I have the frequency dependency of a signal and want to know the time dependency. Therefore I need to evaluate
$$ \int_{-\infty}^{\infty} \frac{e^{i\omega t} }{\sqrt{1 + c^2 \left(\frac{\omega \cdot \omega_0}{\omega^2-\omega_0^2} \right)^2}} d\omega $$
It looks sort of similiar to a Cauchy-distribution to me, but not quite. All variables are real, $t>0$ and $\omega_0 > 0$ can be assumed. If it helps, the approximation $\omega \approx \omega_0$ can also be made.
Alternatively, I could also work with the square of my signal:
$$ \int_{-\infty}^{\infty} \frac{e^{i\omega t} }{1 + c^2 \left(\frac{\omega \cdot \omega_0}{\omega^2-\omega_0^2} \right)^2} d\omega $$
Is this even analytically solvable? I tried contour integration, but the last time I did this is a few years back.
Let's go with the second integral. By rearranging the denominator we get that
$$\omega^4 + (c^2-2)\omega_0^2\omega^2 + \omega_0^4 = 0 \implies \frac{\omega^2}{\omega_0^2} = 1-\frac{c^2}{2}\pm\sqrt{\left(1-\frac{c^2}{2}\right)^2-1}$$
which gives us $3$ cases:
Case $1$: $c^2 > 4$
This implies $\omega^2 < 0 $ and we get four distinct purely imaginary roots.
Case $2$: $c^2 = 4$
$\omega^2$ is still negative but now there are only two imaginary roots, both with multiplicity two.
Case $3$: $c^2 < 4$
$\omega^2$ is now complex, giving four distinct complex roots.
Case $1$ and Case $3$ are actually not so different, but for simple calculations sake I will assume Case $2$ and let you do the others on your own. The integral becomes
$$I = \int_{-\infty}^\infty \frac{(\omega^2-\omega_0^2)}{(\omega-i\omega_0)^2(\omega+i\omega_0)^2}e^{i\omega t}\:d\omega$$
Now consider the line integral
$$\int_\gamma \frac{(\omega^2-\omega_0^2)}{(\omega-i\omega_0)^2(\omega+i\omega_0)^2}e^{i\omega t}\:d\omega$$
over the semicircular contour in the upper half plane of radius $R$. This gives us two integrals:
$$\int_\gamma = \int_{-R}^R + \int_{|\omega|=R\:\cap\:\operatorname{Im}(\omega) > 0}$$
treating $\omega$ as a complex variable. The second integral on the arc will go to zero as we take the limit $R\to\infty$ because on the upper half plane $e^{i\omega t} \to 0$ if $t>0$. Thus only the integral on the left remains, which happens to be the integral we want, and it equals the sum of the residues on the inside.
The only pole within the contour is a second order one at $\omega = i\omega_0$. Calculating the residue we get that
$$\operatorname{Res}(f(\omega),\omega = i\omega_0) = \lim_{\omega\to i\omega_0} \frac{d}{d\omega} \frac{(\omega^2-\omega_0^2)}{(\omega+i\omega_0)^2}e^{i\omega t} = \frac{it}{2}e^{-\omega_0 t}$$
Thus we have that the integral equals
$$I = \int_{-\infty}^\infty \frac{(\omega^2-\omega_0^2)}{(\omega-i\omega_0)^2(\omega+i\omega_0)^2}e^{i\omega t}\:d\omega = 2\pi i\left(\frac{it}{2}e^{-\omega_0 t}\right) = -\pi t e^{-\omega_0 t}$$
The other integrals will be easier because even though there are more residues, all of the poles will be first order, hence no derivatives that need to be taken. (Hint: there will only ever be two poles in the upper half plane).