I was trying to show that when $\xi\in\mathbb{R}$, $$\int_{-\infty}^\infty \frac{e^{-2\pi i x\xi}}{(1 + x^2)^2}\,\mathrm{d}x = \frac{\pi}{2}(1+2\pi|\xi|)e^{-2\pi|\xi|}$$ with contour integration. I was able to solve it except the absolute sign around $\xi$.
My solution:
Let $f(z) = e^{-2\pi i z\xi}/(1+z^2)^2$. Set the contour $\gamma$ to be the semicircle with radius $R$ above the $x$-axis, i.e. $\gamma = \gamma_1 \cup \gamma_2$, where $\gamma_1 = [-R, R]$ and $\gamma_2 = Re^{i\theta}, \theta\in[0,\pi]$.
Within $\gamma$, there is a double pole of $f$ at $z = i$. By the residue theorem, $$\int_\gamma f(z)\,\mathrm{d}z = 2\pi i \mathrm{Res}(f, i) = \frac{\pi}{2}(1-2\pi\xi)e^{2\pi\xi}.$$
As $R\rightarrow\infty$, $$\left|\int_{\gamma_2} f(z)\,\mathrm{d}z\right| \leq \int_{0}^{\pi}\frac{e^{-2\pi i R e^{i\theta}\xi}iRe^{i\theta}}{(1+R^2e^{2i\theta})^2}\,\mathrm{d}\theta\leq \int_0^\pi \frac{Re^{2\pi\xi\sin\theta}}{(R^2 - 1)^2}\leq \frac{\pi Re^{2\pi\xi}}{(R^2-1)^2} \rightarrow 0.$$
Therefore, I get $$\int_{-\infty}^\infty \frac{e^{-2\pi i x \xi}}{(1 + x^2)^2}\,\mathrm{d}x = \frac{\pi}{2}(1-2\pi\xi)e^{2\pi\xi}.$$
I could not figure out where the absolute value should come into place.