Show that $e^{1/x^2}$ is not a distribution in $\mathbb{R}\backslash\{0\}$

Question

In homework I have to do prove that, but $e^{1/x^2}\in L^1_\text{loc}(\mathbb{R}\backslash\{0\})$ and therefore a distribution, isn't it? I am confused.

Answer 1

According to the most common convention, a locally integrable function induces a distribution. Now pick any compact subset $K$ of $\mathbb{R} \setminus \{0\}$. The function $x \mapsto \exp(1/x^2)$ is of class $C^\infty(K)$, and therefore integrable on $K$. It defines a distribution in a canonical way, and we say that it is a distribution.