求助，所有等号全部对齐，代码怎么调？？？？

永远

求助，所有等号全部对齐，代码怎么调？？？？

1. \[\begin{gathered}
   
2.   \quad \quad \quad C = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\overline {{P_{k - 1}}{P_k}} }  \\
   
3.    = \mathop {\lim }\limits_{n \to \infty } \frac{\pi }{{2n}}\sum\limits_{k = 1}^n {\sqrt {{a^2}{{\sin }^2}\frac{{(2k - 1)\pi }}{{4n}} + {b^2}{{\cos }^2}\frac{{(2k - 1)\pi }}{{4n}}} }  \\
   
4.    = \int_0^{\frac{\pi }{2}} {\sqrt {{a^2}{{\sin }^2}\theta  + {b^2}{{\cos }^2}\theta } \;d\theta }  \\
   
5.    = \int_0^{\frac{\pi }{2}} {\sqrt {{a^2}\left( {1 - {{\cos }^2}\theta } \right) + {b^2}{{\cos }^2}\theta } } d\theta  \\
   
6.    = \int_0^{\frac{\pi }{2}} {\sqrt {{a^2} - {a^2}{{\cos }^2}\theta  + {b^2}{{\cos }^2}\theta } } d\theta  \\
   
7.    = \int_0^{\frac{\pi }{2}} {\sqrt {{a^2} - {a^2}\tfrac{{1 + \cos 2\theta }}{2} + {b^2}\tfrac{{1 + \cos 2\theta }}{2}} } d\theta  \\
   
8.    = \int_0^{\frac{\pi }{2}} {\sqrt {{a^2} - \tfrac{{{a^2}}}{2} - \tfrac{{{a^2}}}{2}\cos 2\theta  + \tfrac{{{b^2}}}{2} + \tfrac{{{b^2}}}{2}\cos 2\theta } } d\theta  \\
   
9.    = \int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{{a^2}}}{2} - \tfrac{{{a^2}}}{2}\cos 2\theta  + \tfrac{{{b^2}}}{2} + \tfrac{{{b^2}}}{2}\cos 2\theta } } d\theta  \\
   
10.    = \int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{{a^2} + {b^2}}}{2} - \tfrac{{{a^2} - {b^2}}}{2}\cos 2\theta } } d\theta  \\
    
11.    = \int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{2{a^2} + 2{b^2}}}{4} - \tfrac{{2{a^2} - 2{b^2}}}{4}\cos 2\theta } } d\theta  \\
    
12.    = \frac{1}{2}\int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2}}}{4} - \tfrac{{2\left( {a - b} \right)\left( {a + b} \right)}}{4}\cos 2\theta } } d\theta  \\
    
13.    = \frac{1}{2}\int_0^{\frac{\pi }{2}} {\sqrt {{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2} - 2\left( {a - b} \right)\left( {a + b} \right)\cos 2\theta } } d\theta  \\
    
14.    = \frac{1}{2}\int_0^{\frac{\pi }{2}} {\sqrt {{{\left( {a + b} \right)}^2}\left[ {1 + {{\left( {\tfrac{{a - b}}{{a + b}}} \right)}^2} - 2\tfrac{{a - b}}{{a + b}}\cos 2\theta } \right]} } d\theta  \\
    
15.    = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {{\left( {\tfrac{{a - b}}{{a + b}}} \right)}^2} - 2\tfrac{{a - b}}{{a + b}}\cos 2\theta } } d\theta  \\
    
16.    = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2} - 2\lambda \cos 2\theta } } d\theta  \\
    
17.    = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2} - \lambda \left( {{e^{2i\theta }} + {e^{ - 2i\theta }}} \right)} } d\theta  \\
    
18.    = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2} - \lambda {e^{2i\theta }} - \lambda {e^{ - 2i\theta }}} } d\theta  \\
    
19.    = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2}{e^{2i\theta  - 2i\theta }} - \lambda {e^{2i\theta }} - \lambda {e^{ - 2i\theta }}} } d\theta  \\
    
