Delta function with both limits of the integral set to zero

Question

What is the expression

$$\lim\limits_{t\rightarrow 0}\int_0^t f(s)\,\delta(t-s)\,ds,$$

where $f(s)$ is some "nice" function?

Answer 1

Since $\delta(s)$ is an even function, we can replace $\delta(t-s)$ with $\delta(s-t)$, where we can see that the function has been moved to the right with the amount of $t$. Then we have: $$ \lim_{t \to 0}\int_0^t f(s)\delta(s-t) ds$$ The trick is to know that $\int_0^t f(s)\delta(s-t)ds$ gives back the value of $f(s)$ at $t = s$. (Sampling Property) You can think about the Dirac-Delta as if it had the value $1$ at $0$, and the value $0$ anywhere else. Than it is understandable that $f(s)\delta(s-t)$ has the value $f(t)$ at $t$ and $0$ anywhere else. If you integrate it you get the only available non-zero value. (Note that this argument is obviously not strict, but intuitive.) So the full solution: $$\lim_{t \to 0}\int_0^t f(s)\delta(s-t)ds = \lim_{t \to 0} f(t) = f(0)$$ But in order to use this the integration should take place between $0$ and $t+\varepsilon$ for some $\varepsilon > 0$ as already mentioned in the comments. This way the integral is simply undefined.