How to simplify this expression $\frac{\partial}{\partial t}\left(\text{sgn}(t)\,\delta(t^2-x^2)\right)$ at $t=0$?

Question

Define

$$F(t,x)=\frac{\partial}{\partial t}\bigg(\text{sgn}(t)\,\delta(t^2-x^2)\bigg),$$

where $\text{sgn}(x)$ is the sign function, and $\delta(x)$ is the Dirac $\delta$-function.

Does $F(0,x)$ make sense, and what is it?

Answer 1

Heuristic approach: Before differentiation wrt. $t$, one may rewrite

$$ {\rm sgn}(t) \delta(t^2-x^2)~=~ \frac{1}{2x} \left\{ \delta(t-x) - \delta(t+x)\right\}. \tag{1} $$

Differentiation wrt. $t$ yields

$$F(t,x)~:=~\frac{\partial}{\partial t}\left\{{\rm sgn}(t) \delta(t^2-x^2) \right\} ~=~ \frac{1}{2x} \left\{ \delta^{\prime}(t-x) - \delta^{\prime}(t+x)\right\} .\tag{2}$$

Then $$F(t=0,x)~=~-\frac{1}{x} \delta^{\prime}(x).\tag{3} $$

This can be made mathematically precise by using test functions.