0) r - \frac {f''(0)} 2 r^2 H(1 - r) \right) dr,$$ then $$\mathcal F[r^{-4}] = \frac {\rho^2} 4 \left( \ln \frac \rho 2 + \gamma - 1 \right), \\ \mathcal F \!\left[ \frac {x^3 y} {(x^2 + y^2)^2} \right] = \frac {\partial^4} {\partial \xi^3 \partial \zeta} \mathcal F[r^{-4}] = \frac {\xi \zeta (\xi^2 - 3 \zeta^2)} {(\xi^2 + \zeta^2)^3}.$$ In general, differentiating $r^{-n}$ will produce delta function terms. $\partial \rho^{-2} / \partial \xi$ gives a $\delta^{(1, 0)}(\xi, \zeta)$ term, but $\zeta \delta^{(1, 0)}(\xi, \zeta) = 0$.

It appears that this remarkable relation holds: $$\mathcal F[\sin 2 k \phi] = \frac {2(-1)^k |k|} {\rho^2} \sin 2 k \psi,$$ from which the above transform follows immediately, but I have no idea how to derive it the edit box is too narrow for the proof.

Answer 2

The Fourier transform of $\sin k \phi$ for $k \in \mathbb N$ can be derived in this manner: $$I_k = \int_0^{2 \pi} e^{i \rho r \cos(\phi - \psi)} \sin k \phi \,d\phi = \sin k \psi \int_0^{2 \pi} e^{i \rho r \cos \phi} \cos k \phi \,d\phi = 2 \pi i^k \sin k \psi \,J_k(\rho r), \\ \mathcal F \!\left[ \frac {\sin k \phi} {r^2} \right] = \frac 1 {2 \pi} \int_0^\infty \frac {I_k} r dr = i^k \sin k \psi \int_0^\infty \frac {J_k(r)} r dr = \frac {i^k \sin k \psi} k.$$ Since $$\frac {x^3 y} {(x^2 + y^2)^2} = \frac {2 \sin 2 \phi + \sin 4 \phi} 8, \\ \frac {\xi \zeta(\xi^2 - 3 \zeta^2)} {(\xi^2 + \zeta^2)^3} = \frac {\sin 4 \psi - \sin 2 \psi} {2 \rho^2},$$ the desired result follows.