Fourier Transform of $\frac{1}{(1+x^2)^2}$ I need to find the Fourier Transform of 
$f(x) = \frac{1}{(1+x^2)^2}$ 
Where the Fourier Tranform is of $f$ is denoted as $\hat{f}$, where $\hat{f}$ is defined as $$\hat{f}(y)=\int_\mathbb{R}f(x)e^{-ixy}dx$$
I think I need to use the Fourier Inversion Theorem. From this theorem, I think we know that $\widehat{\widehat{\frac{1}{(1+x^2)^2}}} = (2\pi)\frac{1}{(1+x^2)^2}$
Then from the Fourier Inversion Theorem, I got that $\widehat{\frac{1}{(1+x^2)^2}} = \int_\mathbb{R}\frac{1}{(1+x^2)^2}e^{ixy}dx = \int_\mathbb{R}\frac{cos(xy)}{(1+x^2)^2}dx$
From here, I am unsure about how to calculate the integral, assuming I did everything right so far. Thanks for your help!
 A: Method 1: Residues
Consider the contour integral
$$\oint_C dz \frac{e^{i k z}}{(1+z^2)^2}$$
where $C$ is a semicircle in the upper half plane; here, $k>0$.  Then by the residue theorem and Jordan's lemma:
$$\begin{align}\int_{-\infty}^{\infty} dx  \frac{e^{i k x}}{(1+x^2)^2} &= i 2 \pi \operatorname*{Res}_{z=i} \frac{e^{i k z}}{(1+z^2)^2}\\ &= i 2 \pi \left [\frac{d}{dz} \frac{e^{i k z}}{(z+i)^2} \right ]_{z=i}\\ &= i 2 \pi \left [\frac{i k\, e^{i k z}}{(z+i)^2} - \frac{2 e^{i k z}}{(z+i)^3} \right ]_{z=i}\\ &= \frac{\pi}{2} (k+1) e^{-k}\end{align}$$
For $k \lt 0$, $C$ is a semicircle in the lower half plane; for the same reasons as above, we have:
$$\begin{align}\int_{-\infty}^{\infty} dx  \frac{e^{i k x}}{(1+x^2)^2} &= -i 2 \pi \operatorname*{Res}_{z=-i} \frac{e^{i k z}}{(1+z^2)^2}\\ &= -i 2 \pi \left [\frac{d}{dz} \frac{e^{i k z}}{(z-i)^2} \right ]_{z=-i}\\ &= -i 2 \pi \left [\frac{i k\, e^{i k z}}{(z-i)^2} - \frac{2 e^{i k z}}{(z-i)^3} \right ]_{z=-i}\\ &= \frac{\pi}{2} (-k+1) e^{k}\end{align}$$
Therefore, 
$$\int_{-\infty}^{\infty} dx  \frac{e^{i k x}}{(1+x^2)^2} = \frac{\pi}{2} (1+|k|) e^{-|k|}$$
Method 2: Convolution
Knowing that the FT of $1/(1+x^2)$ is $\pi \, e^{-|k|}$, we may use the convolution theorem to deduce that
$$\int_{-\infty}^{\infty} dx  \frac{e^{i k x}}{(1+x^2)^2} = \frac{\pi^2}{2 \pi} \int_{-\infty}^{\infty} dk' e^{-|k'|} \, e^{-|k-k'|}$$
Again, the way we evaluate the integral on the RHS depends on the sign of $k$.  For $k \gt 0$, this integral is
$$\frac{\pi}{2} \int_{-\infty}^0 dk' \, e^{k'} \, e^{-(k-k')} + \frac{\pi}{2} \int_{0}^k dk' \, e^{-k'} \, e^{-(k-k')}+ \frac{\pi}{2} \int_{k}^{\infty} dk' \, e^{-k'} \, e^{k-k'}$$
For $k \lt 0$, on the other hand, the integral is
$$\frac{\pi}{2} \int_{-\infty}^k dk' \, e^{k'} \, e^{-(k-k')} + \frac{\pi}{2} \int_{k}^0 dk' \, e^{k'} \, e^{k-k'}+ \frac{\pi}{2} \int_{0}^{\infty} dk' \, e^{-k'} \, e^{k-k'}$$
Evaluation of the above integrals reproduces the result derived above.
