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A Feynman integral with three propagators

  1. Nov 13, 2017 #1
    Hi guys,
    So I'm trying to compute this Feynman integral:
    $$ V=\dfrac {-i} {2} \int {\dfrac {d^4 k} {(2\pi)^4}} \dfrac {1} {k^2 - m^2} \dfrac {1} {(k+P_1)^2 -m^2} \dfrac {1} {(k+P_1 +P_2)^2 -m^2}$$
    I have introduced the Feynman parameters and now have the integral:
    $$ V=-i \int dx_1 \int dx_2 \int dx_3 \int {\dfrac {d^4 k} {(2\pi)^4}} \dfrac {\delta (1-x_1 -x_2 -x_3)} {-x_1 (k^2 - m^2) -x_2 ((k+P_1)^2 -m^2) -x_3 ((k+P_1 +P_2)^2 -m^2)}$$
    Now focusing on the denominator I expand it out and need to complete the square and shift the integration variable according to Peskin Schroeder, but i'm not sure how to do this. Here is what I have so far:
    $$ -x_1 k^2 -x_1 m^2 -x_2 k^2 -2x_2 k \cdot P_1 - x_2 P_1^2 -x_2 m^2 -x_3 k^2 -2 x_3 k \cdot P_1 -2 x_3 k \cdot P_2 - x_3 P_1 ^2 -x_3 P_2 ^2 -2x_3 P_1 \cdot P_2 -x_3 m^2 $$

    I know that I need to get rid of the terms with dot products but I'm not sure how, any guidance would be awesome :)
     
  2. jcsd
  3. Nov 13, 2017 #2

    Orodruin

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    First of all, use the ##\delta## to collect all of the ##k^2## terms into a single one with the prefactor 1. Second, collect all of the terms with an inner product with ##k## and write it on the form ##2k\cdot V##, where ##V## is some sum of 4-vectors. Finally, complete the square.
     
  4. Nov 13, 2017 #3
    Hi,
    Thanks for your response :) I'm not sure exactly what you mean by using the delta to collect the k squared terms?
     
  5. Nov 13, 2017 #4

    Orodruin

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    Because of the delta function, ##x_1 + x_2 + x_3 = 1## whenever the integrand is non-zero.
     
  6. Nov 13, 2017 #5
    Oh ok thanks, so I also do that for the mass terms as well. Now I have:
    $$= -k^2 - 2k\cdot (x_2 P_1 +x_3 P_1 +x_3 P_2) - 2 x_3 P_1 \cdot P_2 -m^2 $$
    Now to complete the square I must be making some mistake, I get this:
    $$= (-k^2 - 2k\cdot (x_2 P_1 +x_3 P_1 +x_3 P_2) +(x_2 P_1 +x_3 P_1 +x_3 P_2)^2) - 2 x_3 P_1 \cdot P_2 - (x_2 P_1 +x_3 P_1 +x_3 P_2)^2 -m^2 $$
    This doesn't appear to help me though..
     
  7. Nov 13, 2017 #6

    Orodruin

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    You should have the opposite sign for the last term in the completed square. Once you have completed the square, you can make the variable shift.
     
  8. Nov 13, 2017 #7
    Ah yes, thanks :) so would the appropriate momentum shift be:
    $$ l = k+ x_2 P_1 + x_3 P_1 + x_3 P_2 $$
    I'm not at all clear on how to shift the integration variable.
     
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