In $\mathbb{R}^2\setminus\{(0,0)\}$, let $\theta(x,y)$ denote the branch of polar angle satisfying $-\pi<\theta(x,y)\leq \pi$. Since $\theta \in L^1_{loc}(\mathbb{R}^2)$, it can be regarded as a distribution on $\mathbb{R}^2$. Now the problem asks to show that $$\triangle \theta=2\pi {\partial v \over \partial y}$$ as distribution, where $v \in \mathcal{D}'(\mathbb{R}^2)$ is defined by $$\langle v,\varphi \rangle=\int_{-\infty}^0 \varphi(x,0)\ dx$$
[Problem] The calculation goes as \begin{equation} \begin{split} \langle \triangle \theta,\varphi\rangle &=\langle \theta,\triangle\varphi\rangle\\ &=\int_{\mathbb{R}^2}\theta(x,y)\triangle\varphi(x,y)\ dxdy\\ &=\left(\int_{I+II+III+IV}+\int_{\textrm{x-axis}\cup\textrm{y-axis}}\right)\cdots \end{split} \end{equation} where $I$ denotes the first quadrant, $II$ the second, and likewise. Since $theta(x,y)$ restricted to each quadrant takes form of $$\theta(x,y)=C_1\pi+C_2\arctan{y \over x}$$ where $C_1,C_2 \in \{-1,1\}$, it is simple calculation to see $\triangle \theta=0$ on $I\cup II\cup III\cup IV$ whence the first term of the above integral is $0$. On the other hand, since x-axis and y-axis are both of measure $0$, we must also necessarily have the value $0$ for the second term of the above integral, and hence we conclude $$\langle\triangle\theta,\varphi\rangle=0$$ for all $\varphi \in \mathcal{D}'(\mathbb{R}^2)$. But this is really wield since the problem indicates that the Laplacian of $\theta$ does not vanish. Where am I doing wrong here?