Integration of Dirac Delta Function from -3 to 0 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!


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


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