I have the following integral $$\int d^D l \frac{1}{p^2 (p-l)^2 l^2}$$ which I want to rexpress using feynman parameters. I can write as a first step, $$2 \int_0^1 dx \int_o^{1-x} dy \int d^D l \frac{1}{(p^2(1-x-y) + (p-l)^2x + l^2y)^3}$$ My problem is, when I expand the term in the denominator there I get $$ p^2(1-x-y) + (p-l)^2x + l^2y = l^2(x+y) - 2l \cdot p x$$ I would then proceed by writing this in the form $(l-a)^2 + b$ from which I can use standard dim reg formulae but I can't do that here because of the $x+y$ term attached to the $l^2$.

Any advice?

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    $\begingroup$ $p^2$ is independent of $l$ so it shouldn't get a parameter; just take it out of the integral and combine the other two denominators. $\endgroup$ – Javier Feb 28 '16 at 18:44
  • $\begingroup$ May be it would be helpful to first rewrite the $l$-dependent integrand factors in simpler fractions, and only then transition to Feynman parameters. For instance I get $$\frac{1}{(p-l)^2l^2} = \frac{1}{p^3}\left[ \frac{p}{(p-l)^2} + \frac{p}{l^2} + \frac{2}{p-l} + \frac{2}{l}\right]$$ $\endgroup$ – udrv Feb 28 '16 at 21:07

Note that your integration variable is $l$ so you may as well just consider the following integral: $$\int d^{D}l \frac{1}{l^2(p-l)^2}$$

So it is not necessary to include $p^2$ in the Feynman parametrisation. For $\frac{1}{AB}$, the Feynman parametrisation is given by


Therefore the integral becomes $$\int d^{D}l\int_0^1\frac{1}{(xl^2+(1-x)(p-l)^2)^2}=\int d^Dl\int_0^1\frac{dx}{(l^2-2p(1-x)+(1-x)p^2)^2}$$ and you can change integration variable and do wick rotation and dimensional regularization from here.

One more comment I want to make is that in the question it seems that you want to use Feynman parametrisation for $\frac{1}{ABC}$, although it turns out not to be necessary in this case. However, the one you used does not seem to be correct. See for example, Chapter 14 Exercise 1 in Mark Srednicki's QFT text book for a generalisation for Feynman parametrisation for $n$ variable.


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