RedX
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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?