命令输出超出页面范围

Question

\documentclass{article}

\usepackage{mathtools} % see http://ctan.org/pkg/mathtools
\DeclarePairedDelimiter\norm{\lVert}{\rVert}
\DeclareMathOperator{\tr}{tr}

\begin{document}
\begin{equation}
    \varphi(x)= \varphi(\beta)+ (x-\beta)\varphi^{'}(\alpha)+\int_{\alpha}^{\beta}G_{2}(x,s)\varphi^{''}(s)ds
\end{equation}
    Let us take the integral $\int_{\alpha}^{\beta}G_{2}(x,s)\varphi^{''}(s)ds$ on the right hand side. \\
    \\ Now applying the definition of Green's function, we get
    \begin{eqnarray*}
\int_{\alpha}^{\beta}G_{2}(x,s)\varphi^{''}(s)ds =& \int_{\alpha}^{x}G_{2}(x,s)\varphi^{''}(s)ds + \int_{x}^{\beta}G_{2}(x,s)\varphi^{''}(s)ds.\\
=& \int_{\alpha}^{x}(x-\beta)\varphi^{''}(s)ds + \int_{x}^{\beta}(s-\beta)\varphi^{''}(s)ds.\\
=& (x - \beta ) \int_{\alpha}^{x} \varphi^{''}(s)ds + (s - \beta )\int_{x }^{\beta}\varphi^{''}(s) ds - \left[\int_{x}^{\beta}\left(\frac{d}{ds}(s-\beta ) \int \varphi^{''}(s) ds \right) ds\right]  \\
=& (x -\beta ) \varphi^{'}(s) |_{\alpha}^{x} + (s-\beta)\varphi^{'}(s) |_{x}^{\beta} - \int_{x}^{\beta} 1 \cdot \varphi{'}(s)ds. \\     
=& (x -\beta ) \varphi^{'}(x) - (x -\beta ) \varphi^{'}(\alpha) + (\beta -\beta ) \varphi^{'}(\beta) - (x -\beta ) \varphi^{'}(x) -  \varphi(s) |_{x}^{\beta}\\
=&\varphi^{'}(x)\left[(x-\beta)-(x-\beta)\right] - (x-\beta)\varphi^{'}(\alpha) - \varphi(\beta)+\varphi(x)\\
&= \varphi(x)-\varphi(\beta)-(x-\beta)\varphi^{'}(\alpha)\\         
\Rightarrow \qquad \varphi(x) = \varphi(\beta) +(x- \beta ) \varphi^{'}(\alpha) +  \int_{\alpha }^{\beta} G_{2}(x,s)\varphi^{''}(s)ds.\\    
\text{Which is the required proof of Lemma.}
    \end{eqnarray*} 
\end{document} 

Answer 1

这个怎么样？

\documentclass{article}
\usepackage{fullpage}
\usepackage{mathtools} % see http://ctan.org/pkg/mathtools
\DeclarePairedDelimiter\norm{\lVert}{\rVert}
\DeclareMathOperator{\tr}{tr}

\begin{document}
\begin{equation}
    \varphi(x)= \varphi(\beta)+ (x-\beta)\varphi^{'}(\alpha)+\int_{\alpha}^{\beta}G_{2}(x,s)\varphi^{''}(s)ds
\end{equation}
Let us take the integral $\int_{\alpha}^{\beta}G_{2}(x,s)\varphi^{''}(s)ds$ on the right hand side. \\
    \\ Now applying the definition of Green's function, we get
    \begin{eqnarray*}
\int_{\alpha}^{\beta}G_{2}(x,s)\varphi^{''}(s)ds &=& \int_{\alpha}^{x}G_{2}(x,s)\varphi^{''}(s)ds + \int_{x}^{\beta}G_{2}(x,s)\varphi^{''}(s)ds.\\
&=& \int_{\alpha}^{x}(x-\beta)\varphi^{''}(s)ds + \int_{x}^{\beta}(s-\beta)\varphi^{''}(s)ds.\\
&=& (x - \beta ) \int_{\alpha}^{x} \varphi^{''}(s)ds + (s - \beta )\int_{x }^{\beta}\varphi^{''}(s) ds - \left[\int_{x}^{\beta}\left(\frac{d}{ds}(s-\beta ) \int \varphi^{''}(s) ds \right) ds\right]  \\
&=& (x -\beta ) \varphi^{'}(s) |_{\alpha}^{x} + (s-\beta)\varphi^{'}(s) |_{x}^{\beta} - \int_{x}^{\beta} 1 \cdot \varphi{'}(s)ds. \\     
&=& (x -\beta ) \varphi^{'}(x) - (x -\beta ) \varphi^{'}(\alpha) + (\beta -\beta ) \varphi^{'}(\beta) - (x -\beta ) \varphi^{'}(x) -  \varphi(s) |_{x}^{\beta}\\
&=& \varphi^{'}(x)\left[(x-\beta)-(x-\beta)\right] - (x-\beta)\varphi^{'}(\alpha) - \varphi(\beta)+\varphi(x)\\
&=& \varphi(x)-\varphi(\beta)-(x-\beta)\varphi^{'}(\alpha)\\         
\Rightarrow \qquad \varphi(x) &=& \varphi(\beta) +(x- \beta ) \varphi^{'}(\alpha) +  \int_{\alpha }^{\beta} G_{2}(x,s)\varphi^{''}(s)ds,
\end{eqnarray*}
which is the required proof of Lemma.
\end{document}