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A Feynman's denominator formula

  1. Oct 22, 2016 #1
    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?
     
  2. jcsd
  3. Oct 22, 2016 #2

    vanhees71

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  4. Oct 22, 2016 #3
    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?
     
  5. Oct 22, 2016 #4

    vanhees71

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    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.
     
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