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Your notion and the equality in the book are consistent, since
$$
\delta (a^2 t^2 - |x|^2 ) = \frac{1}{2|at|} (\delta (at-|x|) + \delta (at+|x|) )
$$
(I suppose an assumption is being made that $at\geq 0$$at>0$).
Your notion and the equality in the book are consistent, since
$$
\delta (a^2 t^2 - |x|^2 ) = \frac{1}{2|at|} (\delta (at-|x|) + \delta (at+|x|) )
$$
(I suppose an assumption is being made that $at\geq 0$).
Your notion and the equality in the book are consistent, since
$$
\delta (a^2 t^2 - |x|^2 ) = \frac{1}{2|at|} (\delta (at-|x|) + \delta (at+|x|) )
$$
(I suppose an assumption is being made that $at>0$).
Your notion and the equality in the book are consistent, since
$$
\delta (a^2 t^2 - |x|^2 ) = \frac{1}{2|at|} (\delta (at-|x|) + \delta (at+|x|) )
$$
(I suppose an assumption is being made that $at\geq 0$).
