用方框和颜色画出两行方程

Question

我有以下 mwe。我想只为方程式着色并加框，除了方程式编号。我尝试使用{\颜色{} }和\我的盒子{}但它们不起作用。任何帮助都非常感谢。

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath, amssymb}
\begin{document}
\begin{equation}
\begin{split}
&\sum_{k=-\infty}^{\infty} 
\frac{(-1)^k(\alpha+k+\frac12) \sin \sqrt{b^2+\pi^2(\alpha+k+\frac12)^2}}{(\beta^2+(2k+1)^2)\sqrt{b^2+\pi^2(\alpha+k+\frac12)^2}} \\
&=\frac{\pi(\alpha+\frac12 i \beta)\sin\sqrt{b^2+\pi^2(\alpha+\frac{i}{2}\beta)^2}}{4i\beta\cosh(\frac{\pi}{2}\beta)\sqrt{b^2+\pi^2(\alpha+\frac{i}{2}\beta)^2}}
-
\frac{\pi(\alpha-\frac12 i \beta)\sin\sqrt{b^2+\pi^2(\alpha-\frac{i}{2}\beta)^2}}{4i\beta\cosh(\frac{\pi}{2}\beta)\sqrt{b^2+\pi^2(\alpha-\frac{i}{2}\beta)^2}}.
\end{split}
\end{equation}
\end{document}

Answer 1

像这样？

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{empheq,
            amssymb}
\usepackage[dvipsname]{xcolor}
\newcommand*\widefbox[1]{\colorbox{teal!30}{\qquad#1\qquad}}

\begin{document}
\begin{empheq}[box=\widefbox]{equation}
\begin{multlined}
\sum_{k=-\infty}^{\infty}
    \frac{(-1)^k(\alpha+k+\frac12) \sin \sqrt{b^2+\pi^2(\alpha+k+\frac12)^2}}
         {(\beta^2+(2k+1)^2)\sqrt{b^2+\pi^2(\alpha+k+\frac12)^2}}   \\
    = \frac{\pi(\alpha+\frac12 i \beta)
                \sin\sqrt{b^2+\pi^2(\alpha+\frac{i}{2}\beta)^2}}
         {4i\beta\cosh(\frac{\pi}{2}\beta)
                \sqrt{b^2+\pi^2(\alpha+\frac{i}{2}\beta)^2}}        \\
    - \frac{\pi(\alpha-\frac12 i \beta)
                \sin\sqrt{b^2+\pi^2(\alpha-\frac{i}{2}\beta)^2}}
           {4i\beta\cosh(\frac{\pi}{2}\beta)\sqrt{b^2+\pi^2(\alpha-\frac{i}{2}\beta)^2}}.
\end{multlined}
\end{empheq}
\end{document}

或者您喜欢有框架和彩色的文本？

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{empheq,
            amssymb}
\usepackage[dvipsname]{xcolor}
\newcommand*\widefbox[1]{\fbox{\qquad#1\qquad}}

\begin{document}
\begin{empheq}[box=\widefbox]{equation}
\color{red}
\begin{multlined}
\sum_{k=-\infty}^{\infty}
    \frac{(-1)^k(\alpha+k+\frac12) \sin \sqrt{b^2+\pi^2(\alpha+k+\frac12)^2}}
         {(\beta^2+(2k+1)^2)\sqrt{b^2+\pi^2(\alpha+k+\frac12)^2}}   \\
    = \frac{\pi(\alpha+\frac12 i \beta)
                \sin\sqrt{b^2+\pi^2(\alpha+\frac{i}{2}\beta)^2}}
         {4i\beta\cosh(\frac{\pi}{2}\beta)
                \sqrt{b^2+\pi^2(\alpha+\frac{i}{2}\beta)^2}}        \\
    - \frac{\pi(\alpha-\frac12 i \beta)
                \sin\sqrt{b^2+\pi^2(\alpha-\frac{i}{2}\beta)^2}}
           {4i\beta\cosh(\frac{\pi}{2}\beta)\sqrt{b^2+\pi^2(\alpha-\frac{i}{2}\beta)^2}}.
\end{multlined}
\end{empheq}
\end{document}

