RedX
Jul21-10, 04:00 PM
I was looking at a paper that used dimensional regularization and the following expression was derived:
\int dx \mbox{ }[p^2(1-x)^2-\lambda^2(1-x)]^{\epsilon}
Factoring out p^2(1-x)^2 :
\int dx \mbox{ }[p^2(1-x)^2]^{\epsilon}[1-\frac{\lambda^2}{p^2(1-x)}]^{\epsilon}
The part that I don't understand is that they expanded the rightmost factor in binomial expansion. \lambda^2 is smaller than p^2 (in fact \lambda^2=p^2-m^2 ), but the 1/(1-x) changes all that when x approaches 1, making \frac{\lambda^2}{p^2(1-x)} much greater than 1.
Is it okay to expand the rightmost factor in binomial expansion because when x goes to 1, the factor on the left [p^2(1-x)^2]^{\epsilon} goes to zero, so it doesn't matter what's on the rightmost side at that point? If so, isn't there an error term that needs to be calculated?
\int dx \mbox{ }[p^2(1-x)^2-\lambda^2(1-x)]^{\epsilon}
Factoring out p^2(1-x)^2 :
\int dx \mbox{ }[p^2(1-x)^2]^{\epsilon}[1-\frac{\lambda^2}{p^2(1-x)}]^{\epsilon}
The part that I don't understand is that they expanded the rightmost factor in binomial expansion. \lambda^2 is smaller than p^2 (in fact \lambda^2=p^2-m^2 ), but the 1/(1-x) changes all that when x approaches 1, making \frac{\lambda^2}{p^2(1-x)} much greater than 1.
Is it okay to expand the rightmost factor in binomial expansion because when x goes to 1, the factor on the left [p^2(1-x)^2]^{\epsilon} goes to zero, so it doesn't matter what's on the rightmost side at that point? If so, isn't there an error term that needs to be calculated?