How can we do this computation?
$$\iiint_{R^3} \frac{e^{ik'r}}{r} e^{ik_1x+k_2y+k_3z}dx dy dz$$
where $r=\sqrt{x^2+y^2+z^2}$? I think we must use distributions.
Physically, it's equivalent to find wave vectors $k$ distribution and to write a spherical wave as sum of plane waves.
I know the formula for the inverse problem: write a plane wave as sum of spherical waves. The solution in this case is a serie of spherical harmonics and spherical bessel functions.
Best Answer
From your description, I believe you want to find the Fourier transform of $$ f(\mathbf r) = \frac{e^{ik'r}}r, $$ and the wave can be recovered from the linear superposition of plane waves identified by k $$ f(\mathbf r) = \frac1{(2\pi)^{3/2}}\iiint \mathcal F[f](\mathbf k)e^{i\mathbf k\cdot\mathbf r} d^3 \mathbf k. $$
The spherical wave have spherical symmetry, so what you should do is to perform the integration in spherical coordinates instead of Cartesian. WLOG, assume k is along the z axis, thus $$\begin{aligned} \mathcal F[f](k\hat{\mathbf z}) &= \frac1{(2\pi)^{3/2}} \iiint \frac{e^{ik'r}}r e^{-i\mathbf k\cdot \mathbf r} d^3\mathbf r \\ &= \frac1{(2\pi)^{3/2}} \iiint \frac{e^{ik'r}}r e^{-ikr\cos\theta} r^2 \sin\theta dr d\theta d\phi \\ &= \frac1{(2\pi)^{1/2}} \int_0^\infty \left(re^{ik'r} \int_0^{\pi} e^{-ikr\cos\theta} \sin\theta d\theta\right) dr \\ &= \frac1{(2\pi)^{1/2}} \int_0^\infty r e^{ik'r} \frac{2 \sin kr}{kr} dr \\ &= \frac1k\sqrt{\frac2\pi} \int_0^\infty e^{ik'r} \sin kr dr \\ &= \sqrt{\frac2\pi}\frac1{k^2 - k'^2} \end{aligned}$$