- #1
RedX
- 970
- 3
I was looking at a paper that used dimensional regularization and the following expression was derived:
[tex]\int dx \mbox{ }[p^2(1-x)^2-\lambda^2(1-x)]^{\epsilon} [/tex]
Factoring out [tex]p^2(1-x)^2 [/tex]:
[tex]\int dx \mbox{ }[p^2(1-x)^2]^{\epsilon}[1-\frac{\lambda^2}{p^2(1-x)}]^{\epsilon} [/tex]
The part that I don't understand is that they expanded the rightmost factor in binomial expansion. [tex]\lambda^2[/tex] is smaller than [tex]p^2 [/tex] (in fact [tex]\lambda^2=p^2-m^2 [/tex]), but the 1/(1-x) changes all that when x approaches 1, making [tex]\frac{\lambda^2}{p^2(1-x)} [/tex] 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 [tex][p^2(1-x)^2]^{\epsilon} [/tex] 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?
[tex]\int dx \mbox{ }[p^2(1-x)^2-\lambda^2(1-x)]^{\epsilon} [/tex]
Factoring out [tex]p^2(1-x)^2 [/tex]:
[tex]\int dx \mbox{ }[p^2(1-x)^2]^{\epsilon}[1-\frac{\lambda^2}{p^2(1-x)}]^{\epsilon} [/tex]
The part that I don't understand is that they expanded the rightmost factor in binomial expansion. [tex]\lambda^2[/tex] is smaller than [tex]p^2 [/tex] (in fact [tex]\lambda^2=p^2-m^2 [/tex]), but the 1/(1-x) changes all that when x approaches 1, making [tex]\frac{\lambda^2}{p^2(1-x)} [/tex] 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 [tex][p^2(1-x)^2]^{\epsilon} [/tex] 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?