20.    = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + \lambda {e^{2i\theta }}\lambda {e^{ - 2i\theta }} - \lambda {e^{2i\theta }} - \lambda {e^{ - 2i\theta }}} } d\theta  \\
    
21.    = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right) + \left( {\lambda {e^{2i\theta }}\lambda {e^{ - 2i\theta }} - \lambda {e^{ - 2i\theta }}} \right)} } d\theta  \\
    
22.    = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right) + \lambda {e^{ - 2i\theta }}\left( {\lambda {e^{2i\theta }} - 1} \right)} } d\theta  \\
    
23.    = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right) - \lambda {e^{ - 2i\theta }}\left( {1 - \lambda {e^{2i\theta }}} \right)} } d\theta  \\
    
24.    = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right)\left( {1 - \lambda {e^{ - 2i\theta }}} \right)} } d\theta  \\
    
25.    = \frac{1}{2}\left( {a + b} \right){\int_0^{\frac{\pi }{2}} {\left( {1 - \lambda {e^{2i\theta }}} \right)} ^{\tfrac{1}{2}}}{\left( {1 - \lambda {e^{ - 2i\theta }}} \right)^{\tfrac{1}{2}}}d\theta  \\
    
26.    = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{k = 0}^{ + \infty } {C_{\tfrac{1}{2}}^{\;k}} } {\left( { - \lambda {e^{2i\theta }}} \right)^{\;k}}\sum\limits_{j = 0}^{ + \infty } {C_{\tfrac{1}{2}}^{\;j}} {\left( { - \lambda {e^{ - 2i\theta }}} \right)^{\;j}}d\theta  \\
    
27.    = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{k = 0}^\infty  {(_{\;k}^{1/2}){{( - \lambda )}^k}{e^{2k{\kern 1pt} \theta {\kern 1pt} i}}} \sum\limits_{j = 0}^\infty  {(_{\;j}^{1/2}){{( - \lambda )}^j}{e^{ - 2j{\kern 1pt} \theta {\kern 1pt} i}}} d\theta }  \\
    
28.    = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{k = 0}^\infty  {\sum\limits_{j = 0}^\infty  {(_{\;k}^{1/2})(_{\;j}^{1/2}){{( - \lambda )}^{k + j}}{e^{2k\theta i}}{e^{ - 2j\theta i}}} } d\theta }  \\
    
29.    = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{n = 0}^\infty  {\sum\limits_{k = 0}^n {(_{\;k}^{1/2})(_{\;n - k}^{1/2}){{( - \lambda )}^n}\cos [2(2k - n)\theta ]} } d\theta }  \\
    
30.    = \frac{1}{2}\left( {a + b} \right)\sum\limits_{n = 0}^\infty  {\sum\limits_{k = 0}^n {(_{\;k}^{1/2})(_{\;n - k}^{1/2}){{( - \lambda )}^n}} } \int_0^{\frac{\pi }{2}} {\cos [2(2k - n)\theta ]d\theta }  \\
    
31.    = \frac{1}{2}\left( {a + b} \right)\sum\limits_{n = 0}^\infty  {\sum\limits_{k = 0}^n {(_{\;k}^{1/2})(_{\;n - k}^{1/2}){{( - \lambda )}^n}} } \left\{ {\left. {\frac{1}{{2(2k - n)}}\sin [2(2k - n)\theta ]} \right|_0^{\frac{\pi }{2}}} \right\} \\
    
32. \end{gathered} \]

