tomkeus said:
When I asked him why massive carrier implies finite range, and massless infinite, I didnt get satisfactory (any) explanation.
This is a very important fact. You are right to search for a rigorous answer.
The "bad reason" why this is important is technical. You should be able to calculate the associated potential using Cauchy's theorem.
The "good reason" is conceptual. Gauge invariance imposes the photon to be massless, but this is quite annoying to fix gauge invariance. A useful trick is to set a finite mass, that physically you claim is to small to be observable. Do your calculations, and set m=0 at the very end. It is a quite non-trivial property that this procedure will still respect gauge invariance at the end of the day. But many people use it.
Do you have Zee's "QFT in a nutshell" ? He discusses about all this at the conceptual level. This is a very good introductory book.
So the thing is something like : calculation of the massive scalar potential with a point source at the origin \rho=g\delta(\vec{x})
Klein-Gordon : \partial_{\mu}\partial^{\mu}U+m^{2}U=\rho
Time independence : (-\Delta^{2}+m^2)U=g\delta(\vec{x})
Fourier transform : U(\vec{x})=\frac{g}{2\pi^3}\int d^3k\frac{e^{i\vec{k}\vec{x}}}{\vec{k}^2+m^2}
Go to polar coordinate, trivial angular integration, radial integration done be closing the contour (say) in the positive imaginary part you pick up the pole at k=im. This is the important technical part. Details upon request (try it before !).
Finally
U(\vec{x})=\frac{g}{4\pi}\frac{e^{-mr}}{r}
This is what we mean by finite range, because exponential is decreasing fast. Setting a vanishing mass, you recover the Coulomb potential
