# Using $\frac{1}{A+i\epsilon} = PV\frac{1}{A}-i\pi\delta(A)$ in Feynman Integrals

Is the following operations OK (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

$$\frac{1}{\bigl\vert\partial f/\partial y\bigr\vert_{y=y^{\pm}}}\times[\delta(y-y^-)+\delta(y-y^+)]\,, (1)$$

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 $(1)$?

(Feel free to change the title as I couldn't come up with anything better) Thanks for any comments.

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Those equations would look a lot better and be more readable as displayed equations (which you get by using double dollar signs instead of single dollar signs). –  joriki Mar 11 '13 at 3:58
Is there a particular point in not telling us what $f(x,y)$ is? I don't quite see how anything useful could be said about this without that. –  joriki Mar 11 '13 at 4:02
Thanks for the tip, didn't know that. –  The Noob Mar 11 '13 at 10:26
You're welcome. Rereading my comments they sound a bit unfriendly; maybe I was in a bad mood; sorry about that :-) –  joriki Mar 11 '13 at 10:27
Happens everyone :) –  The Noob Mar 11 '13 at 10:29