# A Feynman integral with three propagators

1. Nov 13, 2017

### 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 :)

2. Nov 13, 2017

### Orodruin

Staff Emeritus
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.

3. Nov 13, 2017

### Milsomonk

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

4. Nov 13, 2017

### Orodruin

Staff Emeritus
Because of the delta function, $x_1 + x_2 + x_3 = 1$ whenever the integrand is non-zero.

5. Nov 13, 2017

### 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..

6. Nov 13, 2017

### Orodruin

Staff Emeritus
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.

7. Nov 13, 2017

### 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.