Feynman trick for linear propagators?

[earth2]

Hey guys,

say i have some standard propagators then I know how to combine them using Feynman's parameter method. But what do I do if one of these propagators is linear?
For instance:

$$\int d^Dk \frac{1}{(k-p)^2(k \cdot q)}$$

where q and p are some momenta. How do I combine them?
Does anyone have an idea about that?

Cheers,
earht2

 

[Simon_Tyler]

I don't see why the the Feynman parameters don't work in this case.

Use the standard formula http://en.wikipedia.org/wiki/Feynman_diagram#Combining_Denominators

$$\frac{1}{AB}= \int_0^1 \frac{1}{( vA+ (1-v)B)^2} dv$$

and complete the square in the denominator as per usual.

$$\frac{1}{(k-p)^2}\frac{1}{k\cdot q} <br /> = \int_0^1 \frac{1}{( v (k-p)^2+ (1-v)k\cdot q)^2} dv<br /> = \int_0^1 \frac{1}{( v (k-(p-q(1-v)/(2v))^2+ f(q,p,v))^2} dv$$

Then shift the variables in the momentum integral to get

$$\int d^dk \frac{1}{(k-p)^2}\frac{1}{k\cdot q} <br /> = \int_0^1 \int d^dk \frac{1}{( v k^2+ f(q,p,v))^2} dv$$

then use the standard formula for integrating the dimensionally regularized momentum integral.