Complex Fourier integral Why is the $\omega$ in the solution for this integral written in absolute value?
$$\int_{-\infty}^{\infty} \frac{x e^{i\omega x}}{(x^2+1)^2}dx = \frac{\pi \omega}{2}e^{-|\omega|}$$
 A: First of all, imagine inverting that transform to get back the original function.  If there were no absolute value, the integral defining the inverse transform would not converge.
As to where the absolute value comes from, we may see this from the computation of the integral using residues.  Consider the integral
$$\oint_C dz \frac{z e^{i \omega z}}{(z^2+1)^2}$$
When $\omega \gt 0$, we may take $C$ to be a semicircle of radius $R$ in the upper half-plane. In this case, the integral about the circular arc vanishes as $R \to \infty$.  When $\omega \lt 0$, however, $C$ cannot be in the upper-half-plane, but the lower half-plane so that the integral about the circular arc vanishes.
In detail, when $C$ is in the upper half-plane, then along the circular arc, $z=R e^{i \theta}$, $\theta \in [0,\pi]$, and
$$e^{i \omega z} = e^{I \omega R \cos{\theta}} e^{-\omega R \sin{\theta}}$$
Note that $\sin{\theta} \ge 0$ here, so the exponential term vanishes as $R \to\infty$ when $\omega \gt 0$.  This is not the case, however, when $\omega \lt 0$.  When $C$ is in the lower half-plane, then $\theta \in [\pi,2 \pi]$, so that $\sin{\theta} \le 0$, and the exponential term vanishes when $\omega \lt 0$.  
Thus, the exponential term will have, after residue calculation, a $\pm \omega$ term, depending on the sign of $\omega$; thus the absolute value.
ADDENDUM
I simply cannot resist working this one out.  It is easier to work with
$$\int_{-\infty}^{\infty} dx \frac{e^{i \omega x}}{(x^2+1)^2}$$
and then take $-i$ times the derivative wrt $\omega$ at the end.  By the arguments above and the residue theorem, we have for $\omega \gt 0$
$$\int_{-\infty}^{\infty} dx \frac{e^{i \omega x}}{(x^2+1)^2} = i 2 \pi \operatorname*{Res}_{z=i} \frac{e^{i \omega z}}{(z^2+1)^2} = i 2 \pi \left [\frac{d}{dz} \frac{e^{i \omega z}}{(z+i)^2} \right ]_{z=i} = \frac{\pi}{2} (1+\omega) e^{-\omega}$$
Similarly, for $\omega \lt 0$:
$$\int_{-\infty}^{\infty} dx \frac{e^{i \omega x}}{(x^2+1)^2} = -i 2 \pi \operatorname*{Res}_{z=-i} \frac{e^{i \omega z}}{(z^2+1)^2} = -i 2 \pi \left [\frac{d}{dz} \frac{e^{i \omega z}}{(z-i)^2} \right ]_{z=-i} = \frac{\pi}{2} (1-\omega) e^{\omega}$$
From this, one may deduce that
$$\int_{-\infty}^{\infty} dx \frac{x e^{i \omega x}}{(x^2+1)^2} = i \frac{\pi}{2} \omega e^{-|\omega|}$$
