Fourier transform of a function of absolute values I need help in computing the Fourier transform of
$$\frac{1}{1+|x|^2}$$
with $x$ scalar. I tried using Cauchy integral formula with contours centred around the origin but I get stacked.
 A: Let $f(x) = \frac{1}{1+x^{2}}$. The definition of the Fourier transform I consider is the following :
$$ 
\begin{eqnarray*}
\hat{f}(\xi) & = & \int_{\mathbb{R}} f(x) e^{-2i \pi x \xi} \: dx \\
& = & \int_{\mathbb{R}} \frac{1}{1+x^{2}} e^{-2i \pi x \xi} \: dx
\end{eqnarray*} $$
for $\xi \in \mathbb{R}$. One idea is to use the residue theorem (see http://en.wikipedia.org/wiki/Residue_theorem). Let be $\xi \in \mathbb{R}$ fixed, you can consider the function
$$ \phi_{\xi}(z) = \frac{1}{1+z^{2}} e^{-2i\pi z \xi}, \, z \in \mathbb{C} $$
$\phi_{\xi}$ is a meromorphic function which has simple poles at $z=i$ and $z=-i$. Use the contour which illustrates the Wikipedia page. The example on the Wikipedia page is almost the same as here.
A: With x being scalar, there is no need of taking absolute value since x is squared. And for taking fourier transform you can just apply duality property of fourier transform as :
            Ff(-s) = invFf(s) ; where invF is inverse Fourier transform. 

Now we know that fourier transform of exp(-a|x|) is 2*a/(a^2 + (2*pi*x)^2) . From this the solution is straight forward that fourier transform of 1/(1 + x^2) would be :
                                  pi*exp(-|2*pi*f|) 

note : exp(-a*|x|) is even function of x.
