Talk:Schwinger parametrization

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

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)

Join with Feynman parametrization
Maybe it is worth to join this article into the article about Feynman parametrization, as all these parametrizations mentioned there are derivable from each other - and the Schwinger parametrization is even named and used in the derivation. This joined article then of course should carry a more general name, e.g. loop integral parametrizations. Stefan Groote (talk) 05:43, 21 September 2022 (UTC)