# Why does a singularity imply the need for a distribution?

I am following Section 11 of Prof. Etingof's MIT OpenCourseWare notes on "Geometry And Quantum Field Theory" in which he says:

...for $$d = 1$$, the Green's function $$G(x)$$ is continuous at $$x = 0$$, for $$d > 1$$ it is singular at $$x = 0$$. Namely, $$G(x)$$ behaves like $$C|x|^{2-d}$$ as $$x \rightarrow 0$$ for $$d > 2$$, and as $$C\ln|x|$$ as $$d = 2$$. Thus for $$d > 1$$, unlike the case $$d = 1$$, the path integral $$\int \phi(x_1)\cdots \phi(x_n) e^{-S(\phi)}D\phi$$ only makes sense if $$x_i \neq x_j$$. In other words, this path integral should be regarded not as a function but rather as a distribution.

Here $$d$$ is the dimension.

Why does the singularity of the Green's function imply that the path integral must be interpreted as a distribution?

Please let me know if this question is better suited for math SE.

1. OP's linked quote mentions that the free massless scalar propagators with Euclidean spacetime signature are singular functions for $$d\geq 2$$.
2. Moreover, if we Wick-rotate to Minkowskian spacetime signature, we should include the Feynman $$i\epsilon$$ prescription. As a result, the propagators become distributions for $$d\geq 2$$.
Since the propagators are building blocks for the path integral, the path integral also become a distribution for $$d\geq 2$$. .