Using $\frac{1}{A+i\epsilon} = PV\frac{1}{A}-i\pi\delta(A)$ in Feynman Integrals 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 $(1)$?
 A: If you are sure that $f$ is continuous and does not vanish in the integration domain, it is by no means necessary making use of regularization theory of distributions. Consider the initial integral:
$$F:= \int_0^1 \mathrm{d}x\int_0^{1-x}\mathrm{d}y \frac{1}{f(x,y)+\mathrm{i}\epsilon}.$$ 
It can be re-written as:
$$F:= \int_{T}  \frac{1}{f(x,y)+\mathrm{i}\epsilon} \mathrm{d}x\mathrm{d}y,$$
where $T$ is the closed triangle:
$$ T := \{(x,y) \in [0,1]\times [0,1] \:|\: 0 \leq y \leq 1-x\}\:.$$
If $f(x,y)$ is continuous on $T$ and does not vanish therein, the function 
$$[0,1]\times T \ni (\epsilon, x,y) \mapsto \left|\frac{1}{f(x,y)+\mathrm{i}\epsilon}\right|$$
is continuous and thus bounded. Let us call $M\geq 0$ its maximum. We can conclude that 
$$\left|\frac{1}{f(x,y)+\mathrm{i}\epsilon}\right| \leq M\quad \mbox{for every $\epsilon \in [0,1]$ and $(x,y)\in T$.} $$
As $T$ has finite measure, the constant function $T \ni (x,y) \mapsto M$ has finite integral. We can thus  apply Lebesgue's dominated convergence theorem, that permits to swap the symbol of limit with that of integral and getting this way:
$$\lim_{\epsilon \to 0^+} \int_0^1 \mathrm{d}x\int_0^{1-x}\mathrm{d}y \frac{1}{f(x,y)+\mathrm{i}\epsilon} = \int_T \lim_{\epsilon \to 0^+}\frac{1}{f(x,y)+\mathrm{i}\epsilon} \mathrm{d}x\mathrm{d}y = \int_T \frac{1}{f(x,y)} \mathrm{d}x\mathrm{d}y
$$
