As a preliminary remark, although $\omega^{(1)}$ is a Kähler form only in a bounded region, it makes sense as a closed 2-form over the entire $(\mathbb{C}^{*})^{2}\times\mathbb{R}_{\mu}$. The integral $\int_{T^{2}}\omega^{(1)}$ is a cohomological invariant, which can be evaluated asymptotically on a $T^{2}$-cycle as $\eta_{1},\eta_{2}$ stay bounded and $\mu\to+\infty$. By choosing the $T^{2}$-cycle on which $\mu,y_{1},y_{2}$ are constants,

$$\begin{split}\int_{T^{2}}\omega^{(1)}=&\frac{\sqrt{-1}}{2}\int_{T^{2}}W^{p\bar{q}}_{(1)}d\eta_{p}\wedge d\bar{\eta}_{q}=\frac{\sqrt{-1}}{2}\int_{T^{2}}(W^{1\bar{2}}_{(1)}-W^{2\bar{1}}_{(1)})dx_{1}\wedge dx_{2}\\ =&\int_{T^{2}}\text{Im}(W^{2\bar{1}}_{(1)})dx_{1}\wedge dx_{2}=\text{Im}(a_{2\bar{1}})+\int_{T^{2}}\gamma_{4}dx_{1}\wedge dx_{2}.\end{split}$$

But by Lemma 4.10 and Proposition 4.13, the quantity $\gamma_{4}\to 0$ as $\mu\to+\infty$, so the only contribution is $\int_{T^{2}}\omega^{(1)}=\text{Im}(a_{2\bar{1}}).$ ∎