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Singularity of function of complex variable

  1. Jan 14, 2009 #1
    1. The problem statement, all variables and given/known data
    I have to proof that this equation:
    [tex]x_r(\omega)=\frac{1}{\pi}*PV \int_{-\infty}^{\infty}\frac{x_i(\omega')}{(\omega'-\omega)}d\omega'[/tex]
    (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:
    [tex]x_r(\omega)=\frac{2}{\pi}* \int_{0}^{\infty}\frac{\omega' x_i(\omega')-\omega x_i(\omega)}{\omega'^2-\omega^2}d\omega'[/tex].

    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:
    [tex]x_r(\omega)=\frac{2}{\pi}*PV \int_{0}^{\infty}\frac{\omega' x_i(\omega')}{(\omega'^2-\omega^2)}d\omega'[/tex]

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

    The definition of PV that I have used in my calculus is the follow:
    [tex]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')[/tex]

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

    [tex]\omega x_i(\omega)[/tex]

    [tex]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'[/tex]
    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
  2. jcsd
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