How to integrate exponential * fraction $$\int_{-\infty}^{\infty}-\frac{1}{2\pi(\omega^2+16)}e^{-i \omega t} d\omega$$
I tried using partial fractions to separate $1/(\omega^2+16)$ into $-1/(8*(4-i\omega))$ and $1/(8*(-4-i\omega))$ but that didn't lead to anything. 
 A: This is a classic piece of complex integration.
1. Replace the integral with a finite $\int_{-R}^R$ for a large number $R$.
2. Add a semicircle in the lower half-plane, radius $R$.  Here, $\Re -iwt\ll0$ so there is a negligible contribution from the semicircle.
3. You now have a closed loop, and the integral equals $-2\pi i$ times the residue where the function becomes infinite - at $\omega=-4i$.
4.  The other point where it becomes infinite - at $\omega=4i$ - doesn't matter as it is outside the loop.
5. The residue at $-4i$ is $$-\frac{e^{-i\omega t}}{2\pi(\omega-4i)}$$ with $\omega=-4i$, so it equals $\frac{e^{-4t}}{2\pi(-8i)}$.
6. Multiply that by $-2\pi i$ because the loop is clockwise, and get $\frac{e^{-4t}}8$
7. Lastly, let $R\to\infty$, to get the full integral from $-\infty$ to $\infty$ and to remove the contribution from the semicircle.
A: I agree with Michael, the usual way to deal with that kind of integrals is to use contour integration and the residue theorem. In probabilistic terms, we are computing the characteristic function of a Cauchy distribution. By using integration by parts, it is not difficult to check that:
$$ \int_{\mathbb{R}} e^{ikx}\,e^{-|x|}\,dx = 2\int_{0}^{+\infty}\cos(kx)e^{-x}\,dx = \frac{2}{k^2+1}, $$
hence by applying an inverse Fourier transform we get:
$$ \int_{\mathbb{R}} \frac{e^{-ikx}}{x^2+1}\,dx = \pi\cdot e^{-|k|}. $$
