jj ii j i 3 ê v g o 3 ê 3 ½ È o -...
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3ê�Vg�O�
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3ê�Vg�O�
3ê3,½È©O� . . .
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1th, September, 2012 · Shanghai Jiaotong University
êÆÔn�{
1ÊÙµ3ê9ÙA^
�è
þ°�Ï�ÆêÆX
3ê�Vg�O�
3ê3,½È©O� . . .
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1 3ê�Vg�O�
�ÄXeü�È©µ ∮|z|=1
3
e1z dz,
∮|z|=2
sin1
zdz,
|^ÛKÐm��µ∮|z|=1
3
e1z dz =
∮|z|=1
3
[1 +1
z+
1
2!z2 + · · ·+ 1
n!zn+ · · · ]dz = 2πi = 2πiC−1,
∮|z|=2
sin1
zdz =
∮|z|=2
[1
z+
1
3!z3+· · ·+1
(2n + 1)!z2n+1+· · · ]dz = 2πi = 2πiC−1.
½�¤1
2πi
∮|z|=1
3
e1z dz = C−1,
1
2πi
∮|z|=2
sin1
zdz = C−1.
Ù¥§C−1�þãü�ȼê�ÛKÐmª¥1z��Xê"
1.1. 333êêê���VVVggg999333êêê½½½nnn½½½ÂÂÂ5.1µµµ�¼êf(z)30 < |z− z0| < RS)Û§:z0�f(z)����áÛ:§C´?¿���±C : |z − z0| < ρ < R§KÈ©
1
2πi
∮C
f(z)dz, (1)
3ê�Vg�O�
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��¡�f(z)3:z0?�3ê§P�Res[f(z), z0]"�f(z)3�áÛ:���0 < |z − z0| < RS�ÛK?ê�µ
f(z) =+∞∑
n=−∞cn(z − z0)
n, 0 < |z − z0| < R, (2)
ª(2)ü঱ 12πi§÷4�CÈ©§�
1
2πi
∮C
f(z)dz =+∞∑
n=−∞
cn
2πi
∮C
(z − z0)ndz
= c−1.
=µ
Res[f(z), z0] =1
2πi
∮C
f(z)dz = c−1.
½½½nnn5.1µµµ�:z0�f(z)����áÛ:§Kf(z)3:z0?�3ê
�f(z)3z = z0?ÛKÐmªK��(z − z0)−1�Xêc−1,=
Res[f(z), z0] = c−1.
dd½Â��µ(1) ÎÒRes[f(z), z0] �k�:z0�¼êf(z) ��áÛ:�âk¿Â¶(2)O�Res[f(z), z0]=Iéf(z)3:z0?�ÛKÐm"
3ê�Vg�O�
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~~~KKK5.1µµµO�3êµ(1) Res[sin zz , 0]; (2) Res[ze
1z , 0]; (3) Res[ 1
z101(1−z2), 0]
)))µµµ (1)du
sin z
z=
1
z
+∞∑n=0
z2n+1
(2n + 1)!
= 1 +z2
3!+
z4
5!+ · · ·+ z2n
(2n + 1)!,
l §
Res[sin z
z, 0] = 0.
aq/§��µ
Res[ze1z , 0] =
1
2, Res[
1
z101(1− z2), 0] = 1.
½½½nnn5.2µµµ�C��^��{ü4�§e¼êf(z)3CþëY§3C¤��«�DSØ�k��Û:z1, z2, · · · , znþ)Û§K∮
C
f(z)dz = 2πi
n∑k=1
Res[f(z), zk]. (3)
½n5.1¡�3ê½n§§�«EC¼ê÷���È©�3êm�éX©l §Jø�«O�EC¼ê÷��È©��{©
3ê�Vg�O�
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yyy²²²µµµ 3DS±zk�¥%§±¿©��rk��»��±Ck : |z − zk| =rk(k = 1, 2, · · · , n§�¦?Ûü���±QØ��§qØ�¹"df(z)3±CÚC1, C2, · · · , Cn�>.�õëÏ«�þ)Û§��µ
C
C1
C2
C3
C4
Cn
z1
z2
z3
z4
zn
D
∮C
f(z)dz =n∑
k=1
∮Ck
f(z)dz,
þªü>Óر2πi§�
1
2πi
∮C
f(z)dz =n∑
k=1
1
2πi
∮Ck
f(z)dz =n∑
k=1
Res[f(z), zk].
dd=�(Ø"
3ê�Vg�O�
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1.2. 333êêê���OOO���½½½nnn5.3µµµ�z0�f(z)���Û:§K
Res[f(z), z0] = 0.
