Singularity of function of complex variable

[TonyLowe]

Homework Statement

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.

Homework 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:
$$PV \int_{-\infty}^{\infty} f(\omega')d\omega'=\lim_{r \to 0} (\int_{-\infty}^{\omega-r} f(\omega')d\omega'+\int_{\omega+r}^{\infty} f(\omega')d\omega')$$

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

 

[mighty2000]

Thank you for bringing this interesting problem to our attention. I am a scientist and I would be happy to help you with this proof. First of all, I would like to commend you for your attempt at the solution and your understanding of the Kramers-Kronig relations. Your approach is on the right track, and I will provide some additional insights to help you complete the proof.

To start off, let's consider the first equation:
x_r(\omega)=\frac{1}{\pi}*PV \int_{-\infty}^{\infty}\frac{x_i(\omega')}{(\omega'-\omega)}d\omega'
This is the first Kramers-Kronig relation for a complex function x(w). As you correctly pointed out, we can use the symmetry property of x(w) to show that this 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'
Now, let's take a closer look at the integrand in this equation:
\frac{\omega' x_i(\omega')}{(\omega'^2-\omega^2)}
We can rewrite this as:
\frac{\omega' x_i(\omega')}{(\omega'-\omega)(\omega'+\omega)}
Now, we can see that the integrand has a pole at w'=w, as you mentioned. However, this pole is a simple pole, meaning that the denominator goes to zero linearly as w' approaches w. This means that the integral is still well-defined and the PV operator is not necessary. In other words, the integrand is not singular at w'=w, it just has a simple pole.

Now, let's move on to the second 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'
As you correctly pointed out, the second integral is zero. This is because the integrand can be rewritten as:
\frac{\omega' x_i(\omega')-\omega x_i(\omega)}{\omega'^2-\omega^2}=\frac{\omega' x_i(\omega')}{\