# Feynman parameter integral

1. May 29, 2010

### Bapelsin

I'm trying to show equation (6.39) on page 189 in Peskin & Schroeder.

1. The problem statement, all variables and given/known data

Show that $$\frac{1}{AB}=\int_0^1dxdy\delta(x+y-1)\frac{1}{(xA+yB)^2}$$.

2. Relevant equations

Defining property of Dirac delta function:

$$\int_{-\infty}^{\infty}dx f(x)\delta(x-x_0)=f(x_0)$$

Also:

$$\int_{x_0-\varepsilon}^{x_0+\varepsilon}dx f(x)\delta(x-x_0)=f(x_0)$$

3. The attempt at a solution

Since the defining properties of the delta function requires the limits of integration to be outside the points where the argument of the delta function is zero, I cannot simply plug my expression into any of these relations. If I instead choose the limits of integration $$1+\varepsilon$$ and $$0-\varepsilon$$ I can use the second
delta function identity from above to "force" x=1-y as I integrate over dx. So

$$\int_{0+\varepsilon}^{1+\varepsilon}dxdy\delta(x+y-1)\frac{1}{(xA+yB)^2} =\int_{1+\varepsilon}^{1+\varepsilon}dy\frac{1}{[(1-y)A+yB]^2}$$. After a few routine steps
the expression I get goes to $$\frac{1}{AB}$$ in the limit $$\varepsilon\rightarrow 0$$, which is the result I want for the integral without the epsilons.

Is this calculation relevant in obtaining the value of $$\int_0^1dx (...)$$ and how are the two integrals related? I'm getting very confused here.