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.

  • $\begingroup$ This is the 3D Fourier transform of $\frac{e^{ik'r}}r$? If so, the formula should be $\iiint_V \frac{e^{ik'r}}r e^{{\color{red}-}i\mathbf k\cdot\mathbf r} d^3\mathbf r$ $\endgroup$
    – kennytm
    Dec 1, 2010 at 18:43
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    $\begingroup$ This question might have a physical motivation but I think it is purely mathematical in nature. You should probably ask for the answer at math.SE. $\endgroup$
    – Marek
    Dec 1, 2010 at 19:34
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    $\begingroup$ @Marek: I think I agree... I mean, the title suggests a physics question, but in essence it is just about how to do an integral. I'm voting to close it (but only because it requires four other people to agree before the question actually gets closed - I wouldn't be comfortable unilaterally closing this if I had the power to do so). $\endgroup$
    – David Z
    Dec 1, 2010 at 19:42
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    $\begingroup$ @David: right, I also wouldn't be comfortable to just close it on my own. @Boy: no need to be sorry, closing a question is no big deal. It's really more about setting boundaries of what should and shouldn't be asked on this site so that people know in the future. $\endgroup$
    – Marek
    Dec 1, 2010 at 20:24
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    $\begingroup$ I know I'm in the minority, but I like seeing these "mathematical" questions here. Physicists have a different culture of math than mathematicians, and are more likely to get a useful answer from other physicists. $\endgroup$
    – nibot
    Dec 2, 2010 at 0:17

2 Answers 2


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 $$\begin{aligned} \mathcal F[f](k\hat{\mathbf z}) &= \frac1{(2\pi)^{3/2}} \iiint \frac{e^{ik'r}}r e^{-i\mathbf k\cdot \mathbf r} d^3\mathbf r \\ &= \frac1{(2\pi)^{3/2}} \iiint \frac{e^{ik'r}}r e^{-ikr\cos\theta} r^2 \sin\theta dr d\theta d\phi \\ &= \frac1{(2\pi)^{1/2}} \int_0^\infty \left(re^{ik'r} \int_0^{\pi} e^{-ikr\cos\theta} \sin\theta d\theta\right) dr \\ &= \frac1{(2\pi)^{1/2}} \int_0^\infty r e^{ik'r} \frac{2 \sin kr}{kr} dr \\ &= \frac1k\sqrt{\frac2\pi} \int_0^\infty e^{ik'r} \sin kr dr \\ &= \sqrt{\frac2\pi}\frac1{k^2 - k'^2} \end{aligned}$$

  • $\begingroup$ thank you! And how have you do the last passage? $1/k \int_0^{+infty} e^{ik'r}sinkrdr= \frac{1}{k^2+k'^2}$ $\endgroup$
    – Boy S
    Dec 1, 2010 at 19:53
  • $\begingroup$ @Boy: What do you mean? $\endgroup$
    – kennytm
    Dec 1, 2010 at 19:56
  • $\begingroup$ I think he wants to know how your last step of the calculation is done, i.e. the integral over $e^{ik' r}\sin kr$. $\endgroup$
    – Lagerbaer
    Dec 1, 2010 at 20:06
  • $\begingroup$ @Boy: Change $\sin kr$ to $(e^{ikr}-e^{-ikr})/2i$. Then notice that $\int_0^\infty e^{iKr} dr = i/K$ (if we ignore convergence stuff. Actually the answer isn't right when $k'$ doesn't have a positive imaginary part since the integral diverges.) $\endgroup$
    – kennytm
    Dec 1, 2010 at 20:06
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    $\begingroup$ The final answer is: $\mathcal F[f](k\hat{\mathbf z})=\sqrt{\frac{2}{\pi}}\frac{1}{k^2-k'^2}+\frac{i}{k}\sqrt{\frac{\pi}{2}}\left(\delta(k-k')-\delta(k+k')\right)$. $\endgroup$
    – rtmd
    Feb 17, 2022 at 22:34

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


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:

$$\boxed{\mathcal{F}[f](k\hat{\mathbf z})=\sqrt{\frac2\pi}\frac1{k^2 - k'^2}+\frac i k\sqrt{\frac\pi2}\big(\delta(k-k')-\delta(k+k')\big).}$$

  • $\begingroup$ I think that the answer is: $$ \mathcal F[f](k\hat{\mathbf z})=\sqrt{\frac{2}{\pi}}\frac{1}{k^2-k'^2}+\frac{i}{k}\sqrt{\frac{\pi}{2}}\left(\delta(k-k')-\delta(k+k')\right). $$ I have added the comments with calculations to kennytm's answer. $\endgroup$
    – rtmd
    Feb 17, 2022 at 22:45
  • $\begingroup$ @rtmd you're right, I've now fixed the answer. $\endgroup$
    – Ruslan
    Feb 17, 2022 at 22:59
  • $\begingroup$ It's quite odd that the sin spherical wave is monochromatic ($|k'| = |k|$), whereas the cos spherical wave is not. The difference between the two appears to just be a phase shift in real space ... I'm surprised that that results in the production of new more k values. Any intuition on that? $\endgroup$
    – ions me
    May 23 at 15:41
  • $\begingroup$ @ionsme note that spherical sin is finite at the origin (the so called first remarkable limit), while spherical cos is singular. $\endgroup$
    – Ruslan
    May 23 at 17:36

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