Integration of Dirac Delta Function from -3 to 0

Question

I'm learning about Dirac Delta functions, and I have a question about when one of the bounds is 0.

An example I'm working on is $\int_{-3}^{0}\,dx\,\delta(x-1)$.

How would I evaluate this at 0?

Would it just be undefined?

Thanks!

Answer 1

In THIS ANSWER and THIS ONE, I provided primers on the Dirac Delta.

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The notation $\int_a^b f(x)\delta(x-c)\,dx$ is interpreted to mean the functional $\langle fp_{ab},\delta_c\rangle$.

Here, $p_{ab}$ is the "rectangular pulse" function, $p_{ab}(x)=u(x-a)-u(x-b)$, and $u$ is the unit step (or Heaviside Function) where

$$u(x)=\begin{cases}1&,x>0\\\\0&,x<0\end{cases}$$

Note that there are various conventions for the value $u(0)$.

Therefore, we have

$$\begin{align} \int_a^b f(x)\delta(x-c)\,dx&=\langle fp_{ab},\delta_c\rangle\\\\ &=\begin{cases}f(c)&,c\in(a,b)\\\\0&,c\notin [a,b]\end{cases} \end{align}$$

In the case at hand, we have $a=-3$, $b=0$, and $c=1$. Therefore, we have

$$\int_{-3}^0 f(x)\delta(x-1)=0$$