We consider the closely related integral

$$\bar{\gamma}(y_{1},y_{2},\mu)=-\frac{1}{8\pi^{2}}\int\frac{1}{(a_{p\bar{q}}\eta_{p}\bar{\eta}_{q}+A\mu^{2})^{3/2}}dx_{1}dx_{2}.$$

After substituting the variables

$$\begin{cases}x_{1}^{\prime}=x_{1}-\frac{\text{Im}(a_{2\bar{1}})}{a_{1\bar{1}}}y_{2}+\frac{\text{Re}(a_{2\bar{1}})}{a_{1\bar{1}}}x_{2},\\ x_{2}^{\prime}=x_{2}+\frac{\text{Im}(a_{2\bar{1}})\text{Re}(a_{2\bar{1}})}{\mathbb{A}}y_{2}+\frac{a_{1\bar{1}}\text{Im}(a_{2\bar{1}})}{\mathbb{A}}y_{1},\end{cases}$$

we complete the square

$$a_{p\bar{q}}\eta_{p}\bar{\eta}_{q}+A\mu^{2}=a_{1\bar{1}}x_{1}^{\prime 2}+\frac{\mathbb{A}}{a_{1\bar{1}}}x_{2}^{\prime 2}+|(y_{1},y_{2},\mu)|_{a}^{\prime 2}=a_{1\bar{1}}x_{1}^{\prime 2}+\frac{\mathbb{A}}{a_{1\bar{1}}}x_{2}^{\prime 2}+\varrho^{2}.$$

This allows us to evaluate using polar coordinates

$$\bar{\gamma}=-\frac{1}{4\pi\varrho\sqrt{\mathbb{A}}}.$$

For fixed $y_{1},y_{2},\mu$, we can compare the integral $\bar{\gamma}$ with the series $\gamma$, by estimating the difference using the mean value inequality

$$\begin{split}&\frac{1}{|(\eta_{1},\eta_{2},\mu)|_{a}^{3}}-\int_{[x_{1}-\frac{1}{2},x_{1}+\frac{1}{2}]\times[x_{2}-\frac{1}{2},x_{2}+\frac{1}{2}]}\frac{1}{|(s_{1}+\sqrt{-1}y_{1},s_{2}+\sqrt{-1}y_{2},\mu)|_{a}^{3}}ds_{1}ds_{2}\\ \leq&\frac{CA^{1/2}}{|(\eta_{1},\eta_{2},\mu)|_{a}^{5}}.\end{split}$$

Summing over all square regions, and applying Cauchy integral test,

$$\begin{split}|\gamma-\bar{\gamma}|&\leq CA^{1/2}\sum_{n,m}\frac{1}{|(\eta_{1}+n,\eta_{2}+m,\mu)|_{a}^{5}}\\ &\leq CA^{1/2}\int\frac{1}{|(s_{1}+\sqrt{-1}y_{1},s_{2}+\sqrt{-1}y_{2},\mu)|_{a}^{5}}ds_{1}ds_{2}\\ &\leq C\varrho^{-3},\end{split}$$

as required. ∎