Using the same strategy as in Lemma 4.22, we reduce to the special case $x_{1}=x_{2}=0,\eta_{p}=\sqrt{-1}y_{p}$. Pairing $(n_{1},n_{2})$ with $(-n_{1},-n_{2})$ in the series (4.26),

$$(\gamma_{p3}+\gamma_{p4})(\eta_{1},\eta_{2},\mu)=\sum_{(n_{1},n_{2})\in\mathbb{Z}^{2}}\frac{\sqrt{-1}a_{p\bar{q}}y_{q}}{2\pi^{2}A^{1/2}|a_{i\bar{j}}(n_{i}+\sqrt{-1}y_{i})(n_{j}-\sqrt{-1}y_{j})|^{2}}.$$

We compare this series expression of $\gamma_{p3}+\gamma_{p4}$ to the closely related integral (cf. Lemma 4.21)

$$\begin{split}\frac{\sqrt{-1}a_{p\bar{q}}y_{q}}{2\pi^{2}A^{1/2}}I_{03}=\frac{\sqrt{-1}a_{p\bar{q}}y_{q}\sqrt{\mathbb{A}}}{2\pi A^{3/2}|y|_{a}^{2}}.\end{split}$$

The deviation between the series and the integral is bounded by

$$C|a_{p\bar{q}}y_{q}|A^{-1/4}\int_{\mathbb{R}^{2}}\frac{1}{|a_{i\bar{j}}(s_{i}+\sqrt{-1}y_{i})(s_{j}-\sqrt{-1}y_{j})|^{5/2}}ds_{1}ds_{2}\leq\frac{C}{A^{1/2}|y|_{a}^{2}},$$

using the same type of Cauchy integral test argument as Lemma 4.1.

The ‘morever’ statement is a minor variant of the proof of Lemma 4.22, where in the application of the Cauchy integral test we use the mean value inequality to estimate the difference between the series and the integral, similar to the argument in Lemma 4.1. ∎