# Homework Help: 3D Fourier Transform

1. Mar 5, 2012

### Ansatz7

The problem statement, all variables and given/known data
Calculate the Fourier transform of f(x) = (1 + |x|2)-1, x$\in$ℝ3

The attempt at a solution

As far as I can tell, the integral we are supposed to set up is:

Mod note: Fixed your equation. You don't want to mix equation-writing methods. Just stick to LaTeX.
$$\int \frac{e^{-2\pi i (\vec{k}\cdot\vec{x})}}{1+|\vec{x}|^2}dV = \int \frac{e^{-2\pi i r(k_1\sin\theta\cos\phi + k_2 \sin\theta\sin\phi + k_3\cos\theta)}}{1+r^2} r^2\sin\theta\,d\theta\,d\phi\,dr$$but I have no idea how to perform this integral. Any ideas appreciated! (Also, sorry about the fractions - I have no idea why they aren't working because I have no tex experience).

Last edited by a moderator: Mar 5, 2012
2. Mar 5, 2012

### Dick

The first step is to use that your f(x) doesn't depend on the direction of the vector x. So the fourier transform won't depend on the direction of k. So you can choose a k that points along the z-axis. That simplifies thing a lot.

3. Mar 5, 2012

### Ansatz7

*facepalm* Of course, that at least makes the angular part of the integral simple. After the angular integral I ended up with:
$$\frac{2}{k}\int \frac{r\sin2\pi kr}{1+r^2}\,dr$$

I don't think this is integrable, but that makes sense based on the way the question was posed.I think it ought to be square integrable though. I tried to compute this using residues - I was hoping to get something analogous to the 1D Fourier transform

$$f(x) = \frac{1}{1 + x^2}, \hat{f}(k) = e^{-2\pi x|k|}$$

but from the look of the residue I have so far it doesn't seem like it will be so aesthetically pleasing. At any rate, thanks for your help, and thanks vela for editing my equation. I've never used LaTex so I was guessing.

4. Mar 5, 2012

### Dick

I'm not checking the details here, so I hope you are keeping track of all of the signs and factors. But that looks integrable to me. There are poles at i and -i. You'll have to split the sin up into exponentials so you can decide which half-plane to close the contours in, but it looks routine to me.

5. Mar 5, 2012

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