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本帖最后由 ╰☆ヾo.海x 于 2014-11-13 07:08 编辑 回复 8# 爪机专用
其实这个积分只是做题时的一步中间过程，那个展开含sin的级数的方法是我同学的，那天只扫了一眼没有细看，他写了不到一半A4纸啊..我一时想不到真是蠢死...先贴出我的繁琐解法算了，大家有简单的方法再说吧...
原题是这样的：
通过考虑 $ \int_{\left| z \right|=1} \left( z-\dfrac{1}{z} \right)^{2n}\dfrac{\rmd{z}}{z} $ 证明 $ \int _{0}^{2\pi}\sin^{2n}t\rmd{t} =2\pi\dfrac{1\cdot 3\cdot 5\cdots \left( 2n-1 \right)}{2\cdot 4\cdot 6\cdots \left( 2n \right)}$
Proof：
Since $ \left| z \right| =1$, we write $ z=e^{it}=\cos t+i\sin t $, then $\rmd{z}=ie^{it}\rmd{t}$, $ \frac{\rmd{z}}{z}=i\rmd{t} $ ,
and we have
\[ \sin t=\frac{z-\frac{1}{z}}{2i} \]
\[ \left( z-\frac{1}{z} \right)^{2n} =\left( 2i \right)^{2n}\sin^{2n}t\]
\begin{align*}
\int_{\left| z \right|=1} \left( z-\dfrac{1}{z} \right)^{2n}\dfrac{\rmd{z}}{z}& =i\int _{0}^{2\pi}\left( e^{it}-e^{-it} \right)^{2n} \rmd{t}\\
&=i\int _{0}^{2\pi}e^{2nti}\left( 1-e^{-2ti} \right)^{2n}\rmd{t}\\
&=i\int _{0}^{2\pi}e^{2nti}\left( 1-\binom{2n}{1} e^{-2ti}+\binom{2n}{2}\left( e^{-2ti} \right)^{2}-\binom{2n}{3}\left( e^{-2ti} \right)^{3}+\cdots \right)\rmd{t}\\
&=i\int _{0}^{2\pi}e^{2nti}\rmd{t}-i\int_{0}^{2\pi}\binom{2n}{1} e^{(2n-2)ti}\rmd{t}+i\int_{0}^{2\pi}\binom{2n}{2}( e^{(2n-2\cdot 2)ti} \rmd{t}\\
&-i\int _{0}^{2\pi}\binom{2n}{3}e^{(2n-2\cdot 3)ti}\rmd{t} +\cdots +\left( -1 \right)^{n}i\int_{0}^{2\pi}\binom{2n}{n}e^{(2n-2n)ti}\rmd{t}+\cdots
\end{align*}
Since $ i\int _{0}^{2\pi}e^{2nki}\rmd{k}=0$, where $n,k\ne0, k\in\mbb{Z}$, the integrals above are all equal to 0 except for the $(n+1)$th term.
Thus,
\[ \int_{\left| z \right|=1} \left( z-\dfrac{1}{z} \right)^{2n}\dfrac{\rmd{z}}{z}=i\left(-1 \right)^{n}2\pi\binom{2n}{n}=\int _{0}^{2\pi}i\left(2i \right)^{2n}\sin^{2n}t\rmd{t}\]
\begin{align*}
\int _{0}^{2\pi}\sin^{2n}t\rmd{t}&=\dfrac{\left( -1 \right)^{n}2\pi\cdot 2n(2n-1)\cdots (n+1)}{\left( 2i \right)^{2n}n!}\\
&=\frac{2\pi\left( 2n \right)!}{2^{2n}\left( n! \right)^{2}}\\
&=2\pi\frac{\left( 2n \right)!}{\left(2^{n}\cdot n! \right)^2}\\
&=2\pi\frac{\left( 2n \right)!}{\left( 2\cdot 4\cdot 6\cdots \left( 2n \right) \right)^{2}}\\
&=2\pi\frac{1\cdot 3\cdot 5\cdots \left( 2n-1 \right)}{2\cdot 4\cdot 6\cdots \left( 2n \right)}
\end{align*} |
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kuing
发表于 2014-11-10 12:12
战巡
发表于 2014-11-13 09:30
代码写得还可以，点个赞！
