\begin{align} - \nabla^2 G(r) &= \delta(r) \\ G(r) &\rightarrow 0 \mbox{ as } r \rightarrow \infty \end{align} where $ r = \sqrt{x_1^2 + x_2^2 + x_3^2}$
In order to find the Green's function I take the Fourier transform
of the equation. This gives me
\begin{align} \mathrm{k}^2 \hat{G}(\mathrm{k}) = 1 \tag{1} \end{align}
where $\mathrm{k} = \sqrt{k_1^2 + k_2^2 + k_3^2} $
The idea is to divide the equation by $\mathrm{k}^2$ then take the back transform of the equation. This gives me the Green's function $G(r)$. However $\mathrm{k}$ can be equal to $0$. Implying equation $(1)$ to be false.
The question is why is the Green's function $G(r)$ found by this method correct?