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Reallllly dumb question about Feynman Parameters (and simplifying them)

  1. Apr 14, 2012 #1
    I know the generalized formula for Feynman parameters, my problem is in simplifying.

    What I mean is something like this:
    Take the simplest 3 parameter equation
    [itex] \frac{1}{ABC} = 2 \int_0^1 dx \int_0^1 dy \int_0^1 dz \frac{ \delta \left( 1-x-y-z \right)}{(xA+yB+zC)^3} [/itex]
    And you can take this and put move to two integrals by integrating over z, I understand that we use the delta function to get (xA+yB+(1-x-y)C)3 in the denominator of the new integrand...however, I don't understand why we now have:
    [itex] 2 \int_0^1 dx \int_0^{1-x} dy[/itex]
    I've done it out by hand to check it, so I know this is what we need...but why do we substitute in 1-x for the limit of integration?

    This is probably a dumb question, but I need to know so I can generalize to four parameters.

    Thanks for any and all help!
  2. jcsd
  3. Apr 14, 2012 #2
    Alright, so after drawing it by hand once...I feel like it has something to do with the shape of region the delta function sort of cuts out? We're realistically only integrating over a plane now, since the delta function assigns zero to everywhere except x+y+z=1, right? So our integration is really over just that plane...still failing from here though.
  4. Apr 14, 2012 #3


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    If y was allowed to be bigger than 1-x, then x+y would be bigger than 1. But x+y+z=1, and z is between 0 and 1. So x+y cannot be bigger than 1. Thus the upper limit on the y integral. (We could equally well first integrate over x from 0 to 1-y, and then over y from 0 to 1.)
  5. Apr 14, 2012 #4
    That makes sense.

    So for four parameters would we have:
    [itex] \int_0^1 dx \int_0^{1-x} dy \int_0^{1-x-y} dz [/itex] ?
  6. Apr 17, 2012 #5
    Bumping this, I would really like to know still...I feel like an idiot, but I sort of need to understand this.
  7. Apr 17, 2012 #6


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