复制代码

目前效果图，很不美观

\[\begin{gathered}
  \quad \quad \quad C = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\overline {{P_{k - 1}}{P_k}} }  \\
   = \mathop {\lim }\limits_{n \to \infty } \frac{\pi }{{2n}}\sum\limits_{k = 1}^n {\sqrt {{a^2}{{\sin }^2}\frac{{(2k - 1)\pi }}{{4n}} + {b^2}{{\cos }^2}\frac{{(2k - 1)\pi }}{{4n}}} }  \\
   = \int_0^{\frac{\pi }{2}} {\sqrt {{a^2}{{\sin }^2}\theta  + {b^2}{{\cos }^2}\theta } \;d\theta }  \\
   = \int_0^{\frac{\pi }{2}} {\sqrt {{a^2}\left( {1 - {{\cos }^2}\theta } \right) + {b^2}{{\cos }^2}\theta } } d\theta  \\
   = \int_0^{\frac{\pi }{2}} {\sqrt {{a^2} - {a^2}{{\cos }^2}\theta  + {b^2}{{\cos }^2}\theta } } d\theta  \\
   = \int_0^{\frac{\pi }{2}} {\sqrt {{a^2} - {a^2}\tfrac{{1 + \cos 2\theta }}{2} + {b^2}\tfrac{{1 + \cos 2\theta }}{2}} } d\theta  \\
   = \int_0^{\frac{\pi }{2}} {\sqrt {{a^2} - \tfrac{{{a^2}}}{2} - \tfrac{{{a^2}}}{2}\cos 2\theta  + \tfrac{{{b^2}}}{2} + \tfrac{{{b^2}}}{2}\cos 2\theta } } d\theta  \\
   = \int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{{a^2}}}{2} - \tfrac{{{a^2}}}{2}\cos 2\theta  + \tfrac{{{b^2}}}{2} + \tfrac{{{b^2}}}{2}\cos 2\theta } } d\theta  \\
   = \int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{{a^2} + {b^2}}}{2} - \tfrac{{{a^2} - {b^2}}}{2}\cos 2\theta } } d\theta  \\
   = \int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{2{a^2} + 2{b^2}}}{4} - \tfrac{{2{a^2} - 2{b^2}}}{4}\cos 2\theta } } d\theta  \\
   = \frac{1}{2}\int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2}}}{4} - \tfrac{{2\left( {a - b} \right)\left( {a + b} \right)}}{4}\cos 2\theta } } d\theta  \\
   = \frac{1}{2}\int_0^{\frac{\pi }{2}} {\sqrt {{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2} - 2\left( {a - b} \right)\left( {a + b} \right)\cos 2\theta } } d\theta  \\
   = \frac{1}{2}\int_0^{\frac{\pi }{2}} {\sqrt {{{\left( {a + b} \right)}^2}\left[ {1 + {{\left( {\tfrac{{a - b}}{{a + b}}} \right)}^2} - 2\tfrac{{a - b}}{{a + b}}\cos 2\theta } \right]} } d\theta  \\
   = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {{\left( {\tfrac{{a - b}}{{a + b}}} \right)}^2} - 2\tfrac{{a - b}}{{a + b}}\cos 2\theta } } d\theta  \\
   = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2} - 2\lambda \cos 2\theta } } d\theta  \\
   = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2} - \lambda \left( {{e^{2i\theta }} + {e^{ - 2i\theta }}} \right)} } d\theta  \\
   = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2} - \lambda {e^{2i\theta }} - \lambda {e^{ - 2i\theta }}} } d\theta  \\
   = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2}{e^{2i\theta  - 2i\theta }} - \lambda {e^{2i\theta }} - \lambda {e^{ - 2i\theta }}} } d\theta  \\
   = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + \lambda {e^{2i\theta }}\lambda {e^{ - 2i\theta }} - \lambda {e^{2i\theta }} - \lambda {e^{ - 2i\theta }}} } d\theta  \\
   = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right) + \left( {\lambda {e^{2i\theta }}\lambda {e^{ - 2i\theta }} - \lambda {e^{ - 2i\theta }}} \right)} } d\theta  \\
   = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right) + \lambda {e^{ - 2i\theta }}\left( {\lambda {e^{2i\theta }} - 1} \right)} } d\theta  \\
   = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right) - \lambda {e^{ - 2i\theta }}\left( {1 - \lambda {e^{2i\theta }}} \right)} } d\theta  \\
   = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right)\left( {1 - \lambda {e^{ - 2i\theta }}} \right)} } d\theta  \\
   = \frac{1}{2}\left( {a + b} \right){\int_0^{\frac{\pi }{2}} {\left( {1 - \lambda {e^{2i\theta }}} \right)} ^{\tfrac{1}{2}}}{\left( {1 - \lambda {e^{ - 2i\theta }}} \right)^{\tfrac{1}{2}}}d\theta  \\
   = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{k = 0}^{ + \infty } {C_{\tfrac{1}{2}}^{\;k}} } {\left( { - \lambda {e^{2i\theta }}} \right)^{\;k}}\sum\limits_{j = 0}^{ + \infty } {C_{\tfrac{1}{2}}^{\;j}} {\left( { - \lambda {e^{ - 2i\theta }}} \right)^{\;j}}d\theta  \\
   = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{k = 0}^\infty  {(_{\;k}^{1/2}){{( - \lambda )}^k}{e^{2k{\kern 1pt} \theta {\kern 1pt} i}}} \sum\limits_{j = 0}^\infty  {(_{\;j}^{1/2}){{( - \lambda )}^j}{e^{ - 2j{\kern 1pt} \theta {\kern 1pt} i}}} d\theta }  \\
   = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{k = 0}^\infty  {\sum\limits_{j = 0}^\infty  {(_{\;k}^{1/2})(_{\;j}^{1/2}){{( - \lambda )}^{k + j}}{e^{2k\theta i}}{e^{ - 2j\theta i}}} } d\theta }  \\
   = \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{n = 0}^\infty  {\sum\limits_{k = 0}^n {(_{\;k}^{1/2})(_{\;n - k}^{1/2}){{( - \lambda )}^n}\cos [2(2k - n)\theta ]} } d\theta }  \\
   = \frac{1}{2}\left( {a + b} \right)\sum\limits_{n = 0}^\infty  {\sum\limits_{k = 0}^n {(_{\;k}^{1/2})(_{\;n - k}^{1/2}){{( - \lambda )}^n}} } \int_0^{\frac{\pi }{2}} {\cos [2(2k - n)\theta ]d\theta }  \\
   = \frac{1}{2}\left( {a + b} \right)\sum\limits_{n = 0}^\infty  {\sum\limits_{k = 0}^n {(_{\;k}^{1/2})(_{\;n - k}^{1/2}){{( - \lambda )}^n}} } \left\{ {\left. {\frac{1}{{2(2k - n)}}\sin [2(2k - n)\theta ]} \right|_0^{\frac{\pi }{2}}} \right\} \\
\end{gathered} \]
楼主想要达到的实际效果类似如下：

