# Homework Help: Singularity of function of complex variable

1. Jan 14, 2009

### TonyLowe

1. The problem statement, all variables and given/known data
I have to proof that this equation:
$$x_r(\omega)=\frac{1}{\pi}*PV \int_{-\infty}^{\infty}\frac{x_i(\omega')}{(\omega'-\omega)}d\omega'$$
(where P denotes Principal Value Integration of Cauchy, r and i denotes rispectively real and imaginary part of x function)
is equivalent to this equation:
$$x_r(\omega)=\frac{2}{\pi}* \int_{0}^{\infty}\frac{\omega' x_i(\omega')-\omega x_i(\omega)}{\omega'^2-\omega^2}d\omega'$$.

This equation (the first) is the first of the Kramers-Kronig relations for complex function.

2. Relevant equations
Using symmetry property of x(w), I can proof that the first equation is equivalent to:
$$x_r(\omega)=\frac{2}{\pi}*PV \int_{0}^{\infty}\frac{\omega' x_i(\omega')}{(\omega'^2-\omega^2)}d\omega'$$

I also show, using PV definition, that:
$$PV \int_{0}^{\infty}\frac{1}{(\omega'^2-\omega^2)}d\omega'=0$$

The definition of PV that I have used in my calculus is the follow:
$$PV \int_{-\infty}^{\infty} f(\omega')d\omega'=\lim_{r \to 0} (\int_{-\infty}^{\omega-r} f(\omega')d\omega'+\int_{\omega+r}^{\infty} f(\omega')d\omega')$$

3. The attempt at a solution
Starting from:
$$x_r(\omega)=\frac{2}{\pi}*PV \int_{0}^{\infty}\frac{\omega' x_i(\omega')}{(\omega'^2-\omega^2)}d\omega'$$
I have added and removed from the numerator the quantity

$$\omega x_i(\omega)$$

so:
$$x_r(\omega)=\frac{2}{\pi}*PV \int_{0}^{\infty}\frac{\omega' x_i(\omega')-\omega x_i(\omega)}{(\omega'^2-\omega^2)}d\omega'+\frac{2}{\pi}*PV \int_{0}^{\infty}\frac{\omega x_i(\omega)}{(\omega'^2-\omega^2)}d\omega'$$
The second integral is zero; the first integral is the solution of the problem a part of the PV operators. I think that if I justify that the integrands have no singularity at w'=w, the problem is solve (in this case, in fact, I can remove PV operators from first integral).
But I don't understand why that integrands is not singular for w'=w: infact in w there is a pole of the function x(w) (from the complete proof of Kramers-Kronig relations I have understand that w is a pole, with no imaginary part, of x(w)), and the denominator goes to zero when w'=w, as the numerator.
Someone can help me?
Thank very much

Last edited: Jan 14, 2009