Let $f:\mathbb R \rightarrow \mathbb R$ given by $f(x)=0$ if $x\leq 0$ and $f(x)=e^{-x}$ if $x > 0$, I calculated Fourier's transform of $f$, i.e. $F(\omega)$ and I got $F(\omega)=\frac{1}{\sqrt{2\pi}\cdot( 1+i\omega)}$.
I am trying to find out how can I use this fact to calculate $\int_{0}^{\infty} \frac{cos(2\omega)+\omega\sin(2\omega)}{1+\omega^2}d\omega$ but it seems That I am missing something?
HINT:
$$ \frac{e^{i\omega x}}{1+i\omega}=\frac{\cos(\omega x)+\omega \sin(\omega x)}{1+\omega^2}+i\, \frac{\sin(\omega x)-\omega \cos(\omega x)}{1+\omega^2}\tag 1$$
Note that the imaginary part of the right-hand side of $(1)$ is odd.