# Regulating a divergent integral in QED

When we try to regulate a divergent integral, we introduce another parameter, say $$\lambda$$ and then compute the integral. We finally take a limit (either $$\lambda \rightarrow 0, \infty$$) to restore the original situation. I have seen two outcomes of it.

1. On taking this limit, the divergences reappear. This happens in QED a lot. These divergences have then to be absorbed in the definition of physical (measurable) masses and charges in terms of the bare (unphysical) masses and charges.

2. On taking these limits, divergences do not appear and everything is fine. An example would be the Fourier transform of the Coulomb potential worked out here.

I want to know what goes on mathematically which makes these two situations different. Why do infinities reappear in the first case and not in the second? I am aware of Cauchy principal value integration if one wants to explain using this concept.

• The example you give of a Fourier transform of a Coloumb potential is not in QED, is it? As a rule-of-thumb, regularization of integrals in quantum field theory is part of a renormalization process and will involve changing bare quantities into fully-dressed ones.
– user245141
Oct 22 '19 at 15:55
• So you are asking why infinities appear in QED and not in some other calculation? Oct 22 '19 at 19:47
• @puppetsock I want to ask what is the mathematical reason that sometimes the introduction of the parameter $\lambda$ can cure the divergences and sometimes it can't. Oct 22 '19 at 23:05
• In principle, if you introduce a regulator and the take the limit wehere it goes away, you should recover the initial expression. If that was really divergent, the end result will be as well. However, you may have singular terms of the type $0\times\text{infinity}$ (this happens e.g with anomalies). Here, a proper regularisation may give a finite result. Aug 25 at 10:25