A catch with Dirac Delta Function

Question

We know that $$ \int_{\mathbb{R}} f(t)\delta(t) \mathrm{d}t = f(0) $$ if $f$ is continuous. What will it be if $f$ is not continuous? For instance, what is the value of $$ \int_{\mathbb{R}} e^t\mathrm{u}(t)\delta(t) \mathrm{d}t $$

Answer 1

The smoothness of $f$ doesn't play a role, it just makes a non-distribution proof available; see this. If you want to make things rigorous, the best viewpoint is as $\delta_a(x):=\delta(x-a)$ as the distribution which maps $\int_\mathbb{R}f(x)\delta_a(x)\,dx$ to $f(a)$.

Thus, $$\int_\mathbb{R} e^t u(t)\delta(t)\,dt=e^t u(t)\Big|_{t=0}=e^{0}\cdot 1=1$$ by the sifting property of the Dirac delta. The last step depends on how you define $u(t)$ at $t=0$.