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多重比对问题

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多重比对问题

我想要两种对齐方式。

\documentclass{article}
\usepackage{nicefrac,amsmath,amssymb}
\usepackage{relsize}


\begin{document}

\begin{equation}
\begin{aligned}
    &\int \limits_{0}^{2\pi} \hat{B}_{\delta r \mu} \cdot \cos \left( \mu p \alpha - \mu \omega_{el} t \right) \cdot \hat{A}_{\nu} \cdot \sin \left( \nu p \alpha - \omega_{el} t \right) d\alpha \\
    &= \hat{B}_{\delta r \mu} \cdot \hat{A}_{\nu} \cdot \int \limits_{0}^{2\pi} \cos \left( \mu p \alpha - \mu \omega_{el} t \right) \cdot \sin \left( \nu p \alpha - \omega_{el} t \right) d\alpha \\
    &= \frac{\hat{B}_{\delta r \mu} \cdot \hat{A}_{\nu}}{2} \cdot \Bigg( \overbrace{\int \limits_{0}^{2\pi} \sin \left( p \alpha \left( \nu - \mu \right) - \omega_{el} t \left(1 + \mu \right) \right) d\alpha}^{\larger\textcircled{\smaller[2]1}} + \overbrace{\int \limits_{0}^{2\pi} \sin \left( p \alpha \left( \nu + \mu \right) - \omega_{el} t \left(1 - \mu \right) \right) d\alpha}^{\larger\textcircled{\smaller[2]2}} \Bigg) \\
    &{\larger\textcircled{\smaller[2]1}}: &&\int \limits_{0}^{2\pi} \sin \left( p \alpha \left( \nu - \mu \right) - \omega_{el} t \left(1 + \mu \right) \right) d\alpha \\
    & &&= \frac{1}{p \cdot (\nu - \mu)} \cdot \Big[ -\cos \left( p \alpha (\nu - \mu) - \omega_{el} t (1 + \mu) \right) \Big]_{0}^{2\pi} \\
    & &&= \frac{1}{p \cdot (\nu - \mu)} \cdot \Big( -\cos \big( 2\pi \overbrace{ p (\nu - \mu)}^{\in \, \mathbb{Z}} - \omega_{el} t (1 + \mu) \big) + \cos \big( - \omega_{el} t (1 + \mu) \big) \Big) \\
    & &&= \frac{1}{p \cdot (\nu - \mu)} \cdot \Big( -\cos \big( - \omega_{el} t (1 + \mu) \big) + \cos \big( - \omega_{el} t (1 + \mu) \big) \Big) = 0
\end{aligned}
\end{equation}

\end{document}

但属于第二次对齐的部分几乎看不见（太靠右）。它应该位于第 4 行积分的正下方。

答案1

您想要的就是这样的东西吗？我添加了一些改进，例如差分符号为直立的 d。

\documentclass{article}
\usepackage{nicefrac, mathtools,amssymb}
\usepackage{relsize}
\usepackage{showframe}
\renewcommand*\ShowFrameLinethickness{0.3pt}
\newcommand*{\dd}{\mathop{}\!\mathrm{d}}

