Complex integral over $\mathbb{R}$

Question

Compute $$\int_{\mathbb{R}} \dfrac{i \omega - \omega^2}{1+\omega^6} d\omega$$

Consulting with software such as Mathematica yields the neat answer $-\dfrac{\pi}{3}$. How can I show this? Methods in the realms of complex analysis, Fourier transforms or plain old calculus are preferable.

Answer 1

The imaginary part is simple: it's an odd integrable function and therefore its integral is $0$.

And$$\int_{-\infty}^{+\infty}\frac{\omega^2}{1+\omega^6}\,\mathrm d\omega=\left[\frac{\arctan(\omega^3)}3\right]_{-\infty}^{+\infty}=\frac\pi3.$$

Answer 2

Here's a "plain old calculus" solution:

Hint The imaginary part is odd, so it does not contribute to the value of the integral. Thus (up to sign) the integral is $$\int_{-\infty}^\infty \frac{\omega^2 \, d\omega}{1 + \omega^6} ,$$ and this can be handled with a straightforward substitution.