\(\small\begin{gathered}
  \sum\limits_{k = 1}^n {\overline {{P_{k - 1}}{P_k}} }  = \sum\limits_{k = 1}^n {\sqrt {{a^2}{{[\cos \frac{{k\pi }}{{2n}} - \cos \frac{{(k - 1)\pi }}{{2n}}]}^2} + {b^2}{{[\sin \frac{{k\pi }}{{2n}} - \sin \frac{{(k - 1)\pi }}{{2n}}]}^2}} }  \hfill \\
  \quad \quad \quad \quad \quad \;\; = \sum\limits_{k = 1}^n {\sqrt {{a^2}{{[ - 2\sin \frac{{(2k - 1)\pi }}{{4n}}\sin \frac{\pi }{{4n}}]}^2} + {b^2}{{[2\cos \frac{{(2k - 1)\pi }}{{4n}}\sin \frac{\pi }{{4n}}]}^2}} }  \hfill \\
  \quad \quad \quad \quad \quad \quad \quad \quad \quad  = \frac{\pi }{{2n}}\sum\limits_{k = 1}^n {\sqrt {{a^2}{{[\sin \frac{{(2k - 1)\pi }}{{4n}}\sin \frac{\pi }{{4n}}/\frac{\pi }{{4n}}]}^2} + {b^2}{{[\cos \frac{{(2k - 1)\pi }}{{4n}}\sin \frac{\pi }{{4n}}/\frac{\pi }{{4n}}]}^2}} }  \hfill \\ \end{gathered} \)

