# Three Dimensional Fourier Transform of Radial Function without Bessel and Neumann

I am trying to compute the Fourier transform of $\frac1{|\mathbf{x}|^2+1}$ where $\mathbf{x}\in\mathbb{R}^3$.

Just writing out the integral: $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac1{|\mathbf{x}|^2+1}e^{-2\pi i (\mathbf{x}\cdot\mathbf{\xi})}dx_1dx_2dx_3$.

Mathematica was no help with this integral. I realized though that the function is radial, so that in spherical coordinates $f(\rho,\theta,\phi)=\frac1{r^2+1}=f(\rho)$. I thought this would simplify matters because then the limits of integration are just $\int_0^{\infty}\int_0^{2\pi}\int_0^{\pi}$ and $dx_1dx_2dx_3 \mapsto \rho^2\sin\theta d\theta d\phi d\rho$. However, the dot product in the exponent is messing me up a lot. I think it should be of the form $||\mathbf{x}||||\xi||$ times some trigonometric function of the angle between them, and $||\mathbf{x}||$ is just $\rho$. Because of this, I don't think doing this integral directly will be that much cleaner than Cartesian coordinates either.

However, I was wondering if I could follow the methods of this and leverage the symmetry of my function and get the answer this way.

I already know the Fourier Transform must be radial from the link. However, I am not sure about how to use the dilation part to extract information for my problem here.

Or do you think it would be propitious to write $x_1$ as $\rho\sin\theta\cos\phi$ etc. and just work it out that way?

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You can get the right spacing for the norm bars by using \lVert and \rVert to produce $\lVert$. – joriki May 7 '12 at 16:32
Ricky, is there any reason you deleted your question after joriki provided an answer? Please note that this site helps everyone online - even though you've got the answer now, if someone had the same question as you, we'd want them to be able to see the answer here too. – Zev Chonoles May 7 '12 at 21:26
@ZevChonoles After thinking more, I thought my question was poorly formulated. But you're right, I should just leave it up. – Steven-Owen May 7 '12 at 23:50

I don't see how to solve this using the sort of scaling arguments used in that other answer, since your denominator doesn't have the same sort of scale invariance.

To solve this directly, the key point to observe is that $\mathbf x\cdot\xi=\lVert\mathbf x\rVert\lVert\xi\rVert\cos\theta$, where $\theta$ is the angle between $\mathbf x$ and $\xi$. Since the function to be transformed is invariant under rotations, you can choose coordinates such that $\theta$ is also the angle between $\mathbf x$ and the $z$-axis. Then, with $k=\lVert\xi\rVert$, your integral is

\begin{align} \int_0^\infty\int_0^\pi\int_0^{2\pi}\frac1{r^2+1}\mathrm e^{-2\pi\mathrm ikr\cos\theta}r^2\sin\theta\,\mathrm d\phi\,\mathrm d\theta\,\mathrm dr &= 2\pi\int_0^\infty\frac{r^2}{r^2+1}\int_0^\pi\mathrm e^{-2\pi\mathrm ikr\cos\theta}\sin\theta\,\mathrm d\theta\,\mathrm dr \\ &= 2\pi\int_0^\infty\frac{r^2}{r^2+1}\int_{-1}^1\mathrm e^{-2\pi\mathrm ikr\cos\theta}\,\mathrm d(\cos\theta)\,\mathrm dr \\ &= \frac{\mathrm i}k\int_0^\infty\frac{r}{r^2+1}\left(\mathrm e^{-2\pi\mathrm ikr}-\mathrm e^{2\pi\mathrm ikr}\right)\mathrm dr \\ &= \Im\frac2k\int_0^\infty\frac{r}{r^2+1}e^{2\pi\mathrm ikr}\mathrm dr \\ &= \Im\frac1k\int_{-\infty}^\infty\frac{r}{r^2+1}e^{2\pi\mathrm ikr}\mathrm dr\;, \end{align}

where the last step works because the imaginary part of the integrand is even. This integral can be evaluated by closing the contour in the complex plane with a semicircle at infinity in the upper half-plane, enclosing the pole at $r=\mathrm i$ with residue $\frac12\mathrm e^{-2\pi k}$, yielding

$$\Im\frac1k2\pi\mathrm i\,\frac12\mathrm e^{-2\pi k}=\frac\pi ke^{-2\pi k}\;.$$

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I don't see why you can call the angle between $\mathbf{x}$ and $\xi$ $\cos (\theta)$. – Arturo don Juan Apr 27 '15 at 1:22
@ArturoDonJuan: I called that angle $\theta$, not $\cos(\theta)$. Since I can call the angle whatever I want, I'm guessing that your question is more about why I can choose this to also be the angle between $\mathbf x$ and the $z$-axis. This is because the function to be transformed is rotationally invariant, so the answer depends only on $k$ and we can choose $\xi$ along the $z$-axis. – joriki Apr 28 '15 at 2:21
Sorry, my wording was incorrect. I meant that. Would it be more appropriate to say that we can orient the $z$-axis along the $\xi$ vector? – Arturo don Juan Apr 28 '15 at 3:14
@ArturoDonJuan: I think both viewpoints are appropriate. The fact that you can choose how to orient the $z$-axis is equally dependent on the rotational symmetry of the function being transformed. – joriki Apr 28 '15 at 9:35