Langevin equation. (complex integration method). I have the following ODE: 
$$m \ddot{x} = -m\gamma \dot{x} -m\omega_0^2 x +R(t)$$
where $R(t)$ is a random noise.
I found by taking Fourier transform that $X(\omega)= \frac{R(\omega)}{m(\omega_0^2-\omega^2)+im\gamma}$; so the power spectrum of the position variable is:
 $$I_x(\omega)= \frac{I_R(\omega)}{m^2(\omega_0^2-\omega^2)^2+\gamma^2}$$
I am assuming that $I_R(\omega)=constant$, i.e it's constant with respect to $\omega$.
Now, I want to calculate the correlator of the position variable with Khintchin-Weiner Theorem, i.e. I need to compute:
$$\phi_x(t) = \int_{-\infty}^\infty I_x(\omega)e^{i\omega t}d\omega$$
But I don't see how to compute it by residue integration, I am a bit rusty with these methods.
Can someone show me how to do this?
Thanks in advance.
 A: $$
I_R\int_{-\infty}^{\infty}\frac{\mathrm{e}^{i\omega t}}{m^2(\omega_0^2-\omega^2)^2+\gamma^2} d\omega
$$
Lets start by partial fractions
$$
m^2(\omega_0^2-\omega^2)^2+\gamma^2 = (m(\omega_0^2-\omega^2) - i\gamma)(m(\omega_0^2-\omega^2) + i\gamma)
$$
we can split as follows
$$
\frac{1}{m^2(\omega_0^2-\omega^2)^2+\gamma^2 } = \frac{1}{2i\gamma}\left[\frac{1}{m(\omega_0^2-\omega^2) - i\gamma} - \frac{1}{m(\omega_0^2-\omega^2) + i\gamma}\right]
$$
so your integral becomes
$$
\frac{I_R}{2\gamma i}\int_{-\infty}^{\infty}\frac{\mathrm{e}^{i\omega t}}{m(\omega_0^2-\omega^2) - i\gamma} - \frac{\mathrm{e}^{i\omega t}}{m(\omega_0^2-\omega^2) + i\gamma}d\omega
$$
then you have two integrals which you need to work with
$$
m(\omega_0^2-\omega^2) - i\gamma = m\left(\omega_0^2-\frac{i\gamma}{m}\right) - m\omega^2 = m\left(\sqrt{\omega_0^2-\frac{i\gamma}{m}} - \omega\right)\left(\sqrt{\omega_0^2-\frac{i\gamma}{m}} + \omega\right)
$$
and repeat the above steps then find the residue.
Most likely a more efficient way than this. 
To compute the residue you need to choose a domain/contour that you want to compute the integrals over and you have to be consistent with your choice for all four of the resulting integrals.
The roots of a complex equation can be determined using
$$
z = r\mathrm{e}^{i\phi}
$$
so we can have 
$$
r\cos \phi + i \sin \phi = \omega_0^2-\frac{i\gamma}{m}
$$
or
$$
r\cos \phi = \omega_0^2\\
r\sin\phi = -\frac{\gamma}{m}
$$
solve for $r$ and $\phi$ then find the "linear" expression of the root and proceed.