永远

论坛代码高手e老师可否 帮忙调试一下

春风晚霞

永远先生：您看可否如下调整？
\begin{align}
C &= \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\overline {{P_{k - 1}}{P_k}} }  \\
&= \mathop {\lim }\limits_{n \to \infty } \frac{\pi }{{2n}}\sum\limits_{k = 1}^n {\sqrt {{a^2}{{\sin }^2}\frac{{(2k - 1)\pi }}{{4n}} + {b^2}{{\cos }^2}\frac{{(2k - 1)\pi }}{{4n}}} }  \\
&= \int_0^{\frac{\pi }{2}} {\sqrt {{a^2}{{\sin }^2}\theta  + {b^2}{{\cos }^2}\theta } \;d\theta }  \\
&= \int_0^{\frac{\pi }{2}} {\sqrt {{a^2}\left( {1 - {{\cos }^2}\theta } \right) + {b^2}{{\cos }^2}\theta } } d\theta  \\
&= \int_0^{\frac{\pi }{2}} {\sqrt {{a^2} - {a^2}{{\cos }^2}\theta  + {b^2}{{\cos }^2}\theta } } d\theta  \\
&= \int_0^{\frac{\pi }{2}} {\sqrt {{a^2} - {a^2}\tfrac{{1 + \cos 2\theta }}{2} + {b^2}\tfrac{{1 + \cos 2\theta }}{2}} } d\theta  \\
&= \int_0^{\frac{\pi }{2}} {\sqrt {{a^2} - \tfrac{{{a^2}}}{2} - \tfrac{{{a^2}}}{2}\cos 2\theta  + \tfrac{{{b^2}}}{2} + \tfrac{{{b^2}}}{2}\cos 2\theta } } d\theta  \\
  &= \int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{{a^2}}}{2} - \tfrac{{{a^2}}}{2}\cos 2\theta  + \tfrac{{{b^2}}}{2} + \tfrac{{{b^2}}}{2}\cos 2\theta } } d\theta  \\
&= \int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{{a^2} + {b^2}}}{2} - \tfrac{{{a^2} - {b^2}}}{2}\cos 2\theta } } d\theta  \\
&= \int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{2{a^2} + 2{b^2}}}{4} - \tfrac{{2{a^2} - 2{b^2}}}{4}\cos 2\theta } } d\theta  \\
&= \frac{1}{2}\int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2}}}{4} - \tfrac{{2\left( {a - b} \right)\left( {a + b} \right)}}{4}\cos 2\theta } } d\theta  \\
&= \frac{1}{2}\int_0^{\frac{\pi }{2}} {\sqrt {{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2} - 2\left( {a - b} \right)\left( {a + b} \right)\cos 2\theta } } d\theta  \\
&= \frac{1}{2}\int_0^{\frac{\pi }{2}} {\sqrt {{{\left( {a + b} \right)}^2}\left[ {1 + {{\left( {\tfrac{{a - b}}{{a + b}}} \right)}^2} - 2\tfrac{{a - b}}{{a + b}}\cos 2\theta } \right]} } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {{\left( {\tfrac{{a - b}}{{a + b}}} \right)}^2} - 2\tfrac{{a - b}}{{a + b}}\cos 2\theta } } d\theta  \\
  &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2} - 2\lambda \cos 2\theta } } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2} - \lambda \left( {{e^{2i\theta }} + {e^{ - 2i\theta }}} \right)} } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2} - \lambda {e^{2i\theta }} - \lambda {e^{ - 2i\theta }}} } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2}{e^{2i\theta  - 2i\theta }} - \lambda {e^{2i\theta }} - \lambda {e^{ - 2i\theta }}} } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + \lambda {e^{2i\theta }}\lambda {e^{ - 2i\theta }} - \lambda {e^{2i\theta }} - \lambda {e^{ - 2i\theta }}} } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right) + \left( {\lambda {e^{2i\theta }}\lambda {e^{ - 2i\theta }} - \lambda {e^{ - 2i\theta }}} \right)} } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right) + \lambda {e^{ - 2i\theta }}\left( {\lambda {e^{2i\theta }} - 1} \right)} } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right) - \lambda {e^{ - 2i\theta }}\left( {1 - \lambda {e^{2i\theta }}} \right)} } d\theta  \\
  &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right)\left( {1 - \lambda {e^{ - 2i\theta }}} \right)} } d\theta  \\
&= \frac{1}{2}\left( {a + b} \right){\int_0^{\frac{\pi }{2}} {\left( {1 - \lambda {e^{2i\theta }}} \right)} ^{\tfrac{1}{2}}}{\left( {1 - \lambda {e^{ - 2i\theta }}} \right)^{\tfrac{1}{2}}}d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{k = 0}^{ + \infty } {C_{\tfrac{1}{2}}^{\;k}} } {\left( { - \lambda {e^{2i\theta }}} \right)^{\;k}}\sum\limits_{j = 0}^{ + \infty } {C_{\tfrac{1}{2}}^{\;j}} {\left( { - \lambda {e^{ - 2i\theta }}} \right)^{\;j}}d\theta  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{k = 0}^\infty  {(_{\;k}^{1/2}){{( - \lambda )}^k}{e^{2k{\kern 1pt} \theta {\kern 1pt} i}}} \sum\limits_{j = 0}^\infty  {(_{\;j}^{1/2}){{( - \lambda )}^j}{e^{ - 2j{\kern 1pt} \theta {\kern 1pt} i}}} d\theta }  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{k = 0}^\infty  {\sum\limits_{j = 0}^\infty  {(_{\;k}^{1/2})(_{\;j}^{1/2}){{( - \lambda )}^{k + j}}{e^{2k\theta i}}{e^{ - 2j\theta i}}} } d\theta }  \\
&= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{n = 0}^\infty  {\sum\limits_{k = 0}^n {(_{\;k}^{1/2})(_{\;n - k}^{1/2}){{( - \lambda )}^n}\cos [2(2k - n)\theta ]} } d\theta }  \\
&= \frac{1}{2}\left( {a + b} \right)\sum\limits_{n = 0}^\infty  {\sum\limits_{k = 0}^n {(_{\;k}^{1/2})(_{\;n - k}^{1/2}){{( - \lambda )}^n}} } \int_0^{\frac{\pi }{2}} {\cos [2(2k - n)\theta ]d\theta }  \\
\end{align}

