quantum field theory - Propagator of a scalar in position space

In his lecture on Supersymmetry and Grand Unification, Leonard Susskind "derives" the propagator for a scalar field from dimensional analysis. He says for a particle going from $x$ to $y$ (where x and y are four-vectors), the (massless) propagator is:

$$ \qquad\quad \left\langle0\right|\phi(y)\, \phi(x) \left|0\right\rangle \propto \frac{1}{|x-y|^2} \qquad (1) $$

I'm a bit confused about this. I know in momentum space, the propagator is

$$ G(p) = \frac{1}{p^2-m^2+i\epsilon}\,. $$

His expression is probably just in position space, but the expressions I've found for the position space propagator are much more complicated (e.g. Wikipedia). Is equation (1) wrong or am I missing something? (I know the first one is massless, and the second one massive. But isn't he missing at least an integral?)

Background: I am trying to formulate a short explanation of renormalization, and I like his overall argument and would like to follow it (its for the theory part of an experimental thesis, so it can be general and sketched, but should be correct). Susskind looked for a term with the correct units ($L^{-2}$) that you could construct from the available variables, and that makes sense physically (the amplitude for detecting a particle at $y$ is higher the closer it is to where it was emitted, $x$). The result is simple, but seems not correct enough to wright it down (e.g. into a thesis).

Is there a way to replace (1) with something more correct, but still keep it simple and intuitive?

Answer

The form of the propagator is correct. The expressions from your Wikipedia link are complicated because they show the propagator for the massive theory, where Susskind's argument fails because the propagator can involve any function of the dimension zero combination $m^2 |x-y|^2$. The "simple" massless result is recovered in the $m\to0$ limit; for example (Feynman propagator, spacelike separation; from Wikipedia) $$ \lim_{m\to0} - \frac{i m}{4\pi^2 \sqrt{-s}}K_1(m\sqrt{-s})=-\frac{i}{4\pi^2 \sqrt{-s}} \frac{1}{\sqrt{-s}}=\frac{-i}{4 \pi^2|x-y|^2} $$