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投影仪演示文稿中数学方程的对齐问题

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投影仪演示文稿中数学方程的对齐问题 
\documentclass{beamer}
\newcommand\Fontvi{\fontsize{2}{4}\selectfont}
\usepackage[utf8]{inputenc}
\usepackage[compat=1.1.0]{tikz-feynman}
\usepackage{amsmath, amsthm, amssymb,amsfonts}
\usepackage{graphicx}
\usepackage{tikz}
\usepackage{tcolorbox}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{tikz-feynman}
\setcounter{MaxMatrixCols}{20}
%\usepackage{breqn}
\usepackage{tikz-feynman} 
\usepackage{amsmath}
\newtheorem{rules}{Rule}
\usetheme{Antibes}
\usecolortheme{}
\title{Negative Dimensional Integral and Modified Method of Brackets}
\author[Sumit Banik]{\textbf {Work By-Sumit Banik\\ \footnotesize Supervised by: B Ananthanarayan}}
\institute{Indian Institute of Science\\Bangalore- 560012}
\date{}
\begin {document}
\begin {frame}
\begin{rules} \label{Rule2}
\begin{equation} \label{eq:rule-B}
          \begin{split}
            &(A_1 + \cdots + A_J)^\alpha
            \\
            &\quad=
            \frac1{\Gamma(-\alpha)}
            \int_{c_1-i\infty}^{c_1+i\infty}\frac{dz_1}{2\pi i}
            \cdots
            \int_{c_J-i\infty}^{c_J+i\infty}\frac{dz_J}{2\pi i}
            \,
            <z_1 + \cdots + z_J - \alpha>
            \,
            A_1^{z_1}
            \cdots
            A_J^{z_J}
            \,
            \Gamma(-z_1)
            \cdots
            \Gamma(-z_J)\,.
          \end{split}          
        \end{equation}
\end{rules}
\end {frame}
\end {document}

答案1

我会选择一个multline环境。无关：不要加载两次包，而且，如果你加载了amssymb，就不需要加载了amsfonts。

\documentclass{beamer}
\newcommand\Fontvi{\fontsize{2}{4}\selectfont}
\usepackage[utf8]{inputenc}
\usepackage[compat=1.1.0]{tikz-feynman}
\usepackage{mathtools, amsthm, amssymb}
\usepackage{graphicx}
\usepackage{tikz}
\usepackage{tcolorbox}
\usepackage{tikz-feynman}
\setcounter{MaxMatrixCols}{20}
\usepackage{tikz-feynman}
\newtheorem{rules}{Rule}
\usetheme{Antibes}
\usecolortheme{}
\title{Negative Dimensional Integral and Modified Method of Brackets}
\author[Sumit Banik]{\textbf {Work By-Sumit Banik\\ \footnotesize Supervised by: B Ananthanarayan}}
\institute{Indian Institute of Science\\Bangalore- 560012}
\date{}

\begin {document}

\begin {frame}
\begin{rules} \label{Rule2}
\begin{multline}\label{eq:rule-B}
             (A_1 + \cdots + A_J)^\alpha = 
            \\
                            \frac1{\Gamma(-\alpha)}
            \int_{c_1-i\infty}^{c_1+i\infty}\frac{dz_1}{2\pi i}
            \cdots
            \int_{c_J-i\infty}^{c_J+i\infty}\frac{dz_J}{2\pi i}
            \,
             \langle z_1 + \cdots + z_J - \alpha\rangle \times \\
              A_1^{z_1}
            \cdots
            A_J^{z_J}
            \,
            \Gamma(-z_1)
            \cdots
            \Gamma(-z_J)\phantom{\times}
        \end{multline}
\end{rules}
\end {frame}

\end {document}

在此处输入图片描述

答案2

我认为，如果再增加几个换行符，等式会变得更好。

在此处输入图片描述

\documentclass{beamer}
\usetheme{Antibes}
\usepackage{amsmath, amsthm, amssymb}%,amsfonts}
\newtheorem{rules}{Rule}
\usepackage{graphicx}
\usepackage[utf8]{inputenc}

