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Expansion of nonregularized integral

  1. May 19, 2012 #1
    Can anyone help me with the expansion of the integral
    [tex]I_{\mu\nu}=\int d^{4}l\frac{4l_{\mu}l_{\nu}-g_{\mu\nu}l^2}{(l^{2}-B+i\epsilon)^{3}}[/tex].
    I would like to know how it could be expanded into two terms
    [tex]I_{\mu\nu}=\frac{1}{2}\int d^{4}l\frac{\partial^2}{\partial l^\mu \partial l^\nu}\frac{1}{l^{2}-B+i\epsilon}-g_{\mu\nu}\int d^{4}l\frac{B}{(l^{2}-B+i\epsilon)^{3}}[/tex]-.
    Another question is how can i calculate the first term to have result
    [tex]-g_{\mu\nu}i\pi^{2}/2[/tex].
    P.S. I am inexperienced with tensor calculations. I would be grateful for any help.
     
    Last edited: May 19, 2012
  2. jcsd
  3. May 20, 2012 #2
    It is all just a chain rule :)
    [tex]\frac{1}{2}\partial_{\mu}\partial_{\nu}\frac{1}{l^{2}-\triangle+i\epsilon}=\partial_{\mu}\left[\frac{-l_{\nu}}{\left[l^{2}-\triangle+i\epsilon\right]^{2}}\right][/tex]
    [tex]=\left[-g_{\mu\nu}\frac{l^{2}-\triangle}{\left[l^{2}-\triangle+i\epsilon\right]^{3}}+\frac{4l_{\nu}l_{\mu}}{\left[l^{2}-\triangle+i\epsilon\right]^{3}}\right][/tex]
     
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