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I am looking for the Fourier transform on $\mathbb{R}^3$ of $e^{-|x|}$.

I tried in spherical coordinates with $x=(r,\phi,\theta)$ and $\xi=(|\xi|,\phi_2,\theta_2)$:

$$\int_{\mathbb{R}^3} e^{-|x|} e^{-i(\xi,x)}dx= \int e^{-r} e^{-ir|\xi|\cos A} r^2\sin \varphi drd\varphi d\theta $$

where $\cos A=\sin \phi \sin \phi_2 \cos (\theta-\theta_2)+\cos \phi\cos\phi_2$

But from here I'm afraid I'm stuck.

Does anybody have a better mmethod?

Tom
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2 Answers2

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The Fourier transform will be rotation invariant so it suffices to compute it in $(\omega, 0, 0)$:

$$ \begin{eqnarray} \hat{f}(\omega, 0, 0) &=& \int_{\mathbb{R}} e^{-i \omega x_1} \int_{\mathbb{R}^2} e^{-\sqrt{x_1^2 + x_2^2 + x_3^2}} dx_2 dx_3 \, dx_1 \\ &=& \int_{\mathbb{R}} e^{-i \omega x } \int_0^{\infty}2\pi r e^{-\sqrt{x^2+r^2}} dr \, dx \\ &=& 2 \pi \int_{\mathbb{R}} e^{-i \omega x}(|x| + 1)e^{-|x|}dx \\ &=& \frac{8 \pi}{(\omega^2 + 1)^2} \end{eqnarray} $$

WimC
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  • I'm confused by the notations: could you please give the explicit change of variables you are doing? What is $x$? What is $r$? – Tom Aug 05 '13 at 10:39
  • @Tom $x=x_1$, $r^2 = x_2^2+x_3^2$. – WimC Aug 05 '13 at 11:00
  • Alright thanks. Next, how do you retrieve $\hat{f}(\xi_1,\xi_2,\xi_3)$ from $\hat{f}(\omega,0,0)$? Basically, could you be a bit more explicit about the rotation invariant argument? – Tom Aug 05 '13 at 11:05
  • @Tom If a function is rotation invariant then so is its Fourier transform. In that case $\hat{f}(\xi) = \hat{f}(|\xi|, 0, 0)$. – WimC Aug 05 '13 at 11:07
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Note that $f(x) = e^{-|x|}$ is a radial function. That means if $U$ is any unitary matrix, i.e it correspond to a rotation then $ f(Ux) = f(x)$. In order to see this, let $ x = r\gamma $ where $|x| = r$ and $ \gamma = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta) $, then you see that $f(r\gamma) = e^{-r}$ that is independent on $\phi $ and $\theta$. Now you can check that this implies that the fourier transform of $f$ is also a radial function. Note that is $U$ is a unitary matrix then $\det U = 1$ and $U^{-1} = U^*$ (the conjugate transpose).

With this in mind we know that $ \widehat{f} (\xi) = \widehat{f} (\rho \eta)$ where $ \rho = |\xi| $ and $ \eta= (\sin \theta_2 \cos \phi_2, \sin \theta_2 \sin \phi_2, \cos \theta_2) $ and $\widehat{f} (U\rho \eta) = \widehat{f} (\rho \eta)$ so chose $U$ in such a way that it rotate the point $ \eta= (\sin \theta_2 \cos \phi_2, \sin \theta_2 \sin \phi_2, \cos \theta_2) $ on the unit sphere to the point $(0,0,1)$.

Then your integral becomes $$ \int_{0}^{\infty}\int_{S^2}e^{-r}e^{-i \rho r \cos \theta} r^2\sin \theta d\theta d \phi dr $$ Now you can solve this integral by substituting $u = \cos \theta$

Henrik Finsberg
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