When finding the fundamental solution to Laplace's eqn, i.e. $G$ such that $\Delta G = \delta$ a constant has so be solved for. How I have seen this done is by finding $c$ so that
$\int_{D(0,\epsilon)} c \log r dA= \int_{D(0,\epsilon)} \delta(x) dA= 1$
where $D(0,\epsilon)$ is the disk of radius $\epsilon$ centered at the origin. It has been a while since I have dealt with delta functions (pun intended) and I'm a little uncomfortable with the last equality,
$\int_{D(0,\epsilon)} \delta(x) dA= 1$.
In particular I have a tendency to go back and forth between thinking of the delta function as a distribution or as a function. I'm fine with starting with the assumption
$\int_{\mathcal{R}^2} \delta dA=1$
and thinking of it as a distribution. But my problem is that to me it only makes sense that
$\int_{D(0,\epsilon)} \delta(x) dA= 1$
if I'm treating it as a function. In this case, then I would just argue that it is 0 outside of the disk so it's equal to the integral over all space. But treating it as a function isn't good since from measure theory the integral should be 0 then since it is only nonzero on a set of measure 0.
So my question is, can you give an explanation for $\int_{D(0,\epsilon)} \delta(x)dA = 1$ that is grounded in distributions? Should I be concerned about any arguments for this that involve treating delta as a function as opposed to a distribution?
This may be justified in the sense that in the space of distributions there are measures. For example, if $\Omega \subset \mathbb{R}^n$ is an open set, the Dirac measure is defined by $\delta_{x_0}(\Omega)=1$ if $x_0 \in \Omega$ and $\delta_{x_0}(\Omega)=0$ if $x_0 \notin \Omega$. Then
$\displaystyle \delta_{x_0} (\varphi)=\int_{\Omega} \varphi(x) d \delta_{x_0}(x)=\varphi(x_0)$ $\forall \varphi \in \mathcal{D}(\Omega)$
and as a function the Dirac measure is a distribution.