1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dimensional Regularization in Peskin

  1. Jun 19, 2015 #1
    1. The problem statement, all variables and given/known data
    I'm trying to understand dimensional regularization with Peskin. There is a transitions that is not clear.

    2. Relevant equations
    On page 250, the general expression for the d-dimensional integral is given:
    ##\int \frac{d^d l_E}{(2\pi)^d}\frac{1}{(l_E^2+\Delta)^2}=\frac{1}{(4\pi)^{d/2}}\frac{\Gamma(2-\frac{d}{2})}{\Gamma(2)}\left(\frac{1}{\Delta}\right)^{2-\frac{d}{2}}##.
    So far everything is clear. But then, in 7.84 he writes
    ##\int \frac{d^d l_E}{(2\pi)^d}\frac{1}{(l_E^2+\Delta)^2}\rightarrow\frac{1}{(4\pi)^2}\left(\frac{2}{\epsilon}-\log\Delta-\gamma+\log(4\pi)+\mathcal{O}(\epsilon)\right),##
    when ##d\rightarrow4##.
    3. The attempt at a solution
    I understand where the ##\frac{2}{\epsilon}## and ##-\gamma## factors come from, but where did the terms involving the logarithm function came from? Even if I take the eventual integration over the Feynman parameters into account I don't get the correct answer.
     
  2. jcsd
  3. Jun 19, 2015 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Use the fact that ##a^b = \exp(b \ln a)## and expand the exponent in ##\epsilon##. This gives a linear term proportional to the logs, which will be multiplied by the ##1/\epsilon## dependence from the gamma function and therefore give a term constant in ##\epsilon##.
     
  4. Jun 19, 2015 #3
    Thank you! Haven't thought of going that way.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Dimensional Regularization in Peskin
  1. Regular transformation (Replies: 0)

Loading...