3. 函数在 \infty<script type="math/tex">\infty</script> 的留数

(1) 定义: 设 \infty<script type="math/tex">\infty</script> 为 f<script type="math/tex">f</script> 的孤立奇点, 则称 \bex \cfrac{1}{2\pi i}\int_{\vGa^-}f(z)\rd z\quad (\vGa:\ |z|=\rho) \eex<script type="math/tex; mode=display">\bex \cfrac{1}{2\pi i}\int_{\vGa^-}f(z)\rd z\quad (\vGa:\ |z|=\rho) \eex</script> 为 f<script type="math/tex">f</script> 在 \infty<script type="math/tex">\infty</script> 的留数, 记作 \dps{\underset{z=\infty}{\Res}f(z)}<script type="math/tex">\dps{\underset{z=\infty}{\Res}f(z)}</script>.

(2) 若 f<script type="math/tex">f</script> 在 r<|z|<\infty<script type="math/tex">r<|z|<\infty</script> 内有 Laurent 展式 \bex f(z)=\cdots+\cfrac{c_{-n}}{z^n}+\cdots+\cfrac{c_{-1}}{z}+c_0+c_1z+\cdots +c_nz^n+\cdots, \eex<script type="math/tex; mode=display">\bex f(z)=\cdots+\cfrac{c_{-n}}{z^n}+\cdots+\cfrac{c_{-1}}{z}+c_0+c_1z+\cdots +c_nz^n+\cdots, \eex</script> 则 \dps{\underset{z=\infty}{\Res}f(z)=-c_{-1}}<script type="math/tex">\dps{\underset{z=\infty}{\Res}f(z)=-c_{-1}}</script>.

(3) 计算

a. 用 Cauchy 留数定理: \bex \underset{z=\infty}{\Res}f(z) =-\sum_{k=1}^n \underset{z=a_k}{\Res}f(z). \eex<script type="math/tex; mode=display">\bex \underset{z=\infty}{\Res}f(z) =-\sum_{k=1}^n \underset{z=a_k}{\Res}f(z). \eex</script>

b. 转换为在零点的留数: \beex \bea \underset{z=\infty}{\Res}f(z) &=\cfrac{1}{2\pi i}\int_{\vGa^-}f(z)\rd z\\ &=\cfrac{1}{2\pi i}\int_{\sev{\zeta}=\cfrac{1}{\rho}} f\sex{\cfrac{1}{\zeta}}\sex{-\cfrac{1}{\zeta^2}}\rd \zeta\quad \sex{\zeta=\cfrac{1}{z}: z=\rho e^{i\tt}\ra \zeta=\cfrac{1}{\rho} e^{-i\tt}}\\ &=\cfrac{1}{2\pi i}\int_{|\zeta|=\cfrac{1}{\rho}} f\sex{\cfrac{1}{\zeta}}\sex{-\cfrac{1}{\zeta^2}}\rd \zeta\\ &=\underset{\zeta=0}{\Res}\sez{f\sex{\cfrac{1}{\zeta}}\sex{-\cfrac{1}{\zeta^2}}}. \eea \eeex<script type="math/tex; mode=display">\beex \bea \underset{z=\infty}{\Res}f(z) &=\cfrac{1}{2\pi i}\int_{\vGa^-}f(z)\rd z\\ &=\cfrac{1}{2\pi i}\int_{\sev{\zeta}=\cfrac{1}{\rho}} f\sex{\cfrac{1}{\zeta}}\sex{-\cfrac{1}{\zeta^2}}\rd \zeta\quad \sex{\zeta=\cfrac{1}{z}: z=\rho e^{i\tt}\ra \zeta=\cfrac{1}{\rho} e^{-i\tt}}\\ &=\cfrac{1}{2\pi i}\int_{|\zeta|=\cfrac{1}{\rho}} f\sex{\cfrac{1}{\zeta}}\sex{-\cfrac{1}{\zeta^2}}\rd \zeta\\ &=\underset{\zeta=0}{\Res}\sez{f\sex{\cfrac{1}{\zeta}}\sex{-\cfrac{1}{\zeta^2}}}. \eea \eeex</script>

(4) 例: 求 \dps{I=\int_{|z|=4}\cfrac{z^{15}}{(z^2+1)^2(z^4+2)^3}\rd z}<script type="math/tex">\dps{I=\int_{|z|=4}\cfrac{z^{15}}{(z^2+1)^2(z^4+2)^3}\rd z}</script>.

6. 2 用留数计算实积分

1. \dps{I=\int_0^{2\pi}R(\cos \tt,\sin\tt)\rd \tt}<script type="math/tex">\dps{I=\int_0^{2\pi}R(\cos \tt,\sin\tt)\rd \tt}</script> 型 (R<script type="math/tex">R</script>: 有理函数).

(1) 数分: 用万能代换 \tan \cfrac{\tt}{2}=x\ra \sin\tt=\cfrac{2x}{1+x^2}, \cos\tt=\cfrac{1-x^2}{1+x^2},\ \rd \tt=\cdots<script type="math/tex">\tan \cfrac{\tt}{2}=x\ra \sin\tt=\cfrac{2x}{1+x^2}, \cos\tt=\cfrac{1-x^2}{1+x^2},\ \rd \tt=\cdots</script>.

(2) 复变: z=e^{i\tt}\ra \cos\tt=\cfrac{z+z^{-1}}{2},\ \sin\tt=\cfrac{z-z^{-1}}{2i},\ \rd \tt=\cfrac{\rd z}{iz}<script type="math/tex">z=e^{i\tt}\ra \cos\tt=\cfrac{z+z^{-1}}{2},\ \sin\tt=\cfrac{z-z^{-1}}{2i},\ \rd \tt=\cfrac{\rd z}{iz}</script>, 而 \bex I=\int_{|z|=1}R\sex{\cfrac{z+z^{-1}}{2},\cfrac{z-z^{-1}}{2i}}\cfrac{\rd z}{iz}. \eex<script type="math/tex; mode=display">\bex I=\int_{|z|=1}R\sex{\cfrac{z+z^{-1}}{2},\cfrac{z-z^{-1}}{2i}}\cfrac{\rd z}{iz}. \eex</script>

(3) 例: 求 \dps{I=\int_0^{2\pi}\cfrac{\rd \tt}{1-2p\cos\tt+p^2}}<script type="math/tex">\dps{I=\int_0^{2\pi}\cfrac{\rd \tt}{1-2p\cos\tt+p^2}}</script>, (p\neq 0<script type="math/tex">p\neq 0</script>, p\neq \pm 1<script type="math/tex">p\neq \pm 1</script>).

作业: P 262-263 T 1 (1) (3) , T 4 (1) .