I want to calculate the divergent part of a Feynman diagram using the Feynman parameters:

$$\frac{1}{A_1 A_2 \ldots A_n} = \int_0^1 dx_1 ... dx_n \delta (\Sigma x_i -1) \frac{(n-1)!}{[x_1 A_1 + x_2 A_2 + ... x_n A_n]^n} $$

for $n=3$. I end up with some expression then like:

$$\int_0^1 dx dy dz\ \delta(x+y+z-1) \left( \frac{2}{\epsilon} - \log \Delta - \gamma + \log(4\pi) + ...\right)$$ $$ \int_{x=0}^{x=1} \int _{y=0}^{y=(1-x)} \frac{2}{\epsilon} dx dy$$

The divergent part evaluates to $\frac{1}{\epsilon}$ in this case.

However, when I have four Feynman parameters, the integral is $$\int_0^1 dx dy dz dw \ \delta(x+y+z+ w-1) \left( \frac{2}{\epsilon} - \log \Delta - \gamma + \log(4\pi) + ...\right)$$ $$ \int_{w=0}^1 \int_{x=0}^1 \int _{y=0}^{(1-x-w)} \frac{2}{\epsilon} dx dw dy$$

Unless I am mistaken, this comes out to zero. How should the integral with the delta function be handled?


1 Answer 1


You got the limits of integration wrong. $x$ can only go up to $1-w$. $$\int_{w=0}^1 \int_{x=0}^{1-w} \int _{y=0}^{(1-w-x)} dydx dw =\int_{w=0}^1 \int_{x=0}^{1-w} (1-w-x) dx dw =\int_{w=0}^1 \frac{1}{2}(1-w)^2 dw = \frac{1}{6}$$


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