- #1
Elwin.Martin
- 207
- 0
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!
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!