如何防止 \smash 与 \intertext 的组合导致不可接受的情况

Question

& = \frac 1 {2\pi i} \int\limits_C \frac{f(w)}{w-a}
\cdot \sum_{n=0}^\infty \left( \frac{z-a}{w-a} \right)^n \, dw \\[8pt]
& = \sum_{n=0}^\infty \frac 1 {2\pi i} \int\limits_C \frac{f(w)}{(w-a)^{n+1}}
\cdot {} \underbrace{(z-a)^n}_{\begin{smallmatrix} \text{No ``$w$''} \\
\text{appears} \\ \text{here!} \end{smallmatrix}} \, dw
\intertext{The absence of ``$w$'' from the expression over this
$\underbrace{\text{underbrace}}$ justifies the next step.}
& = \sum_{n=0}^\infty \left( \,  \smash[b]{\underbrace{ \frac 1 {2\pi i}
\int\limits_C \frac{f(w)}{(w-a)^{n+1}} \, dw}_{\begin{smallmatrix}
\text{No ``$z$''} \\ \text{appears} \\ \text{here!}
\end{smallmatrix}}} \, \right) (z-a)^n
\intertext{The absence of ``$z$'' from the expression over this
$\underbrace{\text{underbrace}}$ justifies this next step.}
& = \sum_{n=0}^\infty c_n(z-a)^n.

使用的\smash[b]{}目的是为了使分隔符具有适当的大小（如果没有下括号的话），并且它确实做到了这一点。

然而，下面的线\intertext{}直接穿过那个小矩阵，这显然是一个令人憎恶的结果。

我怎样才能让它按预期工作，而又不做任何太复杂的事情或冒犯 LaTeX 之神？

Answer 1

不要使用\left...\right结构，而要使用\biggl...\biggr对：

\documentclass{article}

\usepackage{amsmath}

\begin{document}

\begin{align*}
  & = \frac{1}{2 \pi i} \int\limits_C \frac{f(w)}{w - a}
    \cdot \sum_{n = 0}^\infty \biggl( \frac{z - a}{w - a} \biggr)^n \, \mathrm{d}w \\[8pt]
  & = \sum_{n = 0}^\infty \frac{1}{2 \pi i} \int\limits_C \frac{f(w)}{(w - a)^{n + 1}}
    \cdot \underbrace{(z - a)^n}_{\substack{\text{No ``$w$''} \\ \text{appears} \\ \text{here!} }} \, \mathrm{d}w
\intertext{The absence of ``$w$'' from the expression over this $\underbrace{\text{underbrace}}$ justifies the next step.}
  & = \sum_{n = 0}^\infty \biggl( \underbrace{ \frac{1}{2 \pi i}
    \int\limits_C \frac{f(w)}{(w - a)^{n + 1}} \, \mathrm{d}w}_{\substack{\text{No ``$z$''} \\ \text{appears} \\ \text{here!}}} \biggr) (z - a)^n
\intertext{The absence of ``$z$'' from the expression over this $\underbrace{\text{underbrace}}$ justifies this next step.}
  & = \sum_{n = 0}^\infty c_n (z - a)^n.
\end{align*}

\end{document}