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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 Z Mar 10 '13 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 '13 at 0:50
For readers unfamiliar with the equation in the question title, learn more at this Wikipedia page :-D –  Steve B Mar 11 '13 at 0:56
Sorry i should add that it is the same for both. –  The Noob Mar 11 '13 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 '13 at 17:28
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1 Answer 1

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: $$\lim_{\epsilon \to 0^+} \int_0^1 \mathrm{d}x\int_0^{1-x}\mathrm{d}y \frac{1}{f(x,y)+\mathrm{i}\epsilon} = \int_T \lim_{\epsilon \to 0^+}\frac{1}{f(x,y)+\mathrm{i}\epsilon} \mathrm{d}x\mathrm{d}y = \int_T \frac{1}{f(x,y)} \mathrm{d}x\mathrm{d}y $$

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Obviously, in my answer, I supposed $f$ to be real valued. –  V. Moretti Dec 11 '13 at 20:47
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