Fourier transform of $e^{-|x|}$ 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?
 A: The Fourier transform will be rotation invariant so it suffices to compute it in $(\omega, 0, 0)$:
$$
\begin{eqnarray}
\hat{f}(\omega, 0, 0) &=& \int_{\mathbb{R}} e^{-i \omega x_1} \int_{\mathbb{R}^2} e^{-\sqrt{x_1^2 + x_2^2 + x_3^2}} dx_2 dx_3 \, dx_1 \\
 &=& \int_{\mathbb{R}} e^{-i \omega x } \int_0^{\infty}2\pi r e^{-\sqrt{x^2+r^2}} dr \, dx \\
 &=& 2 \pi \int_{\mathbb{R}} e^{-i \omega x}(|x| + 1)e^{-|x|}dx \\
 &=& \frac{8 \pi}{(\omega^2 + 1)^2}
\end{eqnarray}
$$
A: 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 $u = \cos \theta$
