Undergrad Loop Integral Form: Finding a Workable Solution without Regularization

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The discussion revolves around calculating a divergent integral resembling a loop integral, specifically ∫_x^∞ du (u^2 / (ω - u)). The original poster seeks a workable form without introducing a cutoff, acknowledging the integral's divergence. Participants suggest evaluating the integral as a contour integral on the complex plane, particularly around the pole at u = ω. The conversation emphasizes the importance of the integral's context, noting that it converges in the Cauchy Principal Value sense. Ultimately, the feasibility of evaluating the integral depends on its construction and the specific conditions of the problem.
DeathbyGreen
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Hi,

I'm trying to calculate an integral which looks unfortunately divergent. The structure is similar to a loop integral but the appendix in the Peskin textbook didn't have a useable form. The integral form is (I did a u substitution to make it easier to look at)

<br /> \int_x^{\infty}du \frac{u^2}{\omega - u}<br />

Does anyone know of a workable form for this? Introducing a cutoff is possible but I would prefer not to.

Edit: I know that it is divergent. I was hoping for some sort of regularization technique which would allow for a finite answer under certain conditions.
Thank you!
 
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Yes, the integral is divergent as stated. What is the context of this problem? The title says this is a loop integral. If this is indeed a contour integral, then it can be evaulated on the complex plane about the pole at ##u=w##.
 
The integral occurs in a conductivity calculation I'm working through, it has a imaginary convergence factor in the denominator which I didn't write. So you think I could evaluate this as a standard contour and not need the fancy QFT loop integral forms?
 
It depends on what you are trying to evaluate. Without seeing the problem, I am not sure if the integral has been constructed correctly. If your goal is to evaluate a closed line integral, then the integral is zero if the loop does not enclose the pole and is ##2\pi i\text{Res}(f)## if it encloses the pole.
 
The integral converges in the Cauchy Principal Valued sense. For example:

##\text{P.V.}\displaystyle\int_0^2 \frac{z^2}{1-z}dz=-4##
 
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