Talk:Scaling Property of Dirac Delta Function

To handle the signum properly, you might want to split it into two cases: one for $a > 0$ and one for $a < 0$, and then merge the result. --prime mover (talk) 15:38, 13 December 2015 (UTC)

Okay, I will try --Z423x5c6 (talk) 10:33, 16 December 2015 (UTC)

In the proof, the claim $\int_a^b f(x) dx = \int_a^b g(x) dx \implies f = g$ is not used. Instead, the definition of Dirac delta function is used. The function satisfying the two properties stated in the definition is equal to the Dirac delta function. Correct me if I am wrong.

oh okay