We have $\phi(T_{-})=\phi(T_{+})-\Psi,$ where

(4.169)

$$\Psi=\int_{T_{-}}^{T_{+}}zhdz.$$

Away from $H$ we have

(4.170)

$$d_{D}d_{D}^{c}\Psi=\int_{T_{-}}^{T_{+}}zd_{D}d_{D}^{c}hdz=-\int_{T_{-}}^{T_{+}}z\partial_{z}^{2}\tilde{\omega}dz$$

Integration by parts we get

(4.171)

$$d_{D}d_{D}^{c}\Psi=(-z\partial_{z}\tilde{\omega}+\tilde{\omega})|^{T_{+}}_{T_{-}}=\epsilon_{T}$$

Notice since there is a factor $z$ in the integrand we do not get residue term at $z=0$. Notice $\Psi$ is continuous on $D$, and the right hand side is smooth on $D$, so elliptic regularity implies that $\Psi$ is indeed smooth on $D$, and the equation holds globally on $D$.

On the other hand, we have

(4.172)

$$\int_{D}\Psi\omega_{D}^{n-1}=\int_{T_{-}}^{T_{+}}z\int_{D}h\omega_{D}^{n-1}dz$$

Using (4.14)

	$\displaystyle\frac{\int_{D}\Psi\omega_{D}^{n-1}}{\int_{D}\omega_{D}^{n-1}}=$	$\displaystyle T^{2-n}k_{+}^{-2}[\frac{(k_{+}z+T)^{n+1}}{n+1}-\frac{T(k_{+}z+T)^{n}}{n}]$
(4.173)			$\displaystyle-T^{2-n}k_{-}^{-2}[\frac{(k_{-}z+T)^{n+1}}{n+1}-\frac{T(k_{-}z+T)^{n}}{n}]+T^{-1}\underline{B}_{T},$

where we used the definition of $T_{-}$ and $T_{+}$. (4.171) and (4.173) together yield the conclusion. ∎