Analytic verification of Kramers-Kronig Relations

[Septim]

Homework Statement

Show that the real and imaginary parts of the following susceptibility function satisfy the K-K relationships. Use the residue theorem.
$$ \chi(\omega) = \frac{\omega_{p}^2}{(\omega_0^2-\omega^2)+i\gamma\omega} $$

Homework Equations

The Kramers-Kronig relations are

$$ \chi_r(\omega) = \frac{1}{\pi} P \int_{-\infty}^{\infty} d\bar{\omega} \frac{\chi_i(\bar{\omega})}{\bar{\omega}-\omega} $$
$$ \chi_i(\omega) = -\frac{1}{\pi} P \int_{-\infty}^{\infty} d\bar{\omega} \frac{\chi_r(\bar{\omega})}{\bar{\omega}-\omega} $$

The Attempt at a Solution

The problem is that my complex calculus is pretty rusty and I do not know which poles contribute exactly. There are 5 poles in total 4 from the susceptibility function and 1 from the denominator(see the expressions please). The poles are
$$\pm \gamma \pm i\frac{\sqrt{4\omega_0^2-\gamma}}{2} $$

I calculated the residues that are on the negative imaginary $\omega$ plane on Mathematica and they turned out to be zero except the pole at $\omega$ which is the value of the integral using the principal rule in Mathematica. Is this analytically tractable easily? I appreciate any assistance you provide.

Many thanks,

 

[TSny]

$$ \chi(\omega) = \frac{\omega_{p}^2}{(\omega_0^2-\omega^2)+i\gamma\omega} $$
There are 5 poles in total 4 from the susceptibility function and 1 from the denominator(see the expressions please).

$\chi(\omega)$ has only two poles. If you locate these poles, you will see that you can choose the contour so that it doesn't enclose either of these two poles.