复变函数积分

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

在(Bleft(0,r ight))

[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} ight) ]

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

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

[int_{|z |=R} z^{n} log frac{z-a}{z-b} d z quad(a eq 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} ight|<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} ight)=0) and

[egin{array}{c} F(z)=int_{alpha}^{eta} varphi(z, t) mathrm{d} t \ F^{prime}(z)=int_{alpha}^{eta} 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:$$frac{z^{3}+z{2}+2}{zleft(z^{2}-1
ight){2}}$$

(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}{zleft(z^{2}-1 ight)^{2}} ight)=2 ]

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

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

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

the result

[z^3cosfrac{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} ight) ight)=0 ]

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

分式线性变换：

[B(0,1) ightarrow B(0,1) ]

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