Green's Function for 2D Poisson Equation In two dimensions, Poisson's equation has the fundamental solution,
$$G(\mathbf{r},\mathbf{r'}) = \frac{\log|\mathbf{r}-\mathbf{r'}|}{2\pi}. $$
I was trying to derive this using the Fourier transformed equation, and the process encountered an integral that was divergent.  I was able to extract the correct function eventually, but the math was sketchy at best.  I am hoping someone could look at my work and possibly justify it.  Here goes.
First off, make the assumption that $G$ only depends on the difference $\mathbf{v}=\mathbf{r}-\mathbf{r'}$.  Now, let's write $G$ as an inverse Fourier Transform and take the Laplacian,
$$\nabla^2G(\mathbf{v}) = \int\frac{d^2k}{(2\pi)^2}(-k^2)e^{i\mathbf{k} \cdot \mathbf{v}} \hat{G}(\mathbf{k}) = \delta(\mathbf{v}) $$
For this to be a delta function, we require that $\hat{G}(\mathbf{k}) = -1/k^2$.  Now taking the inverse Fourier Transform of $G$...
\begin{align*}
G(\mathbf{v}) &= -\int\frac{d^2k}{(2\pi)^2} \frac{e^{i\mathbf{k}\cdot\mathbf{v}}}{k^2} = -\int\limits_{0}^{\infty} \int\limits_{0}^{2\pi} \frac{dkd\theta}{(2\pi)^2} \frac{e^{i|\mathbf{k}||\mathbf{v}|\cos\theta}}{k}\\
&= - \int\limits_0^{\infty}\frac{dk}{2\pi}\frac{J_0(kv)}{k}
\end{align*}
Here $J_0$ is a Bessel function of the first kind.  This integral is divergent as far as I can tell, but let's continue onward and take a derivative with respect to $|\mathbf{v}|$.
\begin{align*}
\frac{dG}{dv} &= \int\limits_0^{\infty}\frac{dk}{2\pi} J_1(kv)\\
&= \frac{1}{2\pi v}
\end{align*}
Then integrating this and setting the constant to zero we get the desired result...
$$ G(\mathbf{v}) = \frac{\log v}{2\pi} $$
Clearly this was a lot of heuristics, but I am hoping someone could justify some of this with distributions etc...  Could someone tell me what on earth I have done and why it worked?

It suffices to show
\begin{align}
\int_{\mathbb{R}^2} G(\textbf{r}, \textbf{r}') \nabla^2f(\textbf{r}')\ d^2\textbf{r}' = f(\textbf{r}). 
\end{align}
Observe we have
\begin{align}
\int \log|\textbf{r}-\textbf{r}'|\nabla^2f(\textbf{r}')\ d^2\textbf{r}'&=\
\int_{|\textbf{r}-\textbf{r}'|\leq\ \epsilon} \log|\textbf{r}-\textbf{r}'|\nabla^2f(\textbf{r}')\ d^2\textbf{r}'+\int_{|\textbf{r}-\textbf{r}'|>\epsilon} \log|\textbf{r}-\textbf{r}'|\nabla^2f(\textbf{r}')\ d^2\textbf{r}'\\
&=:\ I_1+I_2. 
\end{align}
Let us first focus on $I_1$. Note that we have the following estimate
\begin{align}
\left| I_1\right| \leq&\ \|f\|_{C^2(B(0, 1))} \int_{|\textbf{r}-\textbf{r}'|\leq\ \epsilon} \left|\log|\textbf{r}-\textbf{r}'|\right|\ d^2\textbf{r}'\\
\leq &\ -C\int^\epsilon_0 r\log r\ dr  \leq C\epsilon
\end{align}
which goes to $0$ as $\epsilon \rightarrow 0$. 
For the $I_2$ term, observe we have that
\begin{align}
I_2 &=\  \int_{|\textbf{r}-\textbf{r}'|=\epsilon} \log|\textbf{r}-\textbf{r}'|\frac{\partial f}{\partial n}(\textbf{r}')\ dS(\textbf{r}')-\int_{|\textbf{r}-\textbf{r}'|>\epsilon} \nabla\log|\textbf{r}-\textbf{r}'|\cdot \nabla f(\textbf{r}')\ d^2\textbf{r}'\\
&=\ \int_{|\textbf{r}-\textbf{r}'|=\epsilon} \log|\textbf{r}-\textbf{r}'|\frac{\partial f}{\partial n}(\textbf{r}')- \frac{\partial \log|\textbf{r}-\textbf{r}'|}{\partial n}f(\textbf{r}')\ dS(\textbf{r}')\\
&=:\ J_1+J_2.
\end{align}
For the $J_1$ term we have the estimate
\begin{align}
|J_1| \leq  |\log \epsilon|  \int_{|\textbf{r}-\textbf{r}'| = \epsilon}
 \left|\frac{\partial f}{\partial n}(\textbf{r}')\right|\ dS(\textbf{r}') \leq C|\epsilon \log \epsilon|
\end{align}
which also goes to $0$ as $\epsilon \rightarrow 0$. 
Lastly, observe
\begin{align}
-\int_{|\textbf{r}'|=\epsilon} \frac{\partial \log|\textbf{r}'|}{\partial n}
f(\textbf{r}-\textbf{r}')\ dS(\textbf{r}')= - \int_{|\textbf{r}'| = \epsilon} \frac{f(\textbf{r}-\textbf{r}')}{|\textbf{r}'|} dS(\textbf{r}')\rightarrow -2\pi f(\textbf{r}').
\end{align}
as $\epsilon \rightarrow 0$. 
Note: If we use Fourier transform for the Poisson equation $\Delta u = \delta$, we get
\begin{align}
-4\pi^2|\xi|^2 \hat u= 1 
\end{align}
which means
\begin{align}
\hat u = \frac{-1}{4\pi^2|\xi|^2}.
\end{align}
Taking the Fourier inverse of $\hat u$ yields the desired result. 
