Graduate Feynman integral with three propagators

[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 :)

 

[Orodruin]

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?

 

[Orodruin]

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..

 

[Orodruin]

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.

 

[Milsomonk]

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.