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First time trying to do something "real" in Mathematica, I am having trouble getting it to calculate this Fourier transform. It runs for a long time, then just prints the input expression.

FourierTransform[ Exp[-Sqrt[Wx x^2 + Wy y^2 + Wz z^2]], 
 {x, y, z}, {u, v, w}, 
 Assumptions -> {x ∈ Reals, y ∈ Reals, 
   z ∈ Reals,
   Wx ∈ Reals, Wy ∈ Reals, Wz ∈ Reals, 
   Wx > 0, Wy > 0, Wz > 0}
 ]

I also tried it as an integral.

I am sure the problem is in me, not in mathematica.

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  • $\begingroup$ I also have had some problems (using Mathematica 10.0.1) on this and related examples. I'll post either an answer to this, or a separate question when I have time to look further. $\endgroup$ – DumpsterDoofus Oct 25 '14 at 22:31
  • $\begingroup$ I'd like to point out that a scaling of variables allows you to reduce this to the Fourier-Bessel Transform of a decaying exponential, which can then be converted into the reciprocal space profile by a scaling. I'll post this as a partial answer in a while. $\endgroup$ – DumpsterDoofus Oct 25 '14 at 22:39
  • $\begingroup$ I had also tried adding a constant 'a', meaning (briefly) f(r)=exp(- W r + a), which would be a suitable answer in my application if 'a' is set to a small value. Thinking that this would fix the singularity in the derivative at the origin. It also did not work. $\endgroup$ – bullwinkle Oct 26 '14 at 0:47
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Mathematica's FourierTransform can be a bit difficult at times, so I'll let someone else address whether your code can be tweaked to work, and instead show how to do it symbolically. Defining $$f(\mathbf{r})=\exp\left(-|W\mathbf{r}|\right)$$ where $$W=\left( \begin{array}{ccc} \sqrt{w_x} & 0 & 0 \\ 0 & \sqrt{w_y} & 0 \\ 0 & 0 & \sqrt{w_z} \\ \end{array} \right)$$ note that $$f\left(W^{-1}\mathbf{r}\right)=\exp\left(-|\mathbf{r}|\right).$$ By the scaling property $$\mathcal{F}\left(f(A\mathbf{r}),\mathbf{r},\mathbf{k}\right)=\frac{1}{\text{det}(A)}\mathcal{F}\left(f(\mathbf{r}),\mathbf{r},\mathbf{k}A^{-1}\right)$$ where $\mathcal{F}$ is the Fourier transform and $A$ is any matrix, it suffices to compute $$\mathcal{F}\left(\exp\left(-|\mathbf{r}|\right),\mathbf{r},\mathbf{k}\right).$$ Since $\exp\left(-|\mathbf{r}|\right)$ is spherically symmetric, we obtain the spherical Bessel transform $$\mathcal{F}\left(\exp\left(-|\mathbf{r}|\right),\mathbf{r},\mathbf{k}\right)=\int_0^\infty 4\pi j_0\left(|\mathbf{k}|r\right)r^2\exp(-r)dr=\frac{8\pi}{\left(|\mathbf{k}|^2+1\right)^2}$$ giving the final desired result $$\mathcal{F}\left(\exp\left(-|W\mathbf{r}|\right),\mathbf{r},\mathbf{k}\right)=\frac{8\pi}{\text{det}(W)\left(|\mathbf{k}W^{-1}|^2+1\right)^2}=\frac{8\pi}{\sqrt{w_xw_yw_z}\left(\frac{k_x^2}{w_x}+\frac{k_y^2}{w_y}+\frac{k_z^2}{w_z}+1\right)^2}.$$

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  • $\begingroup$ Thank you, this is very helpful. $\endgroup$ – bullwinkle Oct 26 '14 at 0:44
  • $\begingroup$ @bullwinkle: One minor note: the Fourier transform definition in the linked article is nonunitary, whereas Mathematica's definition of FourierTransform is unitary, so you should include an extra factor of $(2\pi)^{-3/2}$ (I think). $\endgroup$ – DumpsterDoofus Oct 26 '14 at 0:47
  • $\begingroup$ Doofus did you use Mathematica for the Bessel integral? If so could you post your code? I am trying this... Integrate[ 4 Pi BesselJ[0, Sqrt[u^2 + v^2] r] Exp[-r] r^2, {r, 0, Infinity}, Assumptions -> {r [Element] Reals, u [Element] Reals, v [Element] Reals, u >= 0, v >= 0} ] $\endgroup$ – bullwinkle Oct 26 '14 at 1:20
  • $\begingroup$ @bullwinkle: They're spherical Bessel functions $j_n(z)$, not ordinary Bessel functions $J_n(z)$ (see the article I hyperlinked in my answer). In this case $j_0(z)=\text{sinc}(z)=\sin(z)/z$. I used 4 \[Pi] Integrate[(E^-r r Sin[k r])/k, {r, 0, \[Infinity]}, Assumptions -> k > 0]. $\endgroup$ – DumpsterDoofus Oct 26 '14 at 2:06

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