代码附后：

1. \begin{align}
   
2. C &= \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\overline {{P_{k - 1}}{P_k}} }  \\
   
3. &= \mathop {\lim }\limits_{n \to \infty } \frac{\pi }{{2n}}\sum\limits_{k = 1}^n {\sqrt {{a^2}{{\sin }^2}\frac{{(2k - 1)\pi }}{{4n}} + {b^2}{{\cos }^2}\frac{{(2k - 1)\pi }}{{4n}}} }  \\
   
4. &= \int_0^{\frac{\pi }{2}} {\sqrt {{a^2}{{\sin }^2}\theta  + {b^2}{{\cos }^2}\theta } \;d\theta }  \\
   
5. &= \int_0^{\frac{\pi }{2}} {\sqrt {{a^2}\left( {1 - {{\cos }^2}\theta } \right) + {b^2}{{\cos }^2}\theta } } d\theta  \\
   
6. &= \int_0^{\frac{\pi }{2}} {\sqrt {{a^2} - {a^2}{{\cos }^2}\theta  + {b^2}{{\cos }^2}\theta } } d\theta  \\
   
7. &= \int_0^{\frac{\pi }{2}} {\sqrt {{a^2} - {a^2}\tfrac{{1 + \cos 2\theta }}{2} + {b^2}\tfrac{{1 + \cos 2\theta }}{2}} } d\theta  \\
   
8. &= \int_0^{\frac{\pi }{2}} {\sqrt {{a^2} - \tfrac{{{a^2}}}{2} - \tfrac{{{a^2}}}{2}\cos 2\theta  + \tfrac{{{b^2}}}{2} + \tfrac{{{b^2}}}{2}\cos 2\theta } } d\theta  \\
   
9.   &= \int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{{a^2}}}{2} - \tfrac{{{a^2}}}{2}\cos 2\theta  + \tfrac{{{b^2}}}{2} + \tfrac{{{b^2}}}{2}\cos 2\theta } } d\theta  \\
   
10. &= \int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{{a^2} + {b^2}}}{2} - \tfrac{{{a^2} - {b^2}}}{2}\cos 2\theta } } d\theta  \\
    
