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Is there any general solution of contour integral with higher order poles on real axis?

I have got one but valid for simple pole only.

$\mathbf{P}\int\limits_{-\infty}^{\infty}Q(x)\mathrm{d}x=2\pi\mathrm{i}\sum\limits_{y>0}\mathrm{Res}[Q(z)]+\pi\mathrm{i}\sum\limits_{y=0}\mathrm{Res}[Q(z)]$

I have the integral with form like this:- $\int\limits_{-\infty}^{\infty}\frac{(Az^2+\mathrm{i}Bz+C)\mathrm{e}^{-\mathrm{i}za}}{z^2(A_0z^2+\mathrm{i}B‌​z+C)}\mathrm{d}z\quad$ where, $\quad a,A,A_0,B,C>0$

Soumyajit Roy
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    The principal value integral exists if and only if the principal part at all poles on the real axis contains only odd powers. In that case, what happens if you let the radius of the semicircle around the pole shrink to $0$? – Daniel Fischer Sep 10 '15 at 18:36
  • Unfortunately, not really. Typically, to perform such an integration, one introduces a semicircular detour of radius $\epsilon$ to avoid the pole on the axis. For a simple pole, the integral about the detour converges to a finite limit as $\epsilon \to 0$. For higher order poles, the terms that diverge as $\epsilon \to 0$ do not vanish upon integrating over the half-circle. – Ron Gordon Sep 10 '15 at 18:37
  • I have the integral with form like this:-

    $\int\limits_{-\infty}^{\infty}\frac{Az^2+\mathrm{i}Bz+C}{z^2(A_0z^2+\mathrm{i}Bz+C)}\mathrm{d}z

     – Soumyajit Roy Sep 10 '15 at 19:11
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    I have edited my question by putting the integral form, as I am not so familiar with the theorems behind contour integral. – Soumyajit Roy Sep 10 '15 at 19:17

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