Problem
Consider the function $u(\vec{x})=\ln|\vec{x}|$ as a distribution on $\mathbb{R}^3$ and $\mathbb{R}^2$. We want to determine $\Delta u$ in the distribution sense. First calculate $\Delta u$ as a regular function outside the singularity at the origin, using polar coordinates. Then write down the integral that defines the effect of $\Delta u$ on a test function. On this integral one can use Green's second identity. Choose a bounded set whose complement contains a small sphere around the origin. Let the radius of the small sphere approach 0. Is the distribution $\Delta u$ given by the pointwise values or not?
What I've done so far
Following the problem description pretty much verbatim to begin with, I could reach the conclusion that, in the function sense, $\Delta u = \frac{1}{r^2}$ on $\mathbb{R}^3$ and $\Delta u=0$ on $\mathbb{R}^2$. Now, following the problem description I should calculate $(\Delta u)[\phi]$ on both $\mathbb{R}^3$ and $\mathbb{R}^2$ using the given set, i.e. I have to calculate $(\Delta u)[\phi] = u[\Delta\phi] = \lim_{r\to0}\int_r^R\int_0^{2\pi}r\ln r\Delta\phi\mathrm{d}\theta\mathrm{d}r$ and $(\Delta u)[\phi] = u[\Delta\phi] = \lim_{r\to0}\int_r^R\int_0^{2\pi}\int_0^\pi r^2\sin\theta\ln r\Delta\phi\mathrm{d}\theta\mathrm{d}\varphi\mathrm{d}r.$
Now, this is where I get stuck. I think that, in the $\mathbb{R}^2$ case, the integral actually converges to $0$: $\lim_{r\to0}\int_r^R\int_0^{2\pi}r\ln r\Delta\phi\mathrm{d}\theta\mathrm{d}r = \lim_{r\to0}\int_r^Rr\ln r\int_0^{2\pi}\Delta\phi\mathrm{d}\theta\mathrm{d}r =\\= \lim_{r\to0}\int_r^Rr\ln r\left[\int\Delta\phi\mathrm{d}\theta\right]_0^{2\pi}\mathrm{d}r = \lim_{r\to0}\int_r^Rr\ln r \cdot 0\mathrm{d}r = 0.$ In $\mathbb{R}^3$, I am completely stumped. It seems like I could make it converge to $0$ using the same trick I used in $\mathbb{R}^2$, but that seems wrong (as it doesn't make sense that the distribution behave that differently in $\mathbb{R}^2$ and $\mathbb{R}^3)$, which makes me question doing to in the $\mathbb{R}^2$ case. I would expect that the integral computes to $\frac{1}{r^2}$, but that may be wrong.
Question
How do I compute the two integrals (especially the $\mathbb{R}^3$ one) properly?