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Feynman parameter integral

  1. May 29, 2010 #1
    I'm trying to show equation (6.39) on page 189 in Peskin & Schroeder.

    1. The problem statement, all variables and given/known data

    Show that [tex]\frac{1}{AB}=\int_0^1dxdy\delta(x+y-1)\frac{1}{(xA+yB)^2}[/tex].

    2. Relevant equations

    Defining property of Dirac delta function:

    [tex]\int_{-\infty}^{\infty}dx f(x)\delta(x-x_0)=f(x_0)[/tex]

    Also:

    [tex]\int_{x_0-\varepsilon}^{x_0+\varepsilon}dx f(x)\delta(x-x_0)=f(x_0)[/tex]

    3. The attempt at a solution

    Since the defining properties of the delta function requires the limits of integration to be outside the points where the argument of the delta function is zero, I cannot simply plug my expression into any of these relations. If I instead choose the limits of integration [tex]1+\varepsilon[/tex] and [tex]0-\varepsilon[/tex] I can use the second
    delta function identity from above to "force" x=1-y as I integrate over dx. So

    [tex]\int_{0+\varepsilon}^{1+\varepsilon}dxdy\delta(x+y-1)\frac{1}{(xA+yB)^2}
    =\int_{1+\varepsilon}^{1+\varepsilon}dy\frac{1}{[(1-y)A+yB]^2}[/tex]. After a few routine steps
    the expression I get goes to [tex]\frac{1}{AB}[/tex] in the limit [tex]\varepsilon\rightarrow 0[/tex], which is the result I want for the integral without the epsilons.

    Is this calculation relevant in obtaining the value of [tex]\int_0^1dx (...)[/tex] and how are the two integrals related? I'm getting very confused here.
     
  2. jcsd
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