3

I want to know how to make a proper align where the equalites should be even vertically.

\begin{align*}
\Gamma(\lambda_{1})\Gamma(\lambda_{2})&=\int_{(0,\infty)^{2}}\phi(u,v)\, \mathrm{d}m_{2}(u,v)\\
&\overset{12.15}=\int_{(0,\infty)\times (0,1)}x^{(\lambda_{1}+\lambda_{2})-1}e^{-x}y^{\lambda_{1}-1}(1-y)^{\lambda_{2}-1}\mathrm{d}m_{2}(x,y)\\
&=\int_{0}^{\infty}\int_{0}^{1}x^{(\lambda_{1}+\lambda_{2})-1}e^{-x}y^{\lambda_{1}-1}(1-y)^{\lambda_{2}-1}\mathrm{d}y\mathrm{d}x
\end{align*}`

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4

I would merely go with a \mathclap of the over-set item:

enter image description here

\documentclass{article} 
\usepackage{mathtools}

\begin{document}

\begin{align*}
  \Gamma(\lambda_{1})\Gamma(\lambda_{2})&=\int_{(0,\infty)^{2}}\phi(u,v)\, \mathrm{d}m_{2}(u,v)\\
  &\overset{\mathclap{12.15}}{=} \int_{(0,\infty)\times (0,1)}x^{(\lambda_{1}+\lambda_{2})-1}e^{-x}y^{\lambda_{1}-1}(1-y)^{\lambda_{2}-1}\mathrm{d}m_{2}(x,y)\\
  &=\int_{0}^{\infty}\int_{0}^{1}x^{(\lambda_{1}+\lambda_{2})-1}e^{-x}y^{\lambda_{1}-1}(1-y)^{\lambda_{2}-1}\mathrm{d}y\mathrm{d}x
\end{align*}

\end{document}

\mathclap (provided by mathtools, which loads amsmath by default) eliminates the horizontal width of its argument.

  • #Werner. Thanks. I'm unable to download the mathtools package in MiKTeX (v. 2.9.4248) since it mysteriously doesn't exist (anymore). I have tried looking similar in other pages and I still didn't get what I needed. Is there another way to command in LaTeX? – UnknownW Mar 10 '15 at 4:56
  • @AjmalW: Instead of \mathclap{12.15}, you can use \makebox[0pt]{$\scriptstyle 12.15$}. – Werner Mar 10 '15 at 5:03
  • 1
    @AjmalW: Also see Problems installing mathtools. – Werner Mar 10 '15 at 5:04

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