(3)res[(=),-n]=limn[(z+m)(z) lim (二+m)I(z+n+1) (z+1)…(z+n) T(1 n(-n+1)…(-1)(-1)ynl =(-1) coS Z (res[ctg, k -=KT sIn
| 1 sin' cos (4) [ , ] ! 1 ( 1) ( 1) ! (1) ( 1) ( 1) (1) ( 1) ( ) ( ) ( 1) lim (3) [ ( ), ] lim [( ) ( )] = = = - - G = - - + ××× - G = + ××× + + G + + = G - = + G = ®- ®- p z kp n n z n z n z z res ctgz k n n n n z z z n z n z n res z n z n z
0,∞均不为极点 用展开求 1 (-1)1 sin k=0(zk+1)! 2k+1 111 1<|z|k∞ resf(0)=C-1=0, res(∞)=0
( ) 0 ( 0 ) 0 , 1 | | ] 1 ! 1 1 [ 1 ( 1)! 1 ( 1) sin (5 ) 0 , 1 3 2 1 0 ¥ = \ = = < < ¥ = - + × × ×× × + - = \ = ¥ - + ¥ = å resf resf C z z z z z zk z z z z z k k k 用展开求 Q 均不为极点
围道积分: 例: dz H2(2-3)z3-1) (被积函奇:二=3-阶,23=1 le ,0+2k爪k=0,1,2,3,4 (2)在|z|k2:只有zk(k=0,-4) 2(z-3(=3-1) d=2x∑reyf(=k)
ò å ò = = = = - - \ < = - = + = = - = - - 4 | | 2 0 5 5 | | 2 5 2 ( ) ( 3)( 1) 1 (2 ) | | 2 : ( 0, 4 ) , 0,1,2,3,4 5 0 2 1 (1) 3 1, ( 3)( 1) 1 k k z k i k z dz i resf z z z z z k k k z e z z dz z z p p 在 只有 被积函奇： 阶， 例： 二、围道积分：
又resf(3)+∑resf(zk)+resf(∞)=0 rest (3)=lim(2-3) (2-3)=3-1)(z3-1)24a 又在 1111 f(=) 3(x3-1)1 ∑()∑ 5(n+1) n=0 39 resf(∞)=-C-1=0 丌l =2mi( 24a|z
ò å å å - \ = - = \ ¥ = - = = + + + × × ×× + + × × ×× = × - × - = - × - = < < ¥ = - = - - = - × + + ¥ = - ¥ = + ¥ = ® = | | ) 24 1 2 ( ( ) 0 ] 1 1 ][ 1 3 9 [ 1 ) 3 ( 1 1 1 1 1 1 1 ( 1) 1 3 1 ( ) 3 | | : 24 1 ( 1) 1 ( 3 )( 1) 1 (3 ) lim ( 3 ) (3 ) ( ) ( ) 0 1 2 3 5 10 0 5 ( 1 ) 0 5 3 5 1 5 5 3 4 0 5 z i i resf C z z z z z z z z z z z z f z z z z z resf z resf resf z resf n n k k z z z k k p a p a 又在 又
计算实积分(119P113 2兀 1 3Ⅰ d x dx a+sinx 4·0a+sin2x 令z=e,则sinx= dx=dz 2 2iz 11 dz k=-1a+ (21-) dz a2(1+2c)+
ò ò ò ò = = - - - + + = + \ = = - = - = = + = + = | | 1 4 2 | | 1 (2 ) ( 1) 1 2 2 0 2 0 2 (1 2 ) 1 1 1 4 1 1 , 2 1 2 , sin sin 1 4 1 sin 1 3. 2 2 2 2 z z iz z ix dz z a z a z i dz a iz I dz iz dx iz z i z z z e x dx a x dx a x I 令 则 p p 三、计算实积分（119）P113