%% I've commented out all instructions that are repetitive or 
%% do not affect the body of the sample document.
%\newcommand\Fontvi{\fontsize{2}{4}\selectfont}
%\usepackage[compat=1.1.0]{tikz-feynman}
%\usepackage{tikz}
%\usepackage{tcolorbox}
%\usepackage{amsmath}
%\usepackage{amsthm}
%\usepackage{amssymb}
%\usepackage{tikz-feynman}
%\setcounter{MaxMatrixCols}{20}
%\usepackage{breqn}
%\usepackage{tikz-feynman} 
%\usepackage{amsmath}
%\usecolortheme{}
%\title{Negative Dimensional Integral and Modified Method of Brackets}
%\author[Sumit Banik]{\textbf {Work By-Sumit Banik\\ \footnotesize Supervised by: B Ananthanarayan}}
%\institute{Indian Institute of Science\\Bangalore- 560012}
%\date{}

\begin{document}
\begin{frame}
\begin{rules} \label{Rule2}
\begin{equation} \label{eq:rule-B}
\begin{split}
    &(A_1 + \dots + A_J)^\alpha \\
    &\quad= \frac1{\Gamma(-\alpha)}
    \int_{c_1-i\infty}^{c_1+i\infty}\frac{dz_1}{2\pi i}
    \dotsm
    \int_{c_J-i\infty}^{c_J+i\infty}\frac{dz_J}{2\pi i}
    \\
    &\quad< z_1 + \dots + z_J - \alpha\\
    &\quad> A_1^{z_1}\dotsm A_J^{z_J}
    \,\Gamma(-z_1)\dotsm\Gamma(-z_J)\,.
\end{split}          
\end{equation}
\end{rules}
\end{frame}
\end{document}

附录，灵感来自于@Bernard 的平行回答：如果<和>并不表示“小于”和“大于”，而是应该被解释为尖括号（在 TeX 用语中分别为\langle和\rangle），那么最好将上面显示的答案的第三行和第四行合并为一行：

在此处输入图片描述

\documentclass{beamer}
\usetheme{Antibes}
\usepackage{amsmath, amsthm, amssymb}
\newtheorem{rules}{Rule}
\usepackage{graphicx}
\usepackage[utf8]{inputenc}

\begin{document}
\begin{frame}
\begin{rules} \label{Rule2}
\begin{equation} \label{eq:rule-B}
\begin{split}
    &(A_1 + \dots + A_J)^\alpha \\
    &\quad= \frac1{\Gamma(-\alpha)}
    \int_{c_1-i\infty}^{c_1+i\infty}\frac{dz_1}{2\pi i}
    \dotsm
    \int_{c_J-i\infty}^{c_J+i\infty}\frac{dz_J}{2\pi i} \\
    &\qquad\times 
      \langle z_1 + \dots + z_J - \alpha\rangle
      \,A_1^{z_1}\dotsm A_J^{z_J} 
      \,\Gamma(-z_1)\dotsm\Gamma(-z_J)\,.
\end{split}          
\end{equation}
\end{rules}
\end{frame}
\end{document}

答案3

您应该拆分方程式。请考虑以下代码：

\documentclass{beamer}
\usepackage[compat=1.1.0]{tikz-feynman}
\usepackage{amsmath, amsthm, amssymb,amsfonts}
\usepackage{tikz}
\newtheorem{rules}{Rule}
\usetheme{Antibes}
\begin {document}
\begin {frame}
\begin{rules} \label{Rule2}
\begin{align} \label{eq:rule-B}
            &(A_1 + \cdots + A_J)^\alpha
            \nonumber\\
            \quad=&
            \frac1{\Gamma(-\alpha)}
            \int_{c_1-i\infty}^{c_1+i\infty}\frac{dz_1}{2\pi i}
            \cdots
            \int_{c_J-i\infty}^{c_J+i\infty}\frac{dz_J}{2\pi i}
            \,\nonumber\\
            &<z_1 + \cdots + z_J - \alpha>
            \,
            A_1^{z_1}
            \cdots
            A_J^{z_J}
            \,
            \Gamma(-z_1)
            \cdots
            \Gamma(-z_J)\,.       
        \end{align}
\end{rules}
\end {frame}
\end {document}

在此处输入图片描述

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