The basic strategy is a Liouville theorem argument: we will construct a function with the same distributional $\Delta_{a}$-Laplacian as $\beta_{p3}+\beta_{p4}$, and then argue they must be equal.

We start with the Poincaré-Lelong formula

$$S=\frac{\sqrt{-1}}{2\pi}\partial\bar{\partial}\log|1-e^{2\pi i\eta_{1}}-e^{2\pi i\eta_{1}}|^{2}=\frac{\sqrt{-1}}{2\pi}\partial\bar{\partial}\log|f_{S}|^{2},$$

from which we obtain the equality of measures

$$\begin{split}\int_{S}d\mathcal{A}=&\int_{S}\frac{\sqrt{-1}}{2}a_{p\bar{q}}d\eta_{p}\wedge d\bar{\eta}_{q}\\ =&\frac{\sqrt{-1}}{2\pi}\partial\bar{\partial}\log|f_{S}|^{2}\wedge\frac{\sqrt{-1}}{2}a_{p\bar{q}}d\eta_{p}\wedge d\bar{\eta}_{q}\\ =&\frac{1}{4\pi}(\Delta_{a}\log|f_{S}|^{2})d\text{Vol}_{a}.\end{split}$$

The periodic Newtonian potential on $(\mathbb{C}^{*})^{2}$ with the $g_{a}$-metric is

$$\gamma_{5}(\eta_{1},\eta_{2},\mu)=-\frac{1}{4\pi^{2}}\sum_{(n_{1},n_{2})\in\mathbb{Z}^{2}}\{\frac{1}{a_{p\bar{q}}(\eta_{p}+n_{p})(\bar{\eta}_{q}+n_{q})}-\frac{1}{a_{p\bar{q}}n_{p}n_{q}}\}.$$

Thus for any large cutoff scale $\Lambda$, the Green’s representation

$$\int_{S\cap\{|y^{\prime}|_{a}<\Lambda\}}\gamma_{5}(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime})$$

has the same distributional $\Delta_{a}$-Laplacian as that of $\frac{1}{4\pi}\log|f_{S}|^{2}$ in the large compact region. Taking the $\eta_{p}$ derivative and taking the $\Lambda\to\infty$ limit shows that the $\Delta_{a}$-Laplacian of $\frac{-ie^{2\pi i\eta_{p}}}{2f_{S}}$ agrees with that of the improper integral

$$\int_{S}\frac{\partial}{\partial\eta_{p}}\gamma_{5}(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime}),$$

which by formula (4.26) is the same as the improper integral

$$-\frac{1}{2}A^{1/2}\int_{S}(\gamma_{p3}+\gamma_{p4})(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)d\mathcal{A}(\eta_{1}^{\prime},\eta_{2}^{\prime})=\frac{1}{4\pi}(\beta_{p3}+\beta_{p4}).$$

The upshot is that $\beta_{p3}+\beta_{p4}$ differs from $\frac{-2\pi ie^{2\pi i\eta_{p}}}{f_{S}}$ by a globally smooth $\Delta_{a}$-harmonic function on $(\mathbb{C}^{*})^{2}$. It is also easy to show using techniques in this Section that this difference can have at most log growth in $y_{1},y_{2}$ variables. Thus it has to be a constant.

The rest of this proof is to pin down precisely this constant, by considering the limit $(\eta_{1},\eta_{2})=(\sqrt{-1}b,\sqrt{-1}b)$ for $b\to+\infty$. This uses techniques similar to the proof of Lemma 4.24. Without affecting the limit, we can replace $S$ with $\cup\mathfrak{D}_{i}\times S^{1}$ and replace $(\gamma_{p3}+\gamma_{p4})$ with

$$\frac{\sqrt{-1}a_{p\bar{q}}(b-y_{q}^{\prime})\sqrt{\mathbb{A}}}{2\pi A^{3/2}|y-y^{\prime}|_{a}^{2}}.$$

This leads to an asymptotic expression for $b\gg 1$,

$$(\beta_{p3}+\beta_{p4})(\sqrt{-1}b,\sqrt{-1}b)\sim\int_{\cup\mathfrak{D}_{i}\times S^{1}}\frac{\sqrt{-1}a_{p\bar{q}}(-b+y_{q}^{\prime})\sqrt{\mathbb{A}}}{A|y-y^{\prime}|_{a}^{2}}d\mathcal{A},$$

where the RHS is understood as an improper integral. To evaluate this integral we fix $b$ and calculate the $\Lambda\to+\infty$ asymptotic expression of the integral over the large bounded domain $(\cup\mathfrak{D}_{i}\times S^{1})\cap\{|y^{\prime}|_{a}<\Lambda\}.$ The contribution from the end $\mathfrak{D}_{1}\times S^{1}$ is

$$\begin{split}&\frac{\sqrt{-1}a_{p\bar{2}}\sqrt{\mathbb{A}}}{2A}\log(\frac{\Lambda^{2}}{(a_{1\bar{1}}+a_{1\bar{2}}+a_{2\bar{1}}+a_{2\bar{2}})b^{2}})\\ +&\frac{\sqrt{-1}(a_{p\bar{2}}\text{Re}a_{1\bar{2}}-a_{p\bar{1}}a_{2\bar{2}})}{A}(\frac{\pi}{2}+\arctan(\frac{a_{2\bar{2}}+\text{Re}(a_{1\bar{2}})}{\sqrt{\mathbb{A}}}))+o(1).\end{split}$$

The contribution from $\mathfrak{D}_{2}\times S^{1}$ is

$$\begin{split}&\frac{\sqrt{-1}a_{p\bar{1}}\sqrt{\mathbb{A}}}{2A}\log(\frac{\Lambda^{2}}{(a_{1\bar{1}}+a_{1\bar{2}}+a_{2\bar{1}}+a_{2\bar{2}})b^{2}})\\ +&\frac{\sqrt{-1}(a_{p\bar{1}}\text{Re}a_{1\bar{2}}-a_{p\bar{2}}a_{1\bar{1}})}{A}(\frac{\pi}{2}+\arctan(\frac{a_{1\bar{1}}+\text{Re}(a_{1\bar{2}})}{\sqrt{\mathbb{A}}}))+o(1).\end{split}$$

The contribution from $\mathfrak{D}_{3}\times S^{1}$ is

$$-\frac{\sqrt{-1}\sqrt{\mathbb{A}}(a_{p\bar{1}}+a_{p\bar{2}})}{2A}\log(\frac{\Lambda^{2}}{(a_{1\bar{1}}+a_{1\bar{2}}+a_{2\bar{1}}+a_{2\bar{2}})b^{2}})+o(1).$$

Summing up, the log terms cancel out, so the improper integral

$$\int_{\cup\mathfrak{D}_{i}\times S^{1}}\frac{\sqrt{-1}a_{p\bar{q}}(-b+y_{q}^{\prime})\sqrt{\mathbb{A}}}{A|y-y^{\prime}|_{a}^{2}}d\mathcal{A},$$

is equal to the constant $K_{p}(a)$ defined in the statement of the Lemma. This shows limiting value

$$\lim_{b\to+\infty}(\beta_{p3}+\beta_{p4})(\sqrt{-1}b,\sqrt{-1}b)=K_{p}(a).$$

Comparing this with

$$\lim_{b\to+\infty}\frac{-2\pi ie^{2\pi i\eta_{p}}}{f_{S}}(\sqrt{-1}b,\sqrt{-1}b)=0$$

determines the constant. ∎