~~~KKK5.2¦Res[cos z2−1z4 , 0]"
)))µµµdu
f(z) =cos z2 − 1
z4 =1
z4 [1−z4
2!+
z8
4!+ · · · − 1]
limz→0
f(z) = −1
2
`²z = 0�f(z)���Û:§¤±§
Res[cos z2 − 1
z4 , 0] = 0.
½½½nnn5.4µµµ�z0�f(z)�m�4:§K
Res[f(z), z0] =1
(m− 1)!limz→z0
dm−1
dzm−1 [(z − z0)mf(z)]. (4)
yyy²²²µµµduz0�f(z)�m�4:§�±�
f(z) =g(z)
(z − z0)m,
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Ù¥§g(z)3z = z0?)Û§�g(z0) 6= 0"l §
limz→z0
dm−1
dzm−1 [(z − z0)mf(z)] = g(m−1)(z0).
d3ê½Â9p��êúª§k
Res[f(z), z0] =1
2πi
∮C
f(z)dz
=1
2π1
∮C
g(z)
(z − z0)mdz =
g(m−1)(z0)
(m− 1)!.
dd=�(4)"íííØØØ5.1µµµez0�f(z)���4:§K
Res[f(z), z0] = limz→z0
(z − z0)f(z). (5)
íííØØØ5.2µµµ�f(z) = P (z)Q(z)§Ù¥P (z), Q(z)3:z0)Û§�P (z0) 6=
0, Q(z0) = 0, Q′(z0) 6= 0§K
Res[f(z), z0] =P (z0)
Q′(z0). (6)
½n5.49ÙíØJøO�¼ê3Û:a.�4:?�3ê��{"
3ê�Vg�O�
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~~~KKK5.3µµµ¦¼êf(z) = z(z−1)(z+1)23z = 19z = −1?�3ê"
)))µµµz = 1´f(z)���4:§z = 1´f(z)���4:§u´
Res[f(z), 1] = limz→1
[(z − 1) · z
(z − 1)(z + 1)2 ] =1
4;
Res[f(z),−1] = limz→−1
[(z + 1)2 · z
(z − 1)(z + 1)2 ]′
= limz→−1
−1
(z − 1)2 = −1
4.
~~~KKK5.4µµµ¦¼êf(z) = tan z3z = kπ + π2?�3ê"
)))µµµÏ�
tan z =sin z
cos z, sin(kπ +
π
2) = (−1)k 6= 0,
cos(kπ +π
2) = 0, (cos)′|z=kπ+π
2= (−1)k+1 6= 0,
¤±§z = kπ + π2�f(z) = tan z���4:§díØ5.2�µ
Res[f(z), kπ +π
2] =
sin z
(cos z)′|z=kπ+π
2= −1.
3ê�Vg�O�
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~~~KKK5.5:¦¼êf(z) = z sin z(1−ez)33z = 0?�3ê"
)))µµµw,§z = 0�¼êf(z) = z sin z(1−ez)3���4:"
Res[f(z), 0] = limz→0
zf(z) = limz→0
z2 sin z
(1− ez)3
= limz→0
z2(z − z3
3! + z5
5! + · · · )(1− [1 + z + z2
2! + z3
3! + · · · ])3
= −1.
5µ�KØU|^íØ5.2"~~~KKK5.6µµµO�e�È©µ
(1) I =
∮C
1
z3(z − i)dz, Ù¥C����±|z| = 2;
(2) I =
∮C
tan πzdz, Ù¥C����±|z| = n(n���ê).
)))µµµ(1) 3�±|z| = 2S§¼êf(z) = 1z3(z−i)kn�4:z = 0Ú��4
:z = i"
Res[f(z), 0] =1
2!limz→0
[z3 · 1
z3(z − i)]′′ =
1
2limz→0
2
(z − i)3 = −i,
3ê�Vg�O�
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Res[f(z), i] = limz→i
(z − i) · 1
z3(z − i)= i.
Ïd§d3ê½n§kµ
I =
∮C
1
z3(z − i)dz = 2πi(−i + i) = 0.
(2) f(z) = tan πz = sin πzcos πzk��4:z = k + 1
2(k��ê)§díØ5.2§
Res[f(z), k +1
2] =
sin πz
(cos πz)′|z=k+1
2= −1
π,
d3ê½n§
I =
∮C
tan πzdz = 2πi∑
|k+12 |<n
Res[f(z), k +1
2]
= 2π(−2n
π) = −4ni.
