I'm trying to compute the analytical FT of a bidimensional exponential decay as follows:

$f_E(\vec{x}) = \exp \left( - \sqrt{ (\vec{x} - \vec{\mu})^T \Sigma^{-1} (\vec{x} - \vec{\mu}) } \right)$

with $\Sigma^{-1} \in \mathbb{R}^{2 \times 2}$ a symmetric positive-definite matrix, with the following code:

A = {{a, b}, {b, c}}
x = {x1, x2}
μ = {μ1, μ2}
z = {z1, z2}
FourierTransform[Exp[-Sqrt[(x - μ).A.(x - μ)]], x, z]

But the notebook's kernel keeps running and dies after a while. I also tried using Assumptions to make explicit that coefficients involved are real and A is positive-definite with no luck. Also tried

M = {{d, e}, {f, g}}
FourierTransform[Exp[-Norm[M.(x - μ)]], x, z]

I'm new to Mathematica so I don't know if there's an error with the code, a conceptual (mathematical) mistake or if it is just related to computing resources.


1 Answer 1


I don't know why Mathematica does not get it right. At least, it can compute

FourierTransform[Exp[-Sqrt[x.x]], x, z]

1/(1 + z1^2 + z2^2)^(3/2)

From there we can continue on paper:

Let $\varPhi \colon \mathbb{R}^n \to \mathbb{R}^n$ be a diffeomorpism of class at least $C^1$ and denote by $\mathcal{F}$ the Fourier transform. By the transformation formular for integrals (and by certain denseness and continuous extension arguments) one obtains

$$\begin{aligned}\mathcal{F}(f \circ \varPhi)(\xi) &= C \int_{\mathbb{R}^n} f(\varPhi(x)) \, \exp(- \mathrm{i} \, \langle\xi , x \rangle) \, \mathrm{d} x\\&= C \int_{\mathbb{R}^n} f(y) \, \exp(- \mathrm{i} \, \langle\xi , \varPhi^{-1}(y) \rangle) \, |\det(D\varPhi(y))|^{-1}\, \mathrm{d} y\end{aligned}$$

with a constant $C$ that depends on some convention; the precise value won't matter in the end.

In case of an affine mapping $\varPhi(x) = L \, (x - \mu) = L\, x - L \, \mu$ with $L^T \, L =A$, we obtain $|\det(D\varPhi(y))| = \sqrt{\det(A)}$. Thus, assuming that $A$ is positive definite, we obtain

$$\begin{aligned}\mathcal{F}(f \circ \varPhi)(\xi) &= C\,\det(A)^{-\frac{1}{2}} \int_{\mathbb{R}^n} f(y) \, \exp(- \mathrm{i} \, \langle\xi , L^{-1}\,y + \mu \rangle) \, \mathrm{d} y\\&= C \, \det(A)^{-\frac{1}{2}} \, \exp(- \mathrm{i} \, \langle\xi , \mu \rangle)\int_{\mathbb{R}^n} f(y) \, \exp(- \mathrm{i} \, \langle L^{-T}\xi , y\rangle) \, \mathrm{d} y\\&= \det(A)^{-\frac{1}{2}} \, \exp(- \mathrm{i} \, \langle\xi , \mu \rangle) \,\mathcal{F}(f)(L^{-T}\xi).\end{aligned}.$$

With $f(x) = \exp(-\sqrt{\langle x, x\rangle})$ and $$\mathcal{F}(f)(\xi) = (1+\langle \xi, \xi\rangle)^{-\frac{3}{2}},$$ we should obtain $$\mathcal{F}(f \circ \varPhi)(\xi) = \det(A)^{-\frac{1}{2}} \, \exp(- \mathrm{i} \, \langle\xi , \mu \rangle) \, (1+\langle \xi, A^{-1}\xi\rangle)^{-\frac{3}{2}}.$$

But of course, I could have made a mistake. Please check thoroughly.

  • $\begingroup$ I have gone over this a few times now and it checks out, there's just a little typo, inside the sqrt of f(x) should be <x,x>. Thanks! $\endgroup$
    – vlizana
    Feb 21, 2019 at 21:01
  • $\begingroup$ You're welcome! I am glad that this was helpful. $\endgroup$ Feb 21, 2019 at 21:03

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