I am self studying PDEs and have come across the following problem regarding distributions.
Problem Statement: Prove that $$u(x)=\frac{1}{4 \pi} \int_{\mathbb{R}^3} f(y) \frac{e^{-i k |x -y|}}{|x-y|} d^3 y$$ obeys $$L u(x) = -f(x)$$ for $$L=\Delta + k^2$$.
Attempt: I recall to mind the definition of the spherically symmetric Laplacian in polar coordinates for $\mathbb{R}^3$. $$\Delta = \partial_r^2 + \frac{2}{r} \partial_r$$
Then, $f(y)$ is truly a coefficient function, so what is of principal significance remains in exhibiting
$$(\Delta + k^2) \frac{e^{-ikr}}{r} = -4 \pi \delta(r)$$
in the distributional sense.
$$ \begin{align} (\Delta + k^2) \int_{\mathbb{R}^3} \frac{e^{-ikr}}{r} d^3 y &= 4 \pi \int_{\mathbb{R}^3} (\Delta + k^2) r e^{-ikr} dr \\ &= 4 \pi \int_{0}^{\infty} \left(\partial_r^2 + \frac{2}{r} \partial_r + k^2 \right) r e^{-ikr} dr \\ &= 4 \pi \int_{0}^{\infty} \left(-2ik - k^2 r + \frac{2}{r} - 2ik + k^2 r \right) e^{-ikr} dr \\ &= 4 \pi \int_{0}^{\infty} \left(\frac{2}{r} - 4ik \right) e^{-ikr} dr \end{align} $$
Where I'm Stuck: I am not sure how to proceed as my final integrand appears ill-conditioned? Did I make an algebraic mistake?
Other Thoughts: I was curious as to if I might use the following identity in my answer.
For any $k \in \mathbb{R}$
$$ \int_{-\infty}^{\infty} e^{ikx} dx = 2 \pi \delta(x) $$
Regarding the $\frac{2}{r}$ term, could I employ the Cauchy principle value of the integral or another such means of ascribing a value to it? If memory serves, the Fourier transform of $\frac{1}{r}$ is not defined, so I believe that I must have taken a wrong path.
Ultimately, am I correct in my hope that proving
$$ (\Delta + k^2) \frac{e^{-ikr}}{r} = - 4 \pi \delta(r) $$
will allow me to show
$$ \begin{align} (\Delta + k^2) c_0 \int_{\mathbb{R}^3} f(y) \frac{e^{-ikr}}{r} dy &= -4 \pi c_0 \int_{\mathbb{R}^3} f(y) \delta(x -y) d^3 y \\ &= -4 \pi c_0 f(x) \end{align} $$
Then, $c_0= \frac{1}{4 \pi}$ would prove the desired result.