# Analytic verification of Kramers-Kronig Relations

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1. Dec 9, 2014

### Septim

1. The problem statement, all variables and given/known data
Show that the real and imaginary parts of the following susceptibility function satisfy the K-K relationships. Use the residue theorem.
$$\chi(\omega) = \frac{\omega_{p}^2}{(\omega_0^2-\omega^2)+i\gamma\omega}$$
2. Relevant equations

The Kramers-Kronig relations are

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

3. 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,

2. Dec 14, 2014

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Dec 15, 2014

### TSny

$\chi(\omega)$ 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.