# Reallllly dumb question about Feynman Parameters (and simplifying them)

1. Apr 14, 2012

### Elwin.Martin

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
$\frac{1}{ABC} = 2 \int_0^1 dx \int_0^1 dy \int_0^1 dz \frac{ \delta \left( 1-x-y-z \right)}{(xA+yB+zC)^3}$
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:
$2 \int_0^1 dx \int_0^{1-x} dy$
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. Apr 14, 2012

### Elwin.Martin

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.

3. Apr 14, 2012

### Avodyne

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.)

4. Apr 14, 2012

### Elwin.Martin

That makes sense.

So for four parameters would we have:
$\int_0^1 dx \int_0^{1-x} dy \int_0^{1-x-y} dz$ ?

5. Apr 17, 2012

### Elwin.Martin

Bumping this, I would really like to know still...I feel like an idiot, but I sort of need to understand this.

6. Apr 17, 2012

Yes.