Feynman integral with three propagators

• A
Milsomonk
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 :)

Staff Emeritus
Homework Helper
Gold Member
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.

Milsomonk
Hi,
Thanks for your response :) I'm not sure exactly what you mean by using the delta to collect the k squared terms?

Staff Emeritus
Homework Helper
Gold Member
Because of the delta function, ##x_1 + x_2 + x_3 = 1## whenever the integrand is non-zero.

Milsomonk
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..

Staff Emeritus
$$l = k+ x_2 P_1 + x_3 P_1 + x_3 P_2$$