\documentclass{article}
\usepackage{...} %packages your code nees
\begin{document}
\begin{align}
T\triangleq &\int_{0}^{\infty}\frac{\gamma^{k+m_2-1}{\rm exp}\Big(-\frac{m_2\gamma}{\bar\gamma_2}\Big)}{\Big1+(\frac{(m_1+s\bar\gamma_1)\gamma}{cm_1}\Big)^{m_1}(1+\frac{1}{c}\gamma_2)^{\lceil m_1 \rceil -m_1}}{\rm d}\gamma.\\
\nonumber
=&C\int_{0}^{\infty}\gamma^{k+m_2-1}{\rm exp}\Bigg(-\frac{m_2\gamma}{\bar\gamma_2}\Bigg)\\
&\texttimes H_{1,1}^{1,1} \left[\begin{gathered} \frac{(m_1+s\bar\gamma_1)\gamma}{cm_1}\end{gathered}\left\vert \begin{gathered} (1-m_1,1) \\ (0,1)\end{gathered}\right.\right]H_{1,1}^{1,1} \left[\begin{gathered} \frac{\gamma}{c}\end{gathered}\left\vert \begin{gathered} (1-\lceil m_1 \rceil +m_1,1) \\ (0,1)\end{gathered}\right.\right]{\rm d}\gamma\\
=&C\Bigg(\frac{\bar\gamma_2}{m_2}\Bigg)^{k+m_2}H_{1,\,[1:\,1],\,0,\,[1:\,1]}^{1,\,1,\,1,\,1,\,1} \left[\begin{gathered}
\frac{(m_1+s\bar{\gamma}_{_1})\bar{\gamma}_{_2}}{c\,m_1m_2} \\ \frac{\bar{\gamma}_{_2}}{c\,m_2}\end{gathered}
\left\vert \begin{gathered} (k+m_2,1) \\ (1-m_1,1);(1-\lceil{m_1}\rceil+m_1,) \\ - \\ (0,1);(0,1)
\end{gathered}\right.\right].\\
=&C\Bigg(\frac{\bar\gamma_2}{m_2}\Bigg)^{k+m_2}G_{1,\,[1:\,1],\,0,\,[1:\,1]}^{1,\,1,\,1,\,1,\,1}\left[\begin{gathered}\frac{(m_1+s\bar{\gamma}_{_1})\bar{\gamma}_{_2}}{c\,m_1m_2}\\\frac{\bar{\gamma}_{_2}}{c\,m_2}\end{gathered}\left\vert \begin{gathered} k+m_2\\1-m_1;1-\lceil{m_1}\rceil+m_1 \\ - \\ 0;0\end{gathered}\right.\right].
\end{align}
\end{document}
我已经成功生成了图中的方程，但是编译 LaTeX 之后总是显示Missing delimiter (.inserted). \end{align}What's the solution?
答案1
您的输入
\Big1+(
应该
1+\Bigl(
其它\Big标记也应更改：\Bigl在开始分隔符前使用 和\Bigr在结束分隔符前使用。对于\Bigg也应如此，应为\Biggl或\Biggr。
使用\exp、 而不是{\rm exp}和\,\mathrm{d}代替{\rm d}。该\rm命令已被弃用 20 年了。
如果希望对齐点位于=（或\triangleeq）之后，则使用={}&或 ，间距会不正确。
命令\texttimes错误，应该是\times。
而不是笨拙的
\left[ <...> \left| <...> \right.\right]
构造，您可以使用
\left[ <...> \;\middle|\; <...> \right]
最后要注意的一点是：你有几个\gamma_{_1}或类似的，这是错误的：如果你想降低下标，请输入
\gamma^{}_{1}
这是修复版本
\documentclass{article}
\usepackage{amsmath,amssymb} %packages your code nees
\begin{document}
\begin{align}
T\triangleq{} &
\int_{0}^{\infty}
\frac
{\gamma^{k+m_2-1}\exp\Bigl(-\frac{m_2\gamma}{\bar\gamma_2}\Bigr)}
{1+\Bigl(\frac{(m_1+s\bar\gamma_1)\gamma}{cm_1}\Bigr)^{m_1}(1+\frac{1}{c}\gamma_2)
^{\lceil m_1 \rceil -m_1}}
\,\mathrm{d}\gamma
\\
\nonumber
={}& C\int_{0}^{\infty}\gamma^{k+m_2-1}\exp\Biggl(-\frac{m_2\gamma}{\bar\gamma_2}\Biggr)
\\
&\times H_{1,1}^{1,1}
\left[
\frac{(m_1+s\bar\gamma_1)\gamma}{cm_1}
\;\middle|\;
\begin{gathered} (1-m_1,1) \\ (0,1)\end{gathered}
\right]
H_{1,1}^{1,1}
\left[
\frac{\gamma}{c}
\;\middle|\;
t \begin{gathered} (1-\lceil m_1 \rceil +m_1,1) \\ (0,1)\end{gathered}
\right]
\mathrm{d}\gamma
\\
={}& C\Biggl(\frac{\bar\gamma_2}{m_2}\Biggr)^{k+m_2}
H_{1,\,[1:\,1],\,0,\,[1:\,1]}^{1,\,1,\,1,\,1,\,1}
\left[
\begin{gathered}
\frac{(m_1+s\bar{\gamma}^{}_{1})\bar{\gamma}^{}_{2}}{c\,m_1m_2} \\
\frac{\bar{\gamma}^{}_{2}}{c\,m_2}
\end{gathered}
\;\middle|\;
\begin{gathered}
(k+m_2,1) \\
(1-m_1,1);(1-\lceil{m_1}\rceil+m_1,) \\
- \\
(0,1);(0,1)
\end{gathered}
\right]
\\
={}& C\Biggl(\frac{\bar\gamma_2}{m_2}\Biggr)^{k+m_2}
G_{1,\,[1:\,1],\,0,\,[1:\,1]}^{1,\,1,\,1,\,1,\,1}
\left[
\begin{gathered}
\frac{(m_1+s\bar{\gamma}^{}_{1})\bar{\gamma}^{}_{2}}{c\,m_1m_2}\\
\frac{\bar{\gamma}_{_2}}{c\,m_2}
\end{gathered}
\;\middle|\;
\begin{gathered}
k+m_2\\1-m_1;1-\lceil{m_1}\rceil+m_1 \\
- \\
0;0\end{gathered}
\right].
\end{align}
\end{document}
