Contour integration with variable epsilon I've been wracking my brain on a particular contour integration which I thought I solved easily enough, but the answer says differently. It asks to prove the following for any $\xi \in \Bbb{R}$:
$$\int_{-\infty}^{\infty} \frac{e^{-2\pi i x \xi}}{(1+x^2)^2}dx=\frac{\pi}{2}(1+2\pi |\xi|)e^{-2\pi |\xi|}$$
My idea was to just use the residue theorem. The numerator is an entire function, and the denomenator has zeroes at $i$ and $-i$ with an order of $2$. Unless I'm mistaken, the residues of these poles are
$$ res_{i}=\frac{i}{4}e^{2\pi \xi}(2\pi \xi - 1)$$
$$ res_{-i}=-\frac{i}{4}e^{-2\pi \xi}(2\pi \xi + 1)$$
I then proceeded to integrate along the semicircle in the top half of the complex plane. Going along the circle portion with radius $R$, you get
\begin{align}
|\int_{0}^{\pi}\frac{e^{-2\pi i Re^{i\theta} \xi}}{(1+R^2e^{2i\theta})^2}Rie^{i\theta}d\theta| \qquad
&\le& 
\int_{0}^{\pi}|\frac{e^{-2\pi i Re^{i\theta} \xi}}{(1+R^2e^{2i\theta})^2}Rie^{i\theta}|d\theta \\
\\
&=&
\int_{0}^{\pi}\frac{Re^{-2\pi R\xi \sin{\theta}}}{|1+R^2e^{2i\theta}|^2}d\theta \\
\\
&\le& 
\int_{0}^{\pi}\frac{Re^{-2\pi R\xi \sin{\theta}}}{R^4-R^2-1}d\theta
\end{align}
which tends to zero as $R\to \infty$. The requested integral is the only one left, with the contour enclosing the $i$ pole, according to the residue theorem you would get
\begin{align}
\int_{-\infty}^{\infty} \frac{e^{-2\pi i x \xi}}{(1+x^2)^2}dx
&=&
2\pi i \cdot res_{i}
\\
\\
&=&
\frac{\pi}{2}(1 - 2\pi \xi)e^{2\pi \xi}
\end{align}
which is obviously not what was given. But I have absolutely no idea what part of my work I'm in the wrong here. The problem also specifies a hint, which is to handle $\xi < 0$ and $\xi \ge 0$ separately, but I don't understand what difference that's supposed to make. Any tips or corrections are greatly appreciated, thanks in advance!
Also, as a bit of a sidenote, I tried googling for help on this specific integral without much luck. However I get the feeling that it's supposedly related to something in statistics, am I right?

Let $z=x+iy$. Note that
$$
|{e^{-2\pi i z \zeta}}|=e^{2 \pi\zeta y}.
$$
Hence if $\zeta\geq 0$, the contour should be the semicircle in the lower plane and if $\zeta< 0$, the contour should be the semicircle in the upper plane to induce the fact that the integral go to zero over the arc.
