Is Brute Force the Best Approach to Proving Feynman's Denominator Formula?

[spaghetti3451]

I would like to prove Feynman's denominator formula:

$\frac{1}{A_{1}\dots A_{n}} = (n-1)!\int_{0}^{1}dx_{1}\dots dx_{n}\delta(x_{1}+\dots+x_{n}-1)(x_{1}A_{1}+\dots+x_{n}A_{n})^{-n}$

I was wondering if you would recommend brute force approach to solving this problem. I proved the formula for $n=1,2,3$, and then attempted for the general case using a brute force but the algebra looks messy.

Would you recommend trying an alternative method, perhaps proof by induction?

 

[vanhees71]

https://en.wikipedia.org/wiki/Feynman_parametrization

 

[spaghetti3451]

I have also been checking out Schwinger paramaterization and I found the following integral:

$\displaystyle{\frac{1}{A}=-i\int^\infty_0 du \, e^{iuA}}$

I would like to prove this formula but am having a hard time proving it.

I think you need to solve the RHS by contour integration, and since there is a factor of $iuA$ in the exponent, you close the contour in the upper half of the complex $u$-plane. But I don't see any branch points, so I am led to guess that the RHS integral is 0.

For the LHS to be equal to RHS, I think that the residue is ${\frac{1}{2\pi A}}$, but I cannot explain this. Am I missing a branch point in the upper half complex $u$-plane?

 

[vanhees71]

You don't need contour integration. For the integral to exist you must only have $\mathrm{Im} A>0$. Then you can simply use the antiderivative of the exponential function (which is the exponential function):
$$F(u)=\int \mathrm{d} u \exp(\mathrm{i} u A)=-\frac{i}{A} \exp(\mathrm{i} u A),$$
and then the Feynman integral is
$$\int_0^{\infty} \mathrm{d} u \exp(\mathrm{i} u A)=\lim_{u \rightarrow \infty} F(u)-F(0)=\mathrm{i}{A},$$
which proves the formula.

It's exactly what you need for the denominators in perturbation theory, because you have time-ordered Green's functions which always have a positive imaginary part in the denominator, which defines the time-ordered propagator in the usual sense, namely
$$\Delta(p)=\frac{1}{p^2-m^2+\mathrm{i} 0^+}.$$
The $\mathrm{i} 0^+$ is crucial here to get the right propagator, namely the time-ordered one.