复变函数积分

\[1+2 z+3 z^{2}+\cdots+n z^{n-1}=0 \]

在\(B\left(0,r\right)\)

\[\int_{-\infty}^{+\infty} \frac{P(x)}{Q(x)} \mathrm{d} x=2 \pi \mathrm{i} \sum_{j=1}^{n} \operatorname{Res}\left(\frac{P(z)}{Q(z)}, z_{j}\right) \]

\[\int_{0}^{\infty} \frac{1+x^{2}}{1+x^{4}} d x \]

\[\int_{0}^{\frac{2}{\pi}} \frac{1}{a+\sin^{2}\theta} d \theta \]

\[\int_{|z |=R} z^{n} \log \frac{z-a}{z-b} d z \quad(a \neq b, \quad R>\max (|a|,|b|\}) \]

The same statement holds for any \(a, b \in \mathbb{C}\). (Note that since \(f\) is an entire function, \(\int_{a}^{b} f(z) d z\) is independent of path, and hence well defined.
Proof:
Fix an arbitrary positive number \(R>\max \{|a|,|b|\} .\) It is easy to check that when \(|w|,\left|w_{0}\right|<R\) (each branch of ) \(\log \frac{z-w}{z-w_{0}}\) is a well defined holomorphic function of \(z\) on \(\{|z| \geq R\},\) so
\(g_{w_{0}}(w):=\frac{1}{2 \pi i} \int_{|z|=R} f(z) \log \frac{z-w}{z-w_{0}} d z\)
defines a holomorphic function of \(w\) on \(\{|w|<R\} .\) Moreover, \(g_{w_{0}}\left(w_{0}\right)=0\) and

\[\begin{array}{c} F(z)=\int_{\alpha}^{\beta} \varphi(z, t) \mathrm{d} t \\ F^{\prime}(z)=\int_{\alpha}^{\beta} \frac{\partial \varphi(z, t)}{\partial z} \mathrm{d} t \end{array}\]

\(g_{w_{0}}^{\prime}(w)=-\frac{1}{2 \pi i} \int_{|z|=R} \frac{f(z)}{z-w} d z=-f(w)\)
That is to say,
\(g_{w_{0}}(w)=-\int_{w_{0}}^{w} f(z) d z\)
Letting \(w_{0}=a, w=b,\) the conclusion follows.

isolated singularity:$$\frac{z^{3}+z+2}{z\left(z^{2}-1\right){2}}$$

(z^{3}+z^{2}+2)\(z*(z^{2}-1)^{2}
Series of (z^{3}+z^{2}+2)\(z*(z^{2}-1)^{2}) at z=-1
residue of (z^{3}+z^{2}+2)\(z*(z^{2}-1)^{2}) at z=0
residue of (z^{3}+z^{2}+2)\(z*(z^{2}-1)^{2}) at z=1
residue of (z^{3}+z^{2}+2)\(z*(z^{2}-1)^{2}) at z=\infty
residue of (z^{3}+z^{2}+2)\(z*(z^{2}-1)^{2}) at z=-1

\[\operatorname{Res}_{z=0}\left(\frac{z^{3}+z^{2}+2}{z\left(z^{2}-1\right)^{2}}\right)=2 \]

\[\operatorname{Res}_{z=1}\left(\frac{z^{3}+z^{2}+2}{z\left(z^{2}-1\right)^{2}}\right)=-\frac{3}{4} \]

\[\operatorname{Res}_{z=\infty}\left(\frac{z^{3}+z^{2}+2}{z\left(z^{2}-1\right)^{2}}\right)=0 \]

\[\operatorname{Res}_{z=-1}\left(\frac{z^{3}+z^{2}+2}{z\left(z^{2}-1\right)^{2}}\right)=-\frac{5}{4} \]

the result

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\[z^3\cos\frac{1}{z-2} \]

residue of z^3*cos(1/(z-2)) at infinity

\[\operatorname{Res}_{z=0}\left(z^{3} \cos \left(\frac{1}{z-2}\right)\right)=0 \]

\[\operatorname{Res}_{z=\infty}\left(z^{3} \cos \left(\frac{1}{z-2}\right)\right)=\frac{143}{24} \]

分式线性变换：

\[B(0,1) \rightarrow B(0,1) \]

\[\frac{1}{2}, 2, \frac{5}{4}+\frac{3}{4} i \rightarrow \frac{1}{2},2,\infty \]