We denote $\vec{\mu}=(\mu_{1},\mu_{2},\eta)$ and $|\vec{\mu}|_{a}=\sqrt{a_{ij}\mu_{i}\mu_{j}+A|\eta|^{2}}$. The Green representation

$$\begin{split}&\frac{-1}{4\pi^{2}}\int_{0}^{\infty}\frac{1}{|\vec{\mu}-(0,s,0)|_{a}^{2}}ds\\ =&\frac{-1}{4\pi^{2}}\int_{0}^{\infty}\frac{1}{a_{11}\mu_{1}^{2}-2a_{12}\mu_{1}(s-\mu_{2})+a_{22}(s-\mu_{2})^{2}+A|\eta|^{2}}ds\\ =&\frac{-1}{4\pi^{2}}\int_{-\mu_{2}}^{\infty}\frac{1}{a_{11}\mu_{1}^{2}-2a_{12}\mu_{1}s+a_{22}s^{2}+A|\eta|^{2}}ds\\ =&\frac{-1}{4\pi}\frac{1}{\sqrt{A}}\frac{1}{\sqrt{\mu_{1}^{2}+a_{22}|\eta|^{2}}}\{\frac{1}{2}+\frac{1}{\pi}\arctan(\frac{a_{22}\mu_{2}+a_{12}\mu_{1}}{\sqrt{A}\sqrt{\mu_{1}^{2}+a_{22}|\eta|^{2}}})\}\\ =&\frac{-1}{2\pi\sqrt{A}}\alpha_{1}\end{split}$$

shows the equality of the two measures

$$(\Delta_{a}\alpha_{1})d\text{Vol}_{a}=-2\pi\sqrt{A}\int_{\mathfrak{D}_{1}}d\mu_{2},$$

and it is easy to check from this integral calculation

$$\frac{\partial\alpha_{1}}{\partial\mu_{2}}=\frac{\sqrt{A}}{2\pi|\vec{\mu}|_{a}^{2}}.$$

To see the singularity structure near $\mathfrak{D}_{1}=\{\mu_{1}=0,\eta=0,\mu_{2}>0\}$ explicitly, we can write

$$\alpha_{1}=\frac{1}{2\sqrt{\mu_{1}^{2}+a_{22}|\eta|^{2}}}\{1-\frac{1}{\pi}\arctan(\frac{\sqrt{A}\sqrt{\mu_{1}^{2}+a_{22}|\eta|^{2}}}{a_{22}\mu_{2}+a_{12}\mu_{1}})\},$$

and Taylor expand the arctan function.

The situations of $\alpha_{2},\alpha_{3}$ are similar. A fast way to derive them by analogy is to remember that $\frac{\partial}{\partial\mu_{2}},\frac{\partial}{\partial\mu_{1}},-\frac{\partial}{\partial\mu_{1}}-\frac{\partial}{\partial\mu_{2}}$ are the directional vectors along $\mathfrak{D}_{1},\mathfrak{D}_{2},\mathfrak{D}_{3}$, and notice

$$\iota_{\frac{\partial}{\partial\mu_{2}}}g_{a}=d(a_{12}\mu_{1}+a_{22}\mu_{2}),\quad g_{a}(\frac{\partial}{\partial\mu_{2}},\frac{\partial}{\partial\mu_{2}})=a_{22}.$$

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