Equation (8.21) shows that the degree of $\mathbb{P}(E)$ is equal to $(n+r)!\operatorname{vol}(\Delta)$. The same equation together with Proposition 8.20(4) gives that the height of $\mathbb{P}(E)$ is equal to :

(8.25)

$$-\frac{(n+r+1)!}{2}\left(\int_{\Delta}\varepsilon_{r}(y)\,\text{\rm d}x\,\text{\rm d}y+\int_{\Delta}L(y)\cdot\varepsilon_{n}(L(y)^{-1}x)\,\text{\rm d}x\,\text{\rm d}y\right).$$

Let $I_{1}$ and $I_{2}$ be the two above integrals. Observe $\Delta=\bigcup_{y\in\Delta^{r}}(\{y\}\times L(y)\cdot\Delta^{n})$. Then

	$\displaystyle\operatorname{vol}(\Delta)$	$\displaystyle=\int_{\Delta^{r}}\left(\int_{L(y)\cdot\Delta^{n}}dx\right)\,\text{\rm d}y=\frac{1}{n!}\int_{\Delta^{r}}L(y)^{n}\,\text{\rm d}y,$
	$\displaystyle I_{1}$	$\displaystyle=\int_{\Delta^{r}}\left(\int_{L(y)\cdot\Delta^{n}}\,\text{\rm d}x\right)\varepsilon_{r}(y)\,\text{\rm d}y=\frac{1}{n!}\int_{\Delta^{r}}L(y)^{n}\varepsilon_{r}(y)\,\text{\rm d}y,$

since $\int_{L(y)\cdot\Delta^{n}}\,\text{\rm d}x=L(y)^{n}/n!$. And, for the second integral,

	$\displaystyle I_{2}$	$\displaystyle=\int_{\Delta^{r}}L(y)\left(\int_{L(y)\cdot\Delta^{n}}\varepsilon_{n}(L(y)^{-1}x)\,\text{\rm d}x\right)\,\text{\rm d}y$
		$\displaystyle=\left(\int_{\Delta^{r}}L(y)^{n+1}\,\text{\rm d}y\right)\cdot\left(\int_{\Delta^{n}}\varepsilon_{n}(x)\,\text{\rm d}x\right)=-\frac{2\operatorname{h}_{{\overline{{\mathcal{O}}(1)}}}(\mathbb{P}^{n})}{(n+1)!}\int_{\Delta^{r}}L(y)^{n+1}\,\text{\rm d}y.$

since $\int_{L(y)\cdot\Delta^{n}}\varepsilon_{n}(L(y)^{-1}x)\,\text{\rm d}x=L(y)^{n}\int_{\Delta^{n}}\varepsilon_{n}(x)\,\text{\rm d}x$ and

$$\int_{\Delta^{n}}\varepsilon_{n}(x)\,\text{\rm d}x=\frac{-1}{(n+1)!}\cdot\sum_{h=1}^{n}\sum_{j=1}^{h}\frac{1}{j}=-\frac{2\operatorname{h}_{{\overline{{\mathcal{O}}(1)}}}(\mathbb{P}^{n})}{(n+1)!}.$$

The expression for $\operatorname{vol}(\Delta)$ gives the formula for the degree. Carrying the expressions of $I_{1}$ and $I_{2}$ in (8.25) concludes the proof of Lemma 8.22. ∎