1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Peskin Eq 11.72, mathematical identity

  1. Feb 7, 2010 #1

    Hao

    User Avatar

    In Eq 11.72 in the QFT text by Peskin, the following equality is stated:

    [tex]i\int\frac{d^{d}k_{E}}{(2\pi)^{d}}\log(k_{E}^{2}+m^{2})=-i\frac{\partial}{\partial\alpha}\int\frac{d^{d}k_{E}}{(2\pi)^{d}}\frac{1}{(k_{E}^{2}+m^{2})^{\alpha}}|_{\alpha=0}[/tex]

    This suggests that

    [tex]\log(k_{E}^{2}+m^{2})=-\frac{\partial}{\partial\alpha}\frac{1}{(k_{E}^{2}+m^{2})^{\alpha}}|_{\alpha=0}[/tex]

    However, I can't see how this identity follows. Differentiating the right hand side gives

    [tex]-\frac{\partial}{\partial\alpha}\frac{1}{(k_{E}^{2}+m^{2})^{\alpha}}|_{\alpha=0}=\frac{\alpha}{(k_{E}^{2}+m^{2})^{\alpha+1}}|_{\alpha=0}\rightarrow\frac{0}{(k_{E}^{2}+m^{2})^{1}}[/tex]

    Any help would be greatly appreciated.
     
  2. jcsd
  3. Feb 7, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    [tex]\partial_\alpha x^\alpha = \partial_\alpha e^{\alpha \log x} = e^{\alpha \log x} \log x= x^\alpha \log x[/tex]
     
  4. Feb 7, 2010 #3

    Hao

    User Avatar

    Awesome!

    Thanks!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook