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大方程问题

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大方程问题

我需要在 Latex 中写下这个等式： 在此处输入图片描述

但情况变得混乱：

 \begin{equation}
\begin{aligned}
H_x(\rho_c,\phi_c,z_c)&=-\frac{1}{2\pi k\eta}\int_0^\infty F_y(k_{\rho c})\Big[\Big(k^2-\frac{1}{2}k^2_{\rho c}\Big)J_0(k_{\rho c} \rho_c)+\frac{1}{2}k^2_{\rho c}\cos{(2\phi_c)}J_2(k_{\rho c}\rho_c)\Big]\\
%
&\times e^{-jk_{zc}z_c}\frac{k_{\rho c}}{k_{zc}}dk_{\rho c}\\
H_x(\rho_c,\phi_c,z_c)=-\frac{\sin{(\phi_c)}}{4\pi k\eta}\int_0^\infty F_y(k_{\rho c})J_2(k_{\rho c} \rho_c)e^{-jk_{zc}z_c}\frac{k^2_{\rho c}}{k_{zc}}dk_{\rho c}\\
%
H_z(\rho_c,\phi_c,z_c)=-\frac{j\cos{(2\phi_c)}}{2\pi k\eta}\int_0^\infty F_y(k_{\rho c})J_1(k_{\rho c} \rho_c)e^{-jk_{zc}z_c}k^2_{\rho c}dk_{\rho c}\\
\end{aligned}
\label{eqch2_55}
\end{equation}

如果我尝试使用“split”环境，它会列举我想要避免的每个方程式。提前致谢。

答案1

它太宽了，正如您&在最后两行中省略的那样。另外使用\Bigland 而\Bigr不是 `\Big。我在 [ 之前而不是 ] 之后断开，但它仍然太宽，所以我最终在两处都断开，并强制进行一些额外的缩进。

在此处输入图片描述

\documentclass[a4paper]{article}

\usepackage{amsmath}

\begin{document}


 \begin{equation}
\begin{aligned}
H_x(\rho_c,\phi_c,z_c)&=-\frac{1}{2\pi k\eta}\int_0^\infty F_y(k_{\rho c})\\
&\qquad  \Bigl[\Bigl(k^2-\frac{1}{2}k^2_{\rho c}\Bigr)J_0(k_{\rho c} \rho_c)+\frac{1}{2}k^2_{\rho c}\cos{(2\phi_c)}J_2(k_{\rho c}\rho_c)\Bigr]\\
&\qquad\qquad e^{-jk_{zc}z_c}\frac{k_{\rho c}}{k_{zc}}dk_{\rho c}\\[\jot]
H_x(\rho_c,\phi_c,z_c)&=-\frac{\sin{(\phi_c)}}{4\pi k\eta}\int_0^\infty F_y(k_{\rho c})J_2(k_{\rho c} \rho_c)e^{-jk_{zc}z_c}\frac{k^2_{\rho c}}{k_{zc}}dk_{\rho c}\\[\jot]
%
H_z(\rho_c,\phi_c,z_c)&=-\frac{j\cos{(2\phi_c)}}{2\pi k\eta}\int_0^\infty F_y(k_{\rho c})J_1(k_{\rho c} \rho_c)e^{-jk_{zc}z_c}k^2_{\rho c}dk_{\rho c}
\end{aligned}
\label{eqch2_55}
\end{equation} 
\end{document}

也许

在此处输入图片描述

\documentclass[a4paper]{article}

\usepackage{amsmath}

\begin{document}


 \begin{equation}
\begin{aligned}
H_x(\rho_c,\phi_c,z_c)&=-\frac{1}{2\pi k\eta}\int_0^\infty F_y(k_{\rho c})
   \Bigl[\Bigl(k^2-\frac{1}{2}k^2_{\rho c}\Bigr)J_0(k_{\rho c} \rho_c)+{}\\
& \qquad\qquad\qquad\frac{1}{2}k^2_{\rho c}\cos{(2\phi_c)}J_2(k_{\rho c}\rho_c)\Bigr]
 e^{-jk_{zc}z_c}\frac{k_{\rho c}}{k_{zc}}dk_{\rho c}\\[\jot]
H_x(\rho_c,\phi_c,z_c)&=-\frac{\sin{(\phi_c)}}{4\pi k\eta}\int_0^\infty F_y(k_{\rho c})J_2(k_{\rho c} \rho_c)e^{-jk_{zc}z_c}\frac{k^2_{\rho c}}{k_{zc}}dk_{\rho c}\\[\jot]
%
H_z(\rho_c,\phi_c,z_c)&=-\frac{j\cos{(2\phi_c)}}{2\pi k\eta}\int_0^\infty F_y(k_{\rho c})J_1(k_{\rho c} \rho_c)e^{-jk_{zc}z_c}k^2_{\rho c}dk_{\rho c}
\end{aligned}
\label{eqch2_55}
\end{equation} 
\end{document}

答案2

这是一个解决方案，它 (a) 填补了&第二和第三个方程中缺失的对齐粒子，并且 (b) 使用环境aligned来格式化跨越前两行的被积函数。此解决方案还将 的两个实例替换为\frac{1}{2}，这可以显著减少这两行中大括号的大小。最后，也可能是最不重要的，它在 的三个实例之前\tfrac{1}{2}插入(thinspace) 。\,dk_{\rho c}

在此处输入图片描述

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{equation}
\begin{split}
H_x(\rho_c,\phi_c,z_c)
  &=-\frac{1}{2\pi k\eta}\int_0^\infty \!
  \begin{aligned}[t]
  &F_y(k_{\rho c})\bigl[
    (k^2-\tfrac{1}{2}k^2_{\rho c})J_0(k_{\rho c} \rho_c)\\
  &+\tfrac{1}{2}k^2_{\rho c}\cos{(2\phi_c)}J_2(k_{\rho c}\rho_c)\bigr]
  e^{-jk_{zc}z_c}\frac{k_{\rho c}}{k_{zc}}\,dk_{\rho c} 
  \end{aligned}\\
%
H_x(\rho_c,\phi_c,z_c)
  &=-\frac{\sin{(\phi_c)}}{4\pi k\eta}\int_0^\infty \!
  F_y(k_{\rho c})J_2(k_{\rho c} \rho_c)
  e^{-jk_{zc}z_c}\frac{k^2_{\rho c}}{k_{zc}}\,dk_{\rho c}\\
%
H_z(\rho_c,\phi_c,z_c)
  &=-\frac{j\cos{(2\phi_c)}}{2\pi k\eta}\int_0^\infty \!
  F_y(k_{\rho c})J_1(k_{\rho c} \rho_c)
  e^{-jk_{zc}z_c}k^2_{\rho c}\,dk_{\rho c}
\end{split}
\label{eqch2_55}
\end{equation} 

\end{document}

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