Fourier transform of $(\vec{r}- \vec{r}_0)/|\vec{r}- \vec{r}_0|^3$

Question

Could anyone find the Fourier transform of below expression in 3D, where $r_0$ is a constant vector?

$$\frac{\vec{r}- \vec{r}_0}{|\vec{r}- \vec{r}_0|^3}$$

Answer 1

Let $\mathbf F_0 = {\mathbf r}/{|\mathbf r|^3}.$ This has divergence $\nabla \cdot \mathbf F_0 = 4\pi \delta,$ where $\delta$ is the Dirac distribution on $\mathbb R^3$.

Taking the Fourier transform of both sides gives $-i{\mathbf k}\cdot\hat{\mathbf F}_0 = 4\pi.$ Since $\mathbf{F}_0$ is symmetric, so should $\hat{\mathbf F}_0$. Therefore we assert $\hat{\mathbf F}_0 = f(|\mathbf k|) \mathbf{k}$ for some function $f : [0, \infty) \to \mathbb R$.

This gives $-i(\mathbf k\cdot\mathbf k)f(|\mathbf k|) = -i|\mathbf k|^2 f(|\mathbf k|)= 4\pi$ so $f(|\mathbf k|) = i\,4\pi/|\mathbf k|^2,$ i.e. $$\hat{\mathbf F}_0 = i\,4\pi \frac{\mathbf k}{|\mathbf k|^2}.$$

The transform of the translated field $\mathbf F = (\mathbf r - \mathbf r_0)/|\mathbf r - \mathbf r_0|^3$ is given by multiplication by a factor: $$\hat{\mathbf F} = i\,4\pi \frac{\mathbf k}{|\mathbf k|^2} e^{-i \mathbf k \cdot \mathbf r_0}.$$