I am looking for an example of two distribution such that their product is not a distribution. I'm using $f=\frac{1}{\sqrt{|x|}}$ which gives the distribution $\int_{-\infty}^{+\infty}\frac{\varphi(t)}{\sqrt{|t|}}dt$, then $f^2$ gives the "distribution" $\int_{-\infty}^{+\infty}\frac{\varphi(t)}{|t|}dt$.
I think the fact that $f$ is regular integrable gives the convergence of $f=\frac{1}{\sqrt{|x|}}$, but I am not completely sure how to show that $\int_{-\infty}^{+\infty}\frac{\varphi(t)}{|t|}dt$ doesn't converge. Can I simple set $\varphi$ to be the identity?, in which case it follows immediatly from the non ocnvergence of $\frac{1}{x}$
Take $\varphi\in C_{0}^{\infty}$ to be $1$ around a neighbourhood of $0$ then $\displaystyle\int_{-\infty}^{\infty}\dfrac{\varphi(t)}{|t|}dt$ does not converge. It is a standard construction such $\varphi$ must exist.