# 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 $(1)$?

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Actually that's a much better title than we usually get. More informative and specific (to a point) is generally better when it comes to titles. – David Zaslavsky Mar 10 at 23:29
Eq. (1) only works if $|\partial f/\partial y|$ is the same for both $y^\pm$, otherwise you can't pull it out of the brackets. Don't know if this helps your problem. – Michael Brown Mar 11 at 0:50
For readers unfamiliar with the equation in the question title, learn more at en.wikipedia.org/wiki/… :-D – Steve B Mar 11 at 0:56
Sorry i should add that it is the same for both. – The Noob Mar 11 at 10:30
@The Noob: If you would like the community to help solve your apparent paradox, you would have to give the explicit form of $f(x,y)$. Right now, it is hard to conclude anything other than what have already been said by you in the question formulation (v4). – Qmechanic Mar 13 at 17:28