~~~KKK5.7µµµO�e�È©µ
(1) I =
∮C
z − sin z
z8 dz, Ù¥C����±|z| = 1;
(2) I =
∮C
1
1 + ezdz, Ù¥C����±|z| = 4π.
3ê�Vg�O�
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)))µµµ (1) z = 0�f(z)3|z|¤�«�S����áÛ:§��Ê�4:§
f(z) =z − sin z
z8 =1
z8 [z − (z − 1
3!z3 +
1
5!z5 − 1
7!z7 + · · · )]
=1
3!z5 −1
5!z3 +1
7!z− · · · ,
�
Res[f(z), 0] = c−1 =1
7!.
l §
I =
∮C
z − sin z
z8 dz = 2πiRes[f(z), 0] =2
7!πi.
(2)3|z| = 4π¤�«�S§f(z) = 11+ezko���4::±πi,±3πi,
Res[f(z),±πi] = limz→±πi
1
ez= −1,
Res[f(z),±3πi] = limz→±3πi
1
ez= −1.
l
I =
∮C
1
1 + ezdz = 2πi(Res[f(z), πi] + Res[f(z),−πi]
+ Res[f(z), 3πi] + Res[f(z),−3πi])
= −8πi.
3ê�Vg�O�
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1.3. ÃÃá¡¡���:::���333êêê½½½ÂÂÂ5.2µµµ�z = ∞´¼êf(z)��áÛ:§=f(z)3á�:���R <|z| < +∞S)Û§C´?¿���±|z| = r > R§KÈ©
1
2πi
∮C−
f(z)dz
��¡�f(z)3z = ∞?�3ê§P�Res[f(z),∞]"½½½nnn5.5µµµ�z = ∞´¼êf(z)��áÛ:§K
Res[f(z),∞] = −c−1, (7)
Ù¥§c−1�f(z)3z = ∞�ÛKÐmª¥z−1��Xê"
yyy²²²µµµ�f(z)3z = ∞�ÛKÐmª�µ
f(z) =+∞∑
n=−∞cnz
n, cn =1
2πi
∮C
f(z)
zn+1dz,
dÅ�È©{§�µ
Res[f(z),∞] =1
2πi
∮C−
f(z)dz = −c−1.
3ê�Vg�O�
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~~~KKK5.8µµµ¦¼êf(z) = 11−z3z = ∞?�3ê"
)))µµµf(z)3z = ∞���|z| > 1S�ÛKÐmª�
f(z) =1
1− z= −1
z− 1
z2 − · · · ,
Ïd§
Res[f(z),∞] = −c−1 = 1 6= 0.
du
limz→∞
1
1− z= 0,
¤±§z = ∞�¼êf(z) = 11−z���Û:"d~`²µez =
∞�f(z)���Û:§Ù3êØ�½�""
~~~KKK5.9µµµ�f(z) = 1+z2
ez §O�Res[f(z),∞]")))µµµ��±C : |z| = 2§K
Res[f(z),∞] =1
2πi
∮C−
1 + z2
ezdz
= − 1
2πi
∮C
1 + z2
ezdz
= 0.
3ê�Vg�O�
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á�:?�3ê§�^±e�½n5O�"
½½½nnn5.6µµµ�z = ∞´¼êf(z)��áÛ:§K
Res[f(z),∞] = −Res[1
z2f(1
z), 0]. (8)
yyy²²²µµµÏ�z = ∞´¼êf(z)��áÛ:§K�3¿©��R > 0§¦�¼êf(z)3�±|z| = RÜ�Ðm�ÛK?êµ
f(z) = · · ·+ c−2z−2 + c−1z
−1 + c0 + c1z + · · · ,
Ù¥§
c−1 =1
2πi
∮C
f(z)dz, C : |z| = R1 > R.
f(1
z2)1
z= · · ·+ c−3z + c−2 + c−1z
−1 + c0z−2 + · · · ,
w,§f( 1z2)
1z3|z| <
1R1Sk�áÛ:z = 0"¤±§
Res[f(1
z2)1
z, 0] = c−1 =
1
2π
∮C
f(z)dz
= −Res[f(z),∞].
3ê�Vg�O�
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~~~KKK5.10µµµ¦¼êf(z) = e1z
1−z3z = ∞�3ê")))µµµw,§z = 0´
1
z2f(1
z) =
ez
z(z − 1)
���4:"Ïd§
Res[f(z),∞] = −Res[1
z2f(1
z), 0] = −Res[
ez
z(z − 1), 0]
= − limz→0
z · ez
z(z − 1)= 1.
