Q:

The distribution $\Delta u$ (where $u = \ln|\vec{x}|$)

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?

A:

By this time probably you have made this already. On the other hand, multidimensional analytic integration is not my middle name. Anyway, here it comes.
I believe your approach is wrong, because I agree with joriki, that $\int_0^{2 \pi} \Delta u \;d\theta$ is generally not zero.
Let $x \in \mathbb{R}^2$, $u(x) := \log(|x|)$. We want to compute $\Delta u$ in a distributional sense. First we compute $\Delta u(x)$ for $x \neq 0$ to and we obtain $\Delta u(x)=0$ (direct calculus, I omit this). The only "suspicious" point is then $x = 0$, the origin.
Let then $\phi$ be a smooth function with a compact support. We want to compute
\begin{equation}
(\Delta u)[\phi] := u[\Delta \phi] = \int_{\mathbb{R}^2} \log|x| \; \Delta \phi(x) dx = \lim_{r \rightarrow 0} \int_{B(0;R)-B(0;r)}\log|x| \; \Delta \phi(x) dx\;.
\end{equation}
where $R$ is big enough so that $\mathrm{supp}(\phi) \subset B(0;R)$. Then we use the Green's identity, but we do not yet switch to the polar coordinates
\begin{eqnarray}
\int_{B(0;R)-B(0;r)}\log|x| \; \Delta \phi(x) dx
&=&
\int_{B(0;R)-B(0;r)}\Delta \log|x| \; \phi(x) dx
\\&&+
\int_{S(0;R)} 
\left(  \log|x| \; \frac{\partial \phi(x)}{\partial n}
 - 
  \frac{\partial \log|x|}{\partial n} \; \phi(x)\right)dS(x)
\\
& &-
\int_{S(0;r)} \left(
  \log|x| \; \frac{\partial \phi(x)}{\partial n}
 - 
  \frac{\partial \log|x|}{\partial n} \; \phi(x)\right)dS(x)\;.
\end{eqnarray}
Then the first integral is zero, as we know that $\Delta\log|x|=0$ for $x\neq 0$. Also the second integral is zero, as we choosen $R$ big enough for $\phi$ and its derivatives to vanish. Thus there is only third integral left. Then we switch to the polar coordinates, when calculating the spherical integrals 
\begin{equation}
-
\int_{S(0;r)} 
  \log|x| \; \frac{\partial \phi(x)}{\partial n} dS(x)
=
-\int_0^{2\pi} 
  r \log(r) \; \frac{\partial \phi(r,\theta)}{\partial r} d\theta
=
-r \log(r)
   \int_0^{2\pi} \frac{\partial \phi(r,\theta)}{\partial r} d\theta
\rightarrow 0,
\end{equation}
as $r \rightarrow 0$, since $\phi$ and its derivatives are bounded. Still we have
\begin{equation}
\int_{S(0;r)} \frac{\partial \log|x|}{\partial n} \; \phi(x)dS(x)
=
\int_{0}^{2\pi} r \frac{\partial \log(r)}{\partial r} \; \phi(r,\theta)d\theta
=
\int_{0}^{2\pi} r \frac{1}{r} \; \phi(r,\theta)d\theta.
\end{equation}
Then $r$ and $1/r$ cancel and by Lebesgue's dominated convergence theorem
\begin{equation}
\int_{0}^{2\pi} \phi(r,\theta)d\theta
\rightarrow
\int_{0}^{2\pi} \phi(0)d\theta
=
2 \pi \phi(0).
\end{equation}
In conclusion, we obtain $\Delta u [\phi] = 2 \pi \delta [\phi]$. This is, I believe, some standard fact, as $\frac{1}{2 \pi}\log |x|$ is a fundamental solution of Laplacean in $\mathbb{R}^2$.
In $\mathbb{R}^3$ the case is different. I would only denote main differences. First, as you have noted, $\Delta u = |x|^{-2}$, so the first integral does not vanish. Then in the boundary integrals we would have $r^2 \sin\theta$ instead of $r$. So now the third integral vanishes, if my calculus is correct. The second integral vanishes as in $\mathbb{R}^2$. We therefore obtain
\begin{equation}
\int_{B(0;R)-B(0;r)}\log|x| \; \Delta \phi(x) dx
=
\int_{B(0;R)-B(0;r)} |x|^{-2} \; \phi(x) dx.
\end{equation}
Then we pass $r$ to the limit and we conclude that $\Delta u = |x|^{-2}$.

