I know that the "hand-wavy" definition of the $\delta (x)$ function is $$ \delta(x) = \begin{cases} \infty &\quad\ x=0 \\ 0 &\quad\text{otherwise} \end{cases} $$
and the more rigorous definition is that it's the limit of a sequence of functions $f_n$ for which $f_n(x) \rightarrow 0$ for all $x \neq 0$, and $f_n \rightarrow \infty$ for $x=0$, and (edited to add) $\int f_n = 1$ for all $n$. From this perspective, I see why the integral should be one, because the integral of all of the $f_n$ is equal to $1$.
Now, suppose I want to construct a function $f(x,y)$ in the plane for which
$$ \nabla ^2f(x,y) = \begin{cases} a &\quad\ (x,y) \in D \\ 0 &\quad\text{otherwise} \end{cases} $$ where $D$ is some simply connected region.
I can definitely solve $\nabla ^2f(x,y) = \delta(\|(x,y) - (x_0,y_0)\|)$ for any point $(x_0,y_0)$. This is just done by using the fundamental solution $$f(x,y) = \frac{-1}{2\pi} \ln\left( \|(x,y)-(x_0,y_0)\|\right)$$
My question is whether I can do the following:
Because I want the Laplacian of $f$ to be as described above, can I write
$$ f(x,y) = a \int_D \frac{-1}{2\pi} \ln\left( \|(x,y)-(x_0,y_0)\|\right) \,dA \quad ?$$
where $dA$ refers to integration with respect to $(x_0,y_0)$ over the area of $D$.
My confusion is coming from the fact that: The Laplacian of $f$ will be the Laplacian of a sum of (infinitely) many $\delta$ functions, so intuition tells me it should be infinite; on the other hand, integrating a $\delta$ function gives $1$, so the factor of $a$ in front of the integral should give the desired result, no?
You should solve Laplace's equation using a Fourier transform. You can write the actual Delta function as
$$\delta(x)=\int_{-\infty}^{\infty}\!e^{-ikx}\,dk$$
Further, when you Fourier transform the Laplace equation you get
$$\mathcal{F}[\nabla^2f(x,y)]=-k^2\hat{f}(x,y)$$
If you solve the problem in $k$-space for $\hat{f}(x,y)$ and then perform a reverse Fourier transform, you will have your solution.
To be clear: Laplace's equation is NOT solved by integrating a sum of delta functions.