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.
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 $$\begin{aligned} \mathcal F[f](k\hat{\mathbf z}) &= \frac1{(2\pi)^{3/2}} \iiint \frac{e^{ik'r}}r e^{-i\mathbf k\cdot \mathbf r} d^3\mathbf r \\ &= \frac1{(2\pi)^{3/2}} \iiint \frac{e^{ik'r}}r e^{-ikr\cos\theta} r^2 \sin\theta dr d\theta d\phi \\ &= \frac1{(2\pi)^{1/2}} \int_0^\infty \left(re^{ik'r} \int_0^{\pi} e^{-ikr\cos\theta} \sin\theta d\theta\right) dr \\ &= \frac1{(2\pi)^{1/2}} \int_0^\infty r e^{ik'r} \frac{2 \sin kr}{kr} dr \\ &= \frac1k\sqrt{\frac2\pi} \int_0^\infty e^{ik'r} \sin kr dr \\ &= \sqrt{\frac2\pi}\frac1{k^2 - k'^2} \end{aligned}$$
The answer by kennytm is only partially correct. It finds all the values of the Fourier image—where they exist. But the complete Fourier image of a spherical wave is not a function: it's a distribution.
Let's consider a standing wave described in terms of the $0$th order spherical Bessel function (the imaginary part of the OP's function):
$$g(\mathbf r)=k'j_0(k'r)=\frac{\sin(k'r)}r.$$
We can find its Fourier transform similarly to the approach in the kennytm's answer, but with a special treatment of the final integral:
$$I=\int_0^\infty \sin(k'r)\sin kr dr.$$
This integral (up to multiplicative constant) is the sine transform of $\sin(k'r)$, which is equal to
$$I=\delta(k-k')-\delta(k+k'),$$
where $\delta$ is the Dirac delta.
Similarly we can find that the Fourier transform of the second spherical wave—the one with $0$th order spherical Neumann function (the real part of the OP's function):
$$h(\mathbf r)=k'y_0(k'r)=\frac{\cos(k'r)}r.$$
Fourier transforming of this one would reduce (up to constant multiplier) to taking the sine transform of the $\cos(k'r)$, and we'll finally get the same Fourier transform as in kennytm's answer:
$$\mathcal F[h](k\hat{\mathbf z})=\sqrt{\frac2\pi}\frac1{k^2 - k'^2}.$$
Now we can compile the complete Fourier transform of the running wave given in the OP:
$$f(\mathbf r)=\frac{e^{ik'r}}r.$$
It's the combination of the two results found above:
$$\boxed{\mathcal{F}[f](k\hat{\mathbf z})=\sqrt{\frac2\pi}\frac1{k^2 - k'^2}+\frac i k\sqrt{\frac\pi2}\big(\delta(k-k')-\delta(k+k')\big).}$$
