We focus on the $\beta_{p3}$ case, and consider $A^{1/4}\mu\gg|\eta_{1}|+|\eta_{2}|+1$. Using Lemma 4.22, the contribution to $\int\gamma_{p3}d\mathcal{A}$ from the region $\{|\eta_{1}-\eta_{1}^{\prime}|+|\eta_{2}-\eta_{2}^{\prime}|\leq(A^{1/4}\mu)^{1-\epsilon}\}\cap S$ is negligible, where $0<\epsilon\ll 1$ is any small given number. Outside this region $S$ is asymptotic to $\mathfrak{D}_{i}\times S^{1}$ along the three ends up to exponentially small error, and furthermore Lemma 4.23 allows us to replace $\gamma_{p3}$ by $I_{p3}$ without affecting the $\mu\to+\infty$ limit.

We are now left to consider the improper integral

$$\int_{\cup(\mathfrak{D}_{i}\times S^{1})}I_{p3}(y_{1}-y_{1}^{\prime},y_{2}-y_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime}).$$

Using the formula of $I_{p3}$ in Lemma 4.21, we can simplify further by setting $y_{1}=y_{2}=0$ without affecting the $\mu\to+\infty$ limit. Along the $\mathfrak{D}_{1}\times S^{1}$ end,

$$\int_{\mathfrak{D}_{1}\times S^{1}\cap\{y_{2}^{\prime}<\frac{\Lambda}{\sqrt{a_{2\bar{2}}}}\}}I_{p3}(-y_{1}^{\prime},-y_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime})=\int_{0}^{\frac{\Lambda}{\sqrt{a_{2\bar{2}}}}}I_{p3}(0,-y_{2}^{\prime},\mu)a_{2\bar{2}}dy_{2}^{\prime},$$

which we compute as

$$\begin{split}&\frac{-a_{p\bar{2}}\sqrt{-1}}{4\pi A^{1/2}\sqrt{\mathbb{A}}}\int_{0}^{\frac{\Lambda}{\sqrt{a_{2\bar{2}}}}}a_{2\bar{2}}y_{2}^{\prime}dy_{2}^{\prime}\frac{1}{|(0,-y_{2}^{\prime},\mu)|_{a}^{\prime}(|(0,-y_{2}^{\prime},\mu)|_{a}^{\prime}+A^{1/2}\mu)}\\ =&\frac{-a_{p\bar{2}}\sqrt{-1}\sqrt{\mathbb{A}}}{8\pi A^{3/2}}\int_{A\mu^{2}}^{\frac{A}{\mathbb{A}}\Lambda^{2}+A\mu^{2}}\frac{ds}{s^{1/2}(s^{1/2}+A^{1/2}\mu)}\\ =&\frac{-a_{p\bar{2}}\sqrt{-1}\sqrt{\mathbb{A}}}{4\pi A^{3/2}}\log\left(\frac{(\mathbb{A}^{-1}\Lambda^{2}+\mu^{2})^{1/2}+\mu}{2\mu}\right).\end{split}$$

Similarly, the integrals from $\mathfrak{D}_{2}\times S^{1}$ and $\mathfrak{D}_{3}\times S^{1}$ are respectively

$$\frac{-a_{p\bar{1}}\sqrt{-1}\sqrt{\mathbb{A}}}{4\pi A^{3/2}}\log\left(\frac{(\mathbb{A}^{-1}\Lambda^{2}+\mu^{2})^{1/2}+\mu}{2\mu}\right)$$

and

$$\frac{(a_{p\bar{1}}+a_{p\bar{2}})\sqrt{-1}\sqrt{\mathbb{A}}}{4\pi A^{3/2}}\log\left(\frac{(\mathbb{A}^{-1}\Lambda^{2}+\mu^{2})^{1/2}+\mu}{2\mu}\right).$$

Summing over the three contributions and take the limit $\Lambda\to+\infty$,

$$\int_{\cup(\mathfrak{D}_{i}\times S^{1})}I_{p3}(-y_{1}^{\prime},-y_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime})=0.$$

This proves $\lim_{\mu\to+\infty}\beta_{p3}(\eta_{1},\eta_{2},\mu)=0$ . Likewise with the $\beta_{p4}$ case.

The ‘morever’ statement follows from a simpler argument. The key is that higher derivatives of the integrand have faster decay at large distance, so that the divergence issues do not arise. ∎