Disclaimer: My mathematical background is mathematical physics so this question most likely lacks rigor.
I am trying to understand the nature of the singularity of the fundamental solution (aka Green function) of the Laplace equation in an arbitrary number of dimensions. However, to pinpoint my problem, the one-dimensional case is sufficient so I will elaborate on that.
Assume we want to find the solution to the following equation $$\nabla^2 G(x) = \delta(x).$$ In one dimension, the equation reads $$G''(x) = \delta(x)$$ and direct integration yields $$G(x) = (\theta(x) + \alpha) x + \beta,$$ with $\alpha$ and $\beta$ arbitrary constants and $\theta(x)$ the Heaviside theta function. Since the delta function is even, $\delta(-x) = \delta(x)$, we can set $\alpha = -\frac{1}{2}$ to obtain an even Green function $$G(x) = \frac{1}{2} |x| + \beta.$$ Using denoting $r = |x|$ as the "radial" variable, we have $$G(r) = \frac{1}{2} r + \beta$$ for the radially symmetric (aka even) Green function in one dimension.
Now, let us try to solve the same problem in the radial variable from the start. Using $\delta(x) = \frac{1}{2} \delta(r)$, we write $$G''(r) = \frac{1}{2} \delta(r).$$ Integrating this equation from $0$ to $\epsilon$, we have $$G'(\epsilon) - G'(0) = \frac{1}{2},$$ which shows that $G'(r)$ is discontinuous as $r \to 0$. However, we have already calculated $G(r)$ to be a linear function in $r$, and therefore $G'(r)$ is certainly continuous! What went wrong?
My thoughts: The Laplace equation looks the same (except for the factor $\frac{1}{2}$ in front of the delta function) in the $x$ and $r$ variables. However, the $r$ variable is not defined for $r<0$ so this is probably where the problem originates. I, however, am unable to pinpoint the illegal step I have made and how to correct it.
EDIT : ignore this post please
There is a mistake in the integration : if you integrate a Dirac function, remember the following : $\int_{\Omega} \delta (z-z_0) f(z) dz$ = $f(z_0)$, and when $f(z)=1$, this yields $\int_{\Omega} \delta (z-z_0) 1 dz$ =1. (Supposing you are integrating over a domain $\Omega$ that includes the singularity $z_0$, otherwise the integral is zero.)