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跨多行水平对齐方程

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跨多行水平对齐方程

我需要将以下内容对齐，以便第一行（整数）位于顶部左侧的一行上，然后以下各行缩进并相互对齐，但同时两组条件（以\left[\alpha>0,...和开头的行\left[\Real\mu...）将对齐以与页面右侧齐平。我该怎么做？

谢谢

\begin{document}

\begin{multline}
\int_0^\infty e^{-\alpha x}\, Y_\nu \left(\beta x\right)x^{\mu - 1}\,dx\\
= \cot v\pi\frac{\left(\frac{\beta}{2}\right)^{\nu}\Gamma\left(\nu+\mu\right)}{\sqrt{\left(\alpha^{2}+\beta^{2}\right)^{\nu+\mu}}\Gamma\left(\nu+1\right)}F\left(\frac{\nu+\mu}{2},\frac{\nu-\mu+1}{2};\nu+1;\frac{\beta^{2}}{\alpha^{2}+\beta^{2}}\right)\\
= \mathrm{cosec}\, \nu\pi\frac{\left(\frac{\beta}{2}\right)^{-\nu}\Gamma\left(\mu-v\right)}{\sqrt{\left(\alpha ^{2}+\beta^{2}\right)^{\mu-\nu}}\Gamma\left(1-\nu\right)}F\left(\frac{\mu-\nu}{2},\frac{1-\nu-\mu}{2};1-\nu;\frac{\beta^{2}}{\alpha^{2}+\beta^{2}}\right)\\
\left[\Real \mu \geq\lvert \Real \nu\rvert,\quad\Real\left(\alpha\pm \textit{i}\beta\right)>0\right]\\
=-\frac{2}{\pi}\Gamma\left(\nu+\mu\right)\left(\beta^{2}+\alpha^{2}\right)^{-\frac{1}{2}\mu} Q_{\mu-1}^{-\nu}\left[\alpha\left(\alpha^{2}+\beta^{2}\right)^ {-\frac{1}{2}}\right]\\
\left[\alpha>0,\quad\beta>0,\quad\Real\mu>\lvert\Real \nu\rvert\right]
\end{multline}

\end{document}

答案1

像这样？

在此处输入图片描述

我想这\Real意味着\Re......上述方程的 MWE 是

\documentclass{article}
\usepackage{mathtools}
\DeclareMathOperator{\cosec}{cosec}

\begin{document}

\begin{align}
    \MoveEqLeft
\int_0^\infty e^{-\alpha x}\, Y_\nu \left(\beta x\right)x^{\mu - 1}\,dx
                \notag  \\
    & = \cot v\pi\frac{\left(\frac{\beta}{2}\right)^{\nu}
        \Gamma\left(\nu+\mu\right)}{\sqrt{\left(\alpha^{2}+\beta^{2}\right)^{\nu+\mu}}
        \Gamma\left(\nu+1\right)}
    F\left(\frac{\nu+\mu}{2},\frac{\nu-\mu+1}{2};\nu+1;\frac{\beta^{2}}{\alpha^{2}+\beta^{2}}\right)
                \notag  \\
    & = \cosec\, \nu\pi\frac{\left(\frac{\beta}{2}\right)^{-\nu}
        \Gamma\left(\mu-v\right)}{\sqrt{\left(\alpha ^{2}+\beta^{2}\right)^{\mu-\nu}}\Gamma\left(1-\nu\right)}
        F\left(\frac{\mu-\nu}{2},\frac{1-\nu-\mu}{2};1-\nu;\frac{\beta^{2}}{\alpha^{2}+\beta^{2}}\right)
                \notag  \\
\intertext{\raggedleft    
    $\left[\Re \mu \geq\lvert \Re \nu\rvert,
     \quad\Re\left(\alpha\pm \textit{i}\beta\right)>0\right]
    $}
    & =-\frac{2}{\pi}\Gamma\left(\nu+\mu\right)\left(\beta^{2}+\alpha^{2}\right)^{-\frac{1}{2}\mu} 
        Q_{\mu-1}^{-\nu}\left[\alpha\left(\alpha^{2}+\beta^{2}\right)^ {-\frac{1}{2}}\right]
\intertext{\raggedleft
    $\left[\alpha>0,\quad\beta>0,\quad\Re \mu>\lvert\Re \nu\rvert\right]
    $}
                \notag  
\end{align}

\end{document}

答案2

equation您可能希望使用和环境的组合split。（我不会将两个条件行一直推到右边。）

在此处输入图片描述

\documentclass{article}
\usepackage{amsmath,amsfonts}
\DeclareMathOperator{\cosec}{cosec}
\newcommand\Real{\mathfrak{R}} % is this correct?!
\usepackage{mleftright}\mleftright
\begin{document}

\begin{equation}\begin{split}
&\int_0^\infty e^{-\alpha x}\, Y_\nu (x)x^{\mu - 1}\,dx\\
&= \cot \nu\pi\frac{\bigl(\frac{\beta}{2}\bigr)^{\nu}\,\Gamma(\nu+\mu)}{\sqrt{(\alpha^2+\beta^2{)}^{\nu+\mu}}\,\Gamma(\nu+1)}
F\left(\frac{\nu+\mu}{2},\frac{\nu-\mu+1}{2};\nu+1;\frac{\beta^2}{\alpha^2+\beta^2}\right)\\
&= \cosec \nu\pi\frac{\bigl(\frac{\beta}{2}\bigr)^{-\nu}\,\Gamma(\mu-v)}{\sqrt{(\alpha^2+\beta^2{)}^{\mu-\nu}}\,\Gamma(1-\nu)}
F\left(\frac{\mu-\nu}{2},\frac{1-\nu-\mu}{2};1-\nu;\frac{\beta^2}{\alpha^2+\beta^2}\right)\\[1ex]
&\qquad\qquad\bigl[\Real\mu\geq\lvert \Real \nu\rvert,\ 
 \Real(\alpha\pm i\beta)>0\bigr]\\[1ex]
&=-\frac{2}{\pi} \Gamma(\nu+\mu)(\beta^2+\alpha^2{)}^{-(1/2)\mu}\, Q_{\mu-1}^{-\nu}
\bigl[\alpha(\alpha^2+\beta^2{)}^{-1/2}\,\bigr]\\[1ex]
&\qquad\qquad\bigl[\alpha>0,\ \beta>0,\ \Real\mu>\lvert\Real \nu\rvert\bigr]
\end{split}\end{equation}

\end{document}

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