§52利用留数计算实积分 思路:f(x)x 则/(+(=()=2m2e/的) ①f(x)在实轴上无奇点 ②在4上元奇点
§5.2利用留数计算实积分 思路: ( ) ò b a f x dx ( ) ( ) ( ) å ( ) ò ò ò = + = = n k k l l a b f x dx f z dz f z dz i f b 1 2 res 1 则 p 1 l a l b ①f(x)在实轴上无奇点 ②f(z)在 l 1上无奇点
③f()易算 f(xax 无穷积分 f(在实轴上无奇点,Imz>0中有奇点b k=1,2,,n;¥当1→o,:f(=)→0 则。/(x)=2m2>es( Im 2>0 证:考虑 则手()=()+(=2
③òl1 f (z )dz 易算 一、无穷积分 ( ) : ò ¥ - ¥ f x dx ( ) bk f z 在实轴上无奇点 , Im z > 0中有奇点 k = 1,2,K , n;当 z ® ¥, z × f (z) ® 0 ( ) ( ) 1 Im 0 2 res = > ¥ ò- ¥ = å z n k k 则 f x dx pi f b 证:考虑 ( ) ( ) ( ) ( ) l n k k c R l R f z dz f x dx f z dz i f b ò ò ò R å= - = + = 1 则 2p res
当R→a:」f(x)+m=2∑rs( Im 2>0 又[[=( ≤max(z·f(=) R 0 R ∫f()=2n2rs/(b Im 2>0 例 d x +x2+1
( ) ( ) 1 Im 0 : lim 2 res = > ® ¥ ¥ ® ¥ ò- ¥ + ò = å z n k k R c R f x dx i f b R 当 p ( ) ( ) ò ò = × R R c c z dz 又 f z dz z f z £ max ( × ( ))× ¾ ¾¾ ® 0 z ® ¥ R R z f z p ( ) ( ) 1 Im 0 2 res = > ¥ \ ò- ¥ = å z n k b k f x dx p i f ò ¥ - ¥ + + dx x x x 1 : 4 2 2 例
f(=)= +2z2+1-z 分母:+2+1=(+22+1)-:2=(=2+1) +z+11z2-z+1)=0 1±√1-41 ±i √3 上半平面 √3
( ) 4 2 2 2 4 2 2 1 z 2 z 1 z z z z z f z + + - = + + = ( ) ( ) 2 2 4 2 4 2 2 2 分母 : z + z + 1 = z + 2 z + 1 - z = z + 1 - z ( 1)( 1) 0 2 2 令 = z + z + z - z + = 2 3 2 1 2 1 1 4 1, 2 z = - ± i - ± - = 2 3 2 1 , 2 3 2 1 3,4 1,3 z = ± i 上半平面 z = ± + i
f(2)= 2→00 +z2+1 1+ √3 √3 兀lres +ref+ 2Til liml z→21 +lir (-21)(2-2)(2-2(2-=)
( ) 0 1 1 1 1 1 2 4 4 2 2 ¾¾ ®¾ + + = + + × × = z®¥ z z z z z z z z f z ú ú û ù ê ê ë é ÷ ÷ ø ö ç ç è æ + + ÷ ÷ ø ö ç ç è æ \ = - + 2 3 2 1 res 2 3 2 1 I 2pi resf i f i ( ) ( )( )( )( ) ú û ù ê ë é - - - - - êë é = ® 1 2 3 4 2 1 1 2 lim z z z z z z z z z i z z z z p ( ) ( )( )( )( ) ú û ù ê ë é - - - - + - ® 1 2 3 4 2 3 1 lim z z z z z z z z z z z z z
f(x是偶函数 丌 d x x4+x2+12√3 问:① d x 1+x 1+x2a/()= 奇点z=± 1+ f() 2)0 1+z
Q f (x )是偶函数 1 2 3 0 4 2 2 p = + + ò ¥ dx x x x 问: ? 1 1 0 2 = + ò ¥ dx x ① ( ) z i z dx f z x = ± + = + ò ¥ - ¥ ,奇点 1 1 1 1 2 2 ( ) 0 1 1 1 1 2 2 ¾ ¾ ®¾ + = + × = z ® ¥ z z z z z f z
dx=2tiresf(=2Ti 1+x f(x是偶函数 d x 1+ ②下半平面呢? dx=?奇点:e 1+ ④ dx=? x
= p ( ) = p × = p + = ¥ ò- ¥ z i z dx i f i i x 2 1 2 res 2 1 1 2 Q f (x )是偶函数 1 2 1 0 2 p = + ò ¥ dx x ②下半平面呢? p p p 3 4 3 0 3 ? : , , 1 1 i i i dx e e e x ③ = 奇点 + ò ¥ ? 1 1 2 = - ò ¥ -¥ dx x ④
f( cos px dt2|→)o,f(=)→0 sin 则邮(hn1 k=1 f(x)sipr=x∑ sfe Dei I 问①积分区间为什么是而不是 ②为什么要考虑x)的奇偶
( ) , ( ) 0 sin cos 0 ® ¥ ® þ ý ü î í ì ò ¥ dx z f z px px 二、 f x ( ) å [ ( ) ] ò = > ¥ = n k z ipz f x pxdx i f z e 1 Im 0 0 则 cos p res ( ) å [ ( ) ] ò = > ¥ = n k z ipz f x pxdx f z e 1 Im 0 0 sin p res 问:①积分区间为什么是 ò而不是 ¥ 0 ò ¥ -¥ ②为什么要考虑f(x)的奇偶
③为什么f()0而不是f(=)-=0→,0 思路:①若f(x)偶函数, 则(ph 2/。f(x)t 若f(x)奇函数, 则!()mp=2,」/( ②又e= cos px+ 1 Sin px 依照上例考虑f(=)k 沿图中围道积分
( ) 0 ( ) 0 0 0 ¾¾ ®¾ × ¾¾ ®¾ z ® z ® ③为什么 f z 而不是 z f z 思路: ①若 f (x )为偶函数 , ( ) ( ) ò ò ¥ - ¥ ¥ f x pxdx = f x e dx ipx 2 1 cos 0 则 若f (x )为奇函数 , ( ) ( ) ò ò ¥ - ¥ ¥ = f x e dx i f x pxdx ipx 2 1 sin 0 则 e px i px ipx ②又 = cos + sin ( ) 沿图中围道积分 依照上例考虑 ipz \ f z e - R R R c
()d+()dh=2m2rs[/)] 当R→ f(r)e" dx+lim[ f()e"edz 2∑rs/(b)-1l=0 由约旦引理mJ()d=0 若f(x)为偶函数 则2/()k=2nr[(L=0 Ch 5(r)cos prd=Ti2 res [ (bxJe o Im >0
当 R ® ¥ ( ) ( ) å [ ( ) ] ò ò + = - k l ipb k c ipz R R ipx k R f x e dx f z e dz i f b e 内 2p res ( ) ( ) ò ò ® ¥ ¥ - ¥ + R c ipz R ipx f x e dx lim f z e dz å [ ( ) ] > = k z ipb k k i f b e Im 0 2p res lim ( ) = 0 ò ® ¥ R c ipz R 由约旦引理 f z e dz ( ) å [ ( ) ] ò > ¥ = k z ipb k k f x pxdx i f b e Im 0 0 则 2 cos 2p res 若 f (x )为偶函数 , ( ) å [ ( ) ] ò > ¥ \ = k z ipb k k f x pxdx i f b e Im 0 0 cos p res