Talk:Schwinger parametrization

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[edit] easy?

what exactly is this making easy, or more likely easier? --MarSch 12:23, 18 December 2006 (UTC)

[edit] More

Re(Easy): If A(p) is quadratic in momentum p, then the resulting momentum integral is just a gaussian.

This article should also talk about the use of the Schwinger trick to get representations of propagators/greens functions in a classical background via the heat kernel representation.

Also perhaps about how the Schwinger parametrization leads to propertime regularisation... styler 03:49, 13 March 2007 (UTC)