3

I need to write this equation in Latex: enter image description here

but it gets messy:

 \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} 

If I try to use "split" environment it enumerates every equation which I want to avoid. Thanks, in advance.

1
  • I don't follow your claim "If I try to use "split" environment it enumerates every equation which I want to avoid." As the solution in my answer shows, that's not the case unless one has somehow redefined the split environment.
    – Mico
    Jun 20 at 14:27
3

It was over-wide as you have omiteed & in the last two rows. Also use \Bigl and \Bigr not `\Big. I broke before the [ rather than after the ] but it was still too wide so I ended up breaking in both places, and forcing some extra indentation.

enter image description here

\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}

or perhaps

enter image description here

\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}
3

Here's a solution that (a) backfills the missing & alignment particles in the second and third equation and (b) employs an aligned environment to format the integrand that spans the first two rows. This solution also replaces both instances of \frac{1}{2} with \tfrac{1}{2}, which allows a significant reduction in size of the large parentheses in these two rows. Last and probably least, it inserts \, (thinspace) ahead of the three instances of dk_{\rho c}.

enter image description here

\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}
1
  • Thanks! This worked perfectly. Jun 21 at 12:33

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