Are the following operations O.K.? This is related to the Feynman parameter trick.
$$F:= \int_0^1 \mathrm{d}x\int_0^{1-x}\mathrm{d}y \frac{1}{f(x,y)+\mathrm{i}\epsilon}.$$ Now using
$$\frac{1}{A+i\epsilon} = PV\frac{1}{A}-i\pi\delta(A),$$ where $PV$ denotes the Cauchy Principal Value, we get (taking only the imaginary part):
$$\Im{F} = -\pi \int_0^1 \mathrm{d}x\int_0^{1-x}\mathrm{d}y\, \delta(f(x,y)) .$$
The trouble I got is that the zeros of $f(x,y)$ which I call $y^{\pm}$ seems to be outside integration range and hence the delta should yield zero. BUT here's what's funny: when I ignore all this and just perform the formal calculations (assuming I do it correctly) namely; replacing $\delta(f(x,y))$ with
$$\frac{1}{\bigl\vert \partial f/\partial y\bigr\vert_{y=y^{\pm}}}\times(\delta(y-y^-)+\delta(y-y^+)),\ \ \ (1)$$
(where $|\partial f/\partial y|_\pm $ are equal) and assuming that $y^{\pm}\in[0,1-x]$ (which seems to be false) the two deltas just give $1+1 = 2$. Then the result seems to be correct, or at least it agrees with what I have calculated the same thing using a totally different method.
Could this all just be a coincidence? I mean shouldn't the deltas produce zero if $y^{\pm}\notin[0,1-x]$, or I'm I using the wrong formula $(1)$?
