The fundamental solution of the Helmholtz equation in $\mathbb{R}^3$
$$(\Delta+k^2)u=-\delta \tag{1}$$
is well known:
$$u(x)=\frac{e^{\pm ik|x|}}{4\pi |x|}$$
solves the Helmholtz equation in distributional sense. The usual ansatz to obtain fundamental solutions is to Fourier transform both sides. Then $(1)$ becomes
$$(-|x|^2+k^2)\hat{u}(x)=-1 \implies \hat{u}(x)=\frac{1}{k^2-|x|^2}.$$
The problem now arises is that we want to inverse Fourier transform $\hat{u}$ to obtain the solution. But $\hat{u}$ has singularities on the sphere of radius $k$.
How do we proceed in a rigorous distributional way to obtain the above fundamental solution? What do people usually do and is there any book about such problems?
Best Answer
METHODOLGY $(1)$: One way to handle the singularities at $k'=\pm k$ is to let $k$ have a "small," non-zero imaginary part. We will assume that $\text{Im}(k)<0$.
Then, we can write
$$\begin{align} u(\vec r)&=\frac{1}{(2\pi)^3}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{e^{i\vec k'\cdot \vec r}}{k'^2-k^2}\,dk'_x\,dk'_y\,dk'_z\\\\ &=\frac{1}{(2\pi)^3}\int_0^{2\pi}\int_0^\pi\int_0^\infty \frac{e^{ik'r\cos(\theta)}}{k'^2-k^2}\,k'^2\sin(\theta)\,dk'\,d\theta\,d\phi\\\\ &=\frac{1}{(2\pi)^2}\int_0^\pi\int_0^\infty \frac{e^{ik'r\cos(\theta)}}{k'^2-k^2}\,k'^2\sin(\theta)\,dk'\,d\theta\\\\ &=\frac{1}{(2\pi)^2}\int_0^\infty \frac{k'}{k'^2-k^2}\frac{2\sin(k'r)}{r}\,dk'\\\\ &=\frac{1}{(2\pi)^2}\int_{-\infty}^\infty \frac{k'}{k'^2-k^2}\frac{\sin(k'r)}{r}\,dk'\\\\ &=\frac{e^{-ikr}}{4\pi r} \end{align}$$
Had we assumed that $\text{Im}(k)>0$, we would have recovered the solution $\frac{e^{ikr}}{4\pi r}$.
Finally, we can let the imaginary part of $k$ approach $0$ and recover the result for real $k$.
METHODOLGY $(2)$:
We could solve Equation $(1)$ in the OP without the use of integral transformation. Instead, we write
$$\nabla^2 u(\vec r)+k^2u(\vec r)=0$$
for $\vec r\ne 0$. Exploiting the spherical symmetry of the problem, we have
$$\nabla^2 u+k^2 u=\frac1{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial u}{\partial r}\right)+k^2u=0$$
which can be rearranged as
$$\frac{\partial^2}{\partial r^2}(r u)+k^2 (ru)=0 \tag 2$$
Solutions to $(2)$ are trivially seen to be
$$u=C^{\pm}\frac{e^{\pm ikr}}{r}$$
We find the constant $C^\pm$ by enforcing the condition $\int_{|\vec r|\le \epsilon}\nabla^2 u(\vec r)\,dV=-1$.
Applying the Divergence Theorem in the sense of distributions reveals
$$\begin{align} \oint_{|\vec r|=\epsilon}\left.\left(\frac{\partial u(\vec r)}{\partial r}\right)\right|_{|\vec r|=\epsilon}\,\epsilon^2 \sin(\theta)\,d\theta\,d\phi&=-4\pi C^{\pm}\\\\ &=-1\end{align}$$
whereupon solving for $C^{\pm}$ yields $C^\pm=\frac{1}{4\pi}$.