注意：由于方程太宽，其数字无法与方程对齐，我使用multlined环境 from mathtools（由包加载）将其分成三行。如果您喜欢这种格式的方程，您可以像在环境empheq中一样插入方程。split

如果您想更改框线颜色和线条粗细，请使用fcolorbox（从包中cxolor）。例如：

\newcommand*\widefbox[1]%
    {
    \setlength{\fboxrule}{2pt}% if you like to change box lines thicnes
    \fcolorbox{blue}{white}{\qquad#1\qquad}%
    }

您将获得：

（嗯，对我的口味来说，这太花哨了:-()。）

Answer 2

以下是基于的提议这个答案。

\documentclass{article}
\usepackage{amsmath,mathtools}
\usepackage[utf8]{inputenc}
\usepackage{amsmath, amssymb}
\usepackage{tikz}
\usetikzlibrary{fit}
\newcommand{\tikznode}[2]{\tikz[remember picture,baseline=(#1.base)]{\node(#1)[inner sep=0pt]{#2};}}

\definecolor{myblue}{rgb}{.8, .8, 1}
\begin{document}
\begin{equation}
\begin{split}
&\tikznode{1}{\strut}\sum_{k=-\infty}^{\infty} 
\frac{(-1)^k(\alpha+k+\frac12) \sin
\sqrt{b^2+\pi^2(\alpha+k+\frac12)^2}\tikznode{2}{\strut}}{(\beta^2+(2k+1)^2)\sqrt{b^2+\pi^2(\alpha+k+\frac12)^2}}
\\
&=\frac{\pi(\alpha+\frac12 i \beta)\sin\sqrt{b^2+\pi^2(\alpha+\frac{i}{2}\beta)^2}}{4i\beta\cosh(\frac{\pi}{2}\beta)\sqrt{b^2+\pi^2(\alpha+\frac{i}{2}\beta)^2}}
-
\frac{\pi(\alpha-\frac12 i \beta)\sin\sqrt{b^2+\pi^2(\alpha-\frac{i}{2}\beta)^2}}{4i\beta\cosh(\frac{\pi}{2}\beta)\sqrt{b^2+\pi^2(\alpha-\frac{i}{2}\beta)^2}
\tikznode{3}{\strut}}.
\tikznode{4}{\strut}
\end{split}\tikz[remember picture,overlay]{
\node[fit=(1) (2) (3) (4),fill=blue,fill opacity=0.25]{};}
\end{equation}
or
\begin{equation}
{\leavevmode\color{red}
\begin{split}
&\tikznode{5}{\strut}\sum_{k=-\infty}^{\infty} 
\frac{(-1)^k(\alpha+k+\frac12) \sin
\sqrt{b^2+\pi^2(\alpha+k+\frac12)^2}\tikznode{6}{\strut}}{(\beta^2+(2k+1)^2)\sqrt{b^2+\pi^2(\alpha+k+\frac12)^2}}
\\
&=\frac{\pi(\alpha+\frac12 i \beta)\sin\sqrt{b^2+\pi^2(\alpha+\frac{i}{2}\beta)^2}}{4i\beta\cosh(\frac{\pi}{2}\beta)\sqrt{b^2+\pi^2(\alpha+\frac{i}{2}\beta)^2}}
-
\frac{\pi(\alpha-\frac12 i \beta)\sin\sqrt{b^2+\pi^2(\alpha-\frac{i}{2}\beta)^2}}{4i\beta\cosh(\frac{\pi}{2}\beta)\sqrt{b^2+\pi^2(\alpha-\frac{i}{2}\beta)^2}
\tikznode{7}{\strut}}.
\tikznode{8}{\strut}
\end{split}}\tikz[remember picture,overlay]{
\node[fit=(5) (6) (7) (8),draw=blue,fill opacity=0.25]{};}
\end{equation}
\dots and back to black
\end{document}

解释：在一些“极端”位置定义了四个节点，(1)、(2)和。这样就可以制作自定义框而无需改变布局(3)。(4)