\begin{document}

\begin{equation}
\begin{aligned}
    &\int \limits_{0}^{2\pi} \hat{B}_{\delta r \mu} \cdot \cos \left( \mu p \alpha - \mu \omega_{el} t \right) \cdot \hat{A}_{\nu} \cdot \sin \left( \nu p \alpha - \omega_{el} t \right) \dd\alpha \\
    &= \hat{B}_{\delta r \mu} \cdot \hat{A}_{\nu} \cdot \int \limits_{0}^{2\pi} \cos \left( \mu p \alpha - \mu \omega_{el} t \right) \cdot \sin \left( \nu p \alpha - \omega_{el} t \right) \dd\alpha \\
    &= \frac{\hat{B}_{\delta r \mu} \cdot \hat{A}_{\nu}}{2} \cdot \begin{multlined}[t]\Biggl( \overbrace{\int \limits_{0}^{2\pi} \sin \left( p \alpha \left( \nu - \mu \right) - \omega_{el} t \left(1 + \mu \right) \right) \dd\alpha}^{\larger\textcircled{\smaller[2]1}} \\[-2ex] + \overbrace{\int \limits_{0}^{2\pi} \sin \left( p \alpha \left( \nu + \mu \right) - \omega_{el} t \left(1 - \mu \right) \right) \dd\alpha}^{\larger\textcircled{\smaller[2]2}} \Biggr) \end{multlined}\\
 &{\larger\textcircled{\smaller[2]1}}: \begin{aligned}[t] &\int \limits_{0}^{2\pi} \sin \left(p \alpha \left( \nu - \mu \right) - \omega_{el} t \left(1 + \mu \right) \right) d\alpha \\[-1ex]
 &= \frac{1}{p \cdot (\nu - \mu)} \cdot \Bigl[ -\cos \left( p \alpha (\nu - \mu) - \omega_{el} t (1 + \mu) \right) \Bigr]_{0}^{2\pi} \end{aligned}\\
 &= \frac{1}{p \cdot (\nu - \mu)} \cdot \Bigl( -\cos \big( 2\pi \overbrace{ p (\nu - \mu)}^{\in \, \mathbb{Z}} - \omega_{el} t (1 + \mu) \big) + \cos \big( - \omega_{el} t (1 + \mu) \big) \Bigr) \\
&= \frac{1}{p \cdot (\nu - \mu)} \cdot \Bigl( -\cos \big( - \omega_{el} t (1 + \mu) \big) + \cos \bigl( - \omega_{el} t (1 + \mu) \bigr) \Bigr) = 0
\end{aligned}
\end{equation}

\end{document}

在此处输入图片描述

答案2

请查看下面的代码，如果有任何疑问请告诉我。

在此处输入图片描述

\documentclass{article}
\usepackage{nicefrac,amssymb}
\usepackage{mathtools}    % Loads amsmath and also adds \mathrlap{}
\usepackage{relsize}


\newlength\matheqindent   \setlength\matheqindent{3em}   % Length name controlling the indentation

\begin{document}

\noindent%
\begin{align*}
  \mathrlap{
    \int \limits_{0}^{2\pi} \hat{B}_{\delta r \mu} \cdot \cos \left( \mu p \alpha - \mu \omega_{el} t \right) \cdot \hat{A}_{\nu} \cdot \sin \left( \nu p \alpha - \omega_{el} t \right) d\alpha
  }\hspace{\matheqindent} \\
  &= \hat{B}_{\delta r \mu} \cdot \hat{A}_{\nu} \cdot \int \limits_{0}^{2\pi} \cos \left( \mu p \alpha - \mu \omega_{el} t \right) \cdot \sin \left( \nu p \alpha - \omega_{el} t \right) d\alpha \\
  &= \frac{\hat{B}_{\delta r \mu} \cdot \hat{A}_{\nu}}{2} \cdot \Bigg(
  \begin{aligned}[t]
    & \overbrace{\int \limits_{0}^{2\pi} \sin \left( p \alpha \left( \nu - \mu \right) - \omega_{el} t \left(1 + \mu \right) \right) d\alpha}^{\larger\textcircled{\smaller[2]1}} \\
    & + \overbrace{\int \limits_{0}^{2\pi} \sin \left( p \alpha \left( \nu + \mu \right) - \omega_{el} t \left(1 - \mu \right) \right) d\alpha}^{\larger\textcircled{\smaller[2]2}} \Bigg) \\
  \end{aligned} \\
  \mathrlap{
    \larger\textcircled{\smaller[2]1}:
  }\hspace{\matheqindent} & \\
  \mathrlap{
    \int \limits_{0}^{2\pi} \sin \left( p \alpha \left( \nu - \mu \right) - \omega_{el} t \left(1 + \mu \right) \right) d\alpha
  }\hspace{\matheqindent}\\
  &= \frac{1}{p \cdot (\nu - \mu)} \cdot \Big[ -\cos \left( p \alpha (\nu - \mu) - \omega_{el} t (1 + \mu) \right) \Big]_{0}^{2\pi} \\
  &= \frac{1}{p \cdot (\nu - \mu)} \cdot \Big(
  \begin{aligned}[t]
    & - \cos \big( 2\pi \overbrace{ p (\nu - \mu)}^{\in \, \mathbb{Z}} - \omega_{el} t (1 + \mu) \big) \\
    & + \cos \big( - \omega_{el} t (1 + \mu) \big) \Big) \\
  \end{aligned} \\
  &= \frac{1}{p \cdot (\nu - \mu)} \cdot \Big(
  \begin{aligned}[t]
    & - \cos \big( - \omega_{el} t (1 + \mu) \big) \\
    & + \cos \big( - \omega_{el} t (1 + \mu) \big) \Big)
  \end{aligned} \\
  &= 0.
\end{align*}