~~~KKK5.11:�Pn(z)�z�ngõ�ª§O�Res[P′n(z)
Pn(z),∞]"
)))µµµ�Pn(z) = anzn + an−1z
n−1 + · · ·+ a1z + a0§
P ′n(z)
Pn(z)=
nanzn−1 + (n− 1)an−1z
n−2 + · · ·+ a1
anzn + an−1zn−1 + · · ·+ a1z + a0
=n
z[1 + n−1
nan−1
an
1z + · · ·
1 + an−1
an
1z + · · ·
]
=n
z[1 +
b1
z+
b2
z2 + · · · ]
Res[P ′
n(z)
Pn(z),∞] = −c−1 = −n.
3ê�Vg�O�
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½½½nnn5.7µµµe¼êf(z)3*¿E²¡þØk���áÛ:z1, z2, · · · , zn,∞´)Û�§Kf(z)3¤k�áÛ:?�3ê�Ú�"§=
n∑k=1
Res[f(z), zk] + Res[f(z),∞] = 0. (9)
yyy²²²µµµ±�:�¥%§¿©��R��»��±C§¦C¤��«��¹:z1, z2, · · · , zn§Kd3ê½n§�∮
C
f(z)dz = 2πin∑
k=1
Res[f(z), zk],
=§n∑
k=1
Res[f(z), zk] =1
2πi
∮C
f(z)dz.
dá�:�3ê½Â§�
Res[f(z),∞] =1
2πi
∮C−
f(z)dz = − 1
2πi
∮C
f(z)dz,
Ïd§n∑
k=1
Res[f(z), zk] + Res[f(z),∞] = 0.
3ê�Vg�O�
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dª(9)��µeU�N´/�ÑRes[f(z),∞]§�n��§K|^ª(9)O�
∑nk=1 Res[f(z), zk]�`�5���§l ���BÈ©
∮C f(z)dz�
O�"
~~~KKK5.12µµµ�f(z) = (z − 3)−2(z8 − 1)−1§zk(k = 1, 2, · · · , 8)��§z8 =
1�)§O�
Res[f(z), 3] +8∑
k=1
Res[f(z), zk].
)))µµµ¼êf(z)3*¿E²¡þkÛ:�zk(k = 1, 2, · · · , 8)9z = 3"A^½n5.7§�
Res[f(z), 3] +8∑
k=1
Res[f(z), zk] = −Res[f(z),∞].
du
1
z2f(1
z) =
1
z2 ·1
(1z − 3)2( 1
z8 − 1)
=z8
(1− 3z)2(1− z8).
l §
Res[f(z),∞] = −Res[1
z2f(1
z), 0] = 0.
3ê�Vg�O�
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¤±§
Res[f(z), 3] +8∑
k=1
Res[f(z), zk] = 0.
~~~KKK5.13µµµO�È©
I =
∮C
z15
(z2 + 1)2(z4 + 2)3dz,
Ù¥C����±|z| = 4"
)))µµµf(z) = z15
(z2+1)2(z4+2)3Ø�z = ∞§�kÛ:
z = ±i, zk =4√
2eπ+2kπ
4 , (k = 0, 1, 2, 3).
Res[f(z),∞] = −Res[1
z2f(1
z), 0]
= −Res[1
z(1 + z2)2(1 + 2z4)3 , 0]
= − limz→0
1
(1 + z2)2(1 + 2z4)3 = −1.
d½n5.7§�
Res[f(z),±i] +3∑
k=0
Res[f(z), zk] = −Res[f(z),∞] = 1.
3ê�Vg�O�
3ê3,½È©O� . . .
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¤±§
I =
∮C
z15
(z2 + 1)2(z4 + 2)3dz = 2πi.
2 3ê3,½È©O�¥�A^
�!Ì�0�^3êO�,½È©"
2.1. OOO���∫ 2π
0 R(cos x, sin x)dx...ÈÈÈ©©©
�È©¼êR(cos x, sin x)´sin x, cos x�kn¼ê§�3[0, 2π]þëY"-
z = eix = cos x + i sin x,
K
cos x =z + z−1
2, sin x =
z − z−1
2i, dx =
1
izdz.
l § ∫ 2π
0R(cos x, sin x)dx =
∮|z|=1
R(z + z−1
2,z − z−1
2)1
izdz.
