# 留数定理

## 定理

${\displaystyle \oint _{\gamma }f(z)\,dz=2\pi i\sum _{k=1}^{n}\operatorname {I} (\gamma ,a_{k})\operatorname {Res} (f,a_{k}).}$

${\displaystyle \oint _{\gamma }f(z)\,dz=2\pi i\sum _{k=1}^{n}\operatorname {Res} (f,a_{k}).}$

## 例子

### 实轴上的积分

${\displaystyle \int _{-\infty }^{\infty }{e^{itx} \over x^{2}+1}\,dx}$

${\displaystyle \int _{C}{f(z)}\,dz=\int _{C}{e^{itz} \over z^{2}+1}\,dz.}$

 ${\displaystyle {\frac {e^{itz}}{z^{2}+1}}\,\!}$ ${\displaystyle {}={\frac {e^{itz}}{2i}}\left({\frac {1}{z-i}}-{\frac {1}{z+i}}\right)\,\!}$ ${\displaystyle {}={\frac {e^{itz}}{2i}}{\frac {1}{z-i}}-{\frac {e^{itz}}{2i(z+i)}},\,\!}$

f(z)在z = i留数是：

${\displaystyle \operatorname {Res} _{z=i}f(z)={e^{-t} \over 2i}.}$

${\displaystyle \int _{C}f(z)\,dz=2\pi i\cdot \operatorname {Res} _{z=i}f(z)=2\pi i{e^{-t} \over 2i}=\pi e^{-t}.}$

${\displaystyle \int _{\mbox{straight}}+\int _{\mbox{arc}}=\pi e^{-t}\,}$

${\displaystyle \int _{-a}^{a}=\pi e^{-t}-\int _{\mbox{arc}}.}$

${\displaystyle \int _{\mbox{arc}}{e^{itz} \over z^{2}+1}\,dz\leq \int _{\mbox{arc}}\left|{e^{itz} \over z^{2}+1}\right|\,|dz|=\int _{\mbox{arc}}{|e^{itz}| \over |z^{2}+1|}\,|dz|=\int _{\mbox{arc}}{1 \over |z^{2}+1|}\,|dz|\leq \int _{\mbox{arc}}{1 \over a^{2}-1}\,|dz|={\frac {\pi a}{a^{2}-1}}\rightarrow 0\ {\mbox{as}}\ a\rightarrow \infty .}$

${\displaystyle \int _{-\infty }^{\infty }{e^{itz} \over z^{2}+1}\,dz=\pi e^{-t}.}$

${\displaystyle \int _{-\infty }^{\infty }{e^{itz} \over z^{2}+1}\,dz=\pi e^{t},}$

${\displaystyle \int _{-\infty }^{\infty }{e^{itz} \over z^{2}+1}\,dz=\pi e^{-\left|t\right|}.}$

（如果t = 0，这个积分就可以很快用初等方法算出来，它的值为π。）

### 无穷级数

${\displaystyle \sum _{n=-\infty }^{\infty }f(n)}$

${\displaystyle {\frac {1}{2\pi \mathrm {i} }}\int _{\Gamma _{N}}f(z)\pi \cot(\pi z)\mathrm {d} z={\frac {1}{2\pi \mathrm {i} }}\int _{\Gamma _{N}}{\frac {\pi \cot(\pi z)}{z^{2}}}\mathrm {d} z=\operatorname {Res} _{z=0}+{\underset {n\neq 0}{\sum _{n=-N}^{N}}}{\frac {1}{n^{2}}}}$

${\displaystyle N\to \infty }$时，等式左侧由于${\displaystyle O(n^{-2})}$而趋于零；另一方面：

${\displaystyle {\frac {z}{2}}\cot {\frac {z}{2}}=1-{\frac {B_{2}z^{2}}{2!}}+\cdots =1-{\frac {z^{2}}{12}}+\cdots }$

（实际上有${\displaystyle {\frac {z}{2}}\cot {\frac {z}{2}}={\frac {\mathrm {i} z}{1-\mathrm {e} ^{-\mathrm {i} z}}}-{\frac {\mathrm {i} z}{2}}}$）因此，${\displaystyle \operatorname {Res} _{z=0}=-{\frac {\pi ^{2}}{3}}}$，可以得出：

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}}$

## 参考文献

1. ^ 史济怀; 刘太顺. 复变函数. 合肥: 中国科学技术大学出版社. 1998-12-01. ISBN 9787312009990.