\end{document}

根据评论进行更新。没有缩进和多行的公式（有一个例外）。

\documentclass{article}
\usepackage{nicefrac,amssymb}
\usepackage{mathtools}    % Loads amsmath and also adds \mathrlap{}
\usepackage{relsize}


\begin{document}

\noindent%
\begin{align*}
  &\int \limits_{0}^{2\pi} \hat{B}_{\delta r \mu} \cdot \cos \left( \mu p \alpha - \mu \omega_{el} t \right) \cdot \hat{A}_{\nu} \cdot \sin \left( \nu p \alpha - \omega_{el} t \right) d\alpha \\
  &= \hat{B}_{\delta r \mu} \cdot \hat{A}_{\nu} \cdot \int \limits_{0}^{2\pi} \cos \left( \mu p \alpha - \mu \omega_{el} t \right) \cdot \sin \left( \nu p \alpha - \omega_{el} t \right) d\alpha \\
  &= \frac{\hat{B}_{\delta r \mu} \cdot \hat{A}_{\nu}}{2} \cdot \Bigg(
    \begin{aligned}[t]
      & \overbrace{\int \limits_{0}^{2\pi} \sin \left( p \alpha \left( \nu - \mu \right) - \omega_{el} t \left(1 + \mu \right) \right) d\alpha}^{\larger\textcircled{\smaller[2]1}} \\
      & + \overbrace{\int \limits_{0}^{2\pi} \sin \left( p \alpha \left( \nu + \mu \right) - \omega_{el} t \left(1 - \mu \right) \right) d\alpha}^{\larger\textcircled{\smaller[2]2}} \Bigg) \\
    \end{aligned} \\
  &{\larger\textcircled{\smaller[2]1}}:
  \int \limits_{0}^{2\pi} \sin \left(p \alpha \left( \nu - \mu \right) - \omega_{el} t \left(1 + \mu \right) \right) d\alpha
  \\
  &= \frac{1}{p \cdot (\nu - \mu)} \cdot \Big[ -\cos \left( p \alpha (\nu - \mu) - \omega_{el} t (1 + \mu) \right) \Big]_{0}^{2\pi} \\
  &= \frac{1}{p \cdot (\nu - \mu)} \cdot \Big(- \cos \big( 2\pi \overbrace{ p (\nu - \mu)}^{\in \, \mathbb{Z}} - \omega_{el} t (1 + \mu) \big) + \cos \big( - \omega_{el} t (1 + \mu) \big) \Big) \\
  &= \frac{1}{p \cdot (\nu - \mu)} \cdot \Big(- \cos \big( - \omega_{el} t (1 + \mu) \big) + \cos \big( - \omega_{el} t (1 + \mu) \big) \Big) = 0.
\end{align*}

\end{document}

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