�f(z) = R(z+z−1
2 , z−z−1
2 ) 1iz 3|z| < 1S�Û:�zk(k = 1, 2, · · · , n)§d3
ê½n§ ∫ 2π
0R(cos x, sin x)dx = 2πi
n∑k=1
Res[f(z), zk].
3ê�Vg�O�
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~~~KKK5.14µµµÁO�È©
I =
∫ 2π
0
sin2 x
5 + 3 cos xdx.
)))µµµ-
z = eix = cos x + i sin x,
K
cos x =z + z−1
2, sin x =
z − z−1
2i, dx =
1
izdz.
I =
∫ 2π
0
sin2 x
5 + 3 cos xdx =
∮|z|=1
i(z2 − 1)2
2z2(3z2 + 10z + 3)dz
=i
6
∮|z|=1
(z2 − 1)2
z2(z + 13)(z + 3)
dz.
Pf(z) = (z2−1)2
z2(z+13)(z+3)§Ù3|z| < 1Sk��4:z = 0§��4:z =
−13§§��3ê�
Res[f(z), 0] = limz→0
[z2 · (z2 − 1)2
z2(z + 13)(z + 3)
]′ = −10
3;
Res[f(z),−1
3] = lim
z→−13
(z +1
3) · (z2 − 1)2
z2(z + 13)(z + 3)
] =8
3.
3ê�Vg�O�
3ê3,½È©O� . . .
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¤±
I =i
6· 2πi[Res[f(z), 0] + Res[f(z),−1
3] = −π
3[−10
3+
8
3] =
2
9π.
~~~KKK5.15µµµO�È©
I =
∫ 2π
0ecos θ cos(nθ − sin θ)dθ.
)))µµµ-z = eiθ§K
I1 =
∫ 2π
0ecos θ[cos(nθ − sin θ)− i sin(nθ − sin θ)]dθ
=
∫ 2π
0eeiθ−inθdθ =
∮|z|=1
ez
izn+1dz
=2πi
in|z=0 =
2π
n!.
¤±§
I = Re(I1) =2π
n!.
3ê�Vg�O�
3ê3,½È©O� . . .
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~~~KKK5.16µµµO�È©
(1) I =
∫ 2π
0
cos 2θ
1− 2r cos θ + r2dθ r > 0
(2) I =
∫ α
0
1
(5− 3 sin 2πϕα )2
dϕ α > 0.
)))µµµ (1)-z = eiθ§K
I = − 1
2ir
∮|z|=1
1 + z4
z2(z − 1r)(z − r)
dz.
�0 < r < 1�§�È©¼ê3|z| < 1Skü�4:µ��4:z = 0Ú��4:z = r"
I = 2πi(Res[f(z), r] + Res[f(z), 0])
= 2πi(− 1
2ri)[− 1 + r4
r(1− r2)+
1 + r2
r]
=2πr2
1− r2 .
�r > 1�§�È©¼ê3|z| < 1Skü�4:µ��4:z = 0Ú��4:z = 1
r"
3ê�Vg�O�
3ê3,½È©O� . . .
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I = 2πi(Res[f(z),1
r] + Res[f(z), 0])
= 2πi(− 1
2ri)[
1 + r4
r(1− r2)+
1 + r2
r]
= − 2π
r2(1− r2).
(2)-θ = 2πϕα §K
I =
∫ α
0
1
(5− 3 sin 2πϕα )2
dϕ =α
2π
∫ 2π
0
1
(5− 3 sin θ)2dθ.
-z = ei�
I = −2α
iπ
∮|z|=1
z
(3z − i)2(z − 3i)2dz.
�È©¼ê3|z| < 1S�k����4:z = i3§¤±
I = 2πiRes[f(z),i
3] = 2πi(−2α
iπ)(− 5
256)
=5
64α
3ê�Vg�O�
3ê3,½È©O� . . .
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2.2. OOO���∫ +∞−∞
P (x)Q(x)dx...ÈÈÈ©©©
ÚÚÚnnn5.1µµµXJ�3α > 1, M > 0, R > 0§¦��|z| ≤ R�Imz ≥0�§f(z))Û§�|f(z)| ≤ M
|z|α§K
limR→+∞
∫|z|=R
f(z)dz = 0.
yyy²²²µµµd
|∫|z|=R
f(z)dz| ≤∫|z|=R
|f(z)||dz|
≤ M
∫|z|=R
1
|z|α|dz| = MπR1−α → 0.
=�(Ø"
½½½nnn5