11. &= \int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{2{a^2} + 2{b^2}}}{4} - \tfrac{{2{a^2} - 2{b^2}}}{4}\cos 2\theta } } d\theta  \\
    
12. &= \frac{1}{2}\int_0^{\frac{\pi }{2}} {\sqrt {\tfrac{{{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2}}}{4} - \tfrac{{2\left( {a - b} \right)\left( {a + b} \right)}}{4}\cos 2\theta } } d\theta  \\
    
13. &= \frac{1}{2}\int_0^{\frac{\pi }{2}} {\sqrt {{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2} - 2\left( {a - b} \right)\left( {a + b} \right)\cos 2\theta } } d\theta  \\
    
14. &= \frac{1}{2}\int_0^{\frac{\pi }{2}} {\sqrt {{{\left( {a + b} \right)}^2}\left[ {1 + {{\left( {\tfrac{{a - b}}{{a + b}}} \right)}^2} - 2\tfrac{{a - b}}{{a + b}}\cos 2\theta } \right]} } d\theta  \\
    
15. &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {{\left( {\tfrac{{a - b}}{{a + b}}} \right)}^2} - 2\tfrac{{a - b}}{{a + b}}\cos 2\theta } } d\theta  \\
    
16.   &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2} - 2\lambda \cos 2\theta } } d\theta  \\
    
17. &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2} - \lambda \left( {{e^{2i\theta }} + {e^{ - 2i\theta }}} \right)} } d\theta  \\
    
18. &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2} - \lambda {e^{2i\theta }} - \lambda {e^{ - 2i\theta }}} } d\theta  \\
    
19. &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + {\lambda ^2}{e^{2i\theta  - 2i\theta }} - \lambda {e^{2i\theta }} - \lambda {e^{ - 2i\theta }}} } d\theta  \\
    
20. &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {1 + \lambda {e^{2i\theta }}\lambda {e^{ - 2i\theta }} - \lambda {e^{2i\theta }} - \lambda {e^{ - 2i\theta }}} } d\theta  \\
    
21. &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right) + \left( {\lambda {e^{2i\theta }}\lambda {e^{ - 2i\theta }} - \lambda {e^{ - 2i\theta }}} \right)} } d\theta  \\
    
22. &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right) + \lambda {e^{ - 2i\theta }}\left( {\lambda {e^{2i\theta }} - 1} \right)} } d\theta  \\
    
23. &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right) - \lambda {e^{ - 2i\theta }}\left( {1 - \lambda {e^{2i\theta }}} \right)} } d\theta  \\
    
24.   &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sqrt {\left( {1 - \lambda {e^{2i\theta }}} \right)\left( {1 - \lambda {e^{ - 2i\theta }}} \right)} } d\theta  \\
    
25. &= \frac{1}{2}\left( {a + b} \right){\int_0^{\frac{\pi }{2}} {\left( {1 - \lambda {e^{2i\theta }}} \right)} ^{\tfrac{1}{2}}}{\left( {1 - \lambda {e^{ - 2i\theta }}} \right)^{\tfrac{1}{2}}}d\theta  \\
    
26. &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{k = 0}^{ + \infty } {C_{\tfrac{1}{2}}^{\;k}} } {\left( { - \lambda {e^{2i\theta }}} \right)^{\;k}}\sum\limits_{j = 0}^{ + \infty } {C_{\tfrac{1}{2}}^{\;j}} {\left( { - \lambda {e^{ - 2i\theta }}} \right)^{\;j}}d\theta  \\
    
27. &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{k = 0}^\infty  {(_{\;k}^{1/2}){{( - \lambda )}^k}{e^{2k{\kern 1pt} \theta {\kern 1pt} i}}} \sum\limits_{j = 0}^\infty  {(_{\;j}^{1/2}){{( - \lambda )}^j}{e^{ - 2j{\kern 1pt} \theta {\kern 1pt} i}}} d\theta }  \\
    
28. &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{k = 0}^\infty  {\sum\limits_{j = 0}^\infty  {(_{\;k}^{1/2})(_{\;j}^{1/2}){{( - \lambda )}^{k + j}}{e^{2k\theta i}}{e^{ - 2j\theta i}}} } d\theta }  \\
    
29. &= \frac{1}{2}\left( {a + b} \right)\int_0^{\frac{\pi }{2}} {\sum\limits_{n = 0}^\infty  {\sum\limits_{k = 0}^n {(_{\;k}^{1/2})(_{\;n - k}^{1/2}){{( - \lambda )}^n}\cos [2(2k - n)\theta ]} } d\theta }  \\
    
30. &= \frac{1}{2}\left( {a + b} \right)\sum\limits_{n = 0}^\infty  {\sum\limits_{k = 0}^n {(_{\;k}^{1/2})(_{\;n - k}^{1/2}){{( - \lambda )}^n}} } \int_0^{\frac{\pi }{2}} {\cos [2(2k - n)\theta ]d\theta }  \\
    
31. \end{align}

复制代码

永远

春风晚霞 发表于 2021-4-4 23:18
永远先生：您看可否如下调整？
\begin{align}
C &= \mathop {\lim }\limits_{n \to \infty } \sum\limit ...

把编辑好的代码输入到，插入代码对话框，点击确定即可，这个插入代码对话框看图片上笑脸表情图旁边“<>”点击一下即可打开

永远

春风晚霞 发表于 2021-4-4 23:18
永远先生：您看可否如下调整？
\begin{align}
C &= \mathop {\lim }\limits_{n \to \infty } \sum\limit ...

惊呆了！前辈说说看您是怎样把全部等号对的这么齐的，怎么做到的？？？？？？？

这么整齐，有什么诀窍吗，还是用啥软件里做的？？？？

春风晚霞

永远 发表于 2021-4-5 12:22
惊呆了！前辈说说看您是怎样把全部等号对的这么齐的，怎么做到的？？？？？？？

这么整齐，有什么诀 ...

其实也没有什么诀窍，主是论坛所附Latex适合align环境。关于诸多等号对齐（参见《教程》六等号对齐，教程六介绍的split环境我试了一下，设达到预期效果（也许是我不够专心吧）才改用的align环境，该环境的语法格式附后。不知先生手边有没有在长双向箭头上下添加注释的程序，如有能否支援我一个（我也就不再去思考或找寻它了）。这个东西在用元素考察法证两个集合相等时，可少写一半的代码。无论您有还是没有都提前謝谢了。
align语法附后

1. \begin{align}……\end{align｝语法格式为：
   
2. \begin{align}
   
3. C&=第一行右端\\
   
4. &=第二行右端\\
   
5. &=第三行右端\\
   
6. ……
   
7. &=第n行右端\\
   
8. ………
   
9. \end{align}
   
10. 
    
11. (注:其中C为第一行左端)。

复制代码

永远

春风晚霞 发表于 2021-4-5 15:04
其实也没有什么诀窍，主是论坛所附Latex适合align环境。关于诸多等号对齐（参见《教程》六等号对齐，教 ...

现在在工厂车间里干活，回复可能有点慢，巧了貌似mathype软件里有， 晚上下班回去看看，如果合适发个链接共享一下，这个简单

永远

春风晚霞 发表于 2021-4-5 15:04
其实也没有什么诀窍，主是论坛所附Latex适合align环境。关于诸多等号对齐（参见《教程》六等号对齐，教 ...

所谓的长双向箭头，具体指哪一种，有好多种！！！！！！！！

永远

对于不太熟悉LaTeX文本的坛友，在数学公式编辑器MathType编辑好公式，复制粘贴到论坛的MathType预览框，加括号反斜杠等，稍微编辑一下就好了。或者在Mathematica其他数学软件上，复制为LaTeX文本，再到预览框中编辑，省去记那些代码的麻烦，基本语法还是懂一些，

春风晚霞

永远 发表于 2021-4-5 21:28
对于不太熟悉LaTeX文本的坛友，在数学公式编辑器MathType编辑好公式，复制粘贴到论坛的MathType预览框，加 ...

谢谢。我已找到能任意伸缩且能在其上下打字的Latex箭头模块,虽然如此，还是谢谢！