Fourier transform of a Green's function I was studying for an exam and I found this question which has caused me a bit of trouble:
Given the Green's function that satisfies the equation
$$\Box G(\mathbf{r}-\mathbf{r}',t-t')=-4\pi\delta(\mathbf{r}-\mathbf{r}',t-t'),$$ where $\Box$ is the d'Alembertian operator
$$\Box=\Delta-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}$$ show that, given a suitable method of handling the poles, its Fourier transform is
$$g(\mathbf{k},\omega)=\frac{1}{4\pi^3\left(\|\mathbf{k}\|^2-\tfrac{\omega^2}{c^2}\right)}.$$
The aim of the question is obviously to look at ways of solving the wave equation by Fourier transform, but I'm a little bit unsure how to proceed, especially since $g$ looks a bit bizarre to me. The term $\omega$ refers to the angular frequency of the wave, and $\mathbf{k}$ the wave vector. What is the simplest/best way to go about this? Finding $G$ first and then taking its Fourier transform seems like a tedious way to solve this problem.
Thanks for any help.

Generally you would Fourier transform the differential equation. The elements of the D'Alembertian transform to $\vec{k}\cdot\vec{k}$ (the Laplacian) and $-\omega^2$ (the 2nd time derivative), while the delta function transforms to $1$.
Then you get something like
$$ (2\pi)^4 \left(\left\lVert \vec{k}\right\lVert^2 - \frac{\omega^2}{c^2}\right) g= 4\pi $$
Solving for $g$ gives you the answer.
This is obviously greatly simplified and makes many assumptions about certain integrals converging, but it's the general approach.
