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I Generic Feynman parameterisation

  1. Jun 8, 2017 #1

    CAF123

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    I am in the process of reducing tensor integrals down to a sum of scalar ones with the tensor structure factored out in some basis of decomposition. I am able to write some scalar products in terms of appearing propagators but I encountered one where I have something like $$\int_k \frac{k^2-m^2}{A_1(k) A_2(k) A_3(k) A_4(k)}$$ where ##A_i## are all different propagators depending on k and external momenta.

    Now, turns out I can write ##k^2 - m^2 = A_3 - f(p_i) - m^2 - 2k \cdot g(p_i)##. The first three terms pose no problem but the k dependence in the last term does. For only this particular term, I was thinking of using Feynman parameters to write my denominator of four terms in terms of one single propagator as follows $$\frac{1}{A_1 A_2 A_3 A_4} \sim \frac{1}{(k^2 - \Delta)^4}$$ and then argue that under the integral $$\int d^dk \frac{k \cdot g(p_i)}{(k^2 - \Delta)^4} = 0$$ from symmetric integration.

    So, my question is, without doing the lengthy calculation of feynman parameters is it true that I can write $$\frac{1}{A_1 A_2 A_3 A_4} \sim \frac{1}{(k^2 - \Delta)^4}$$ where the ##\sim## accounts for the three integrations over feynman parameters but otherwise crucially the k dependence is simply as shown?

    Thanks!
     
  2. jcsd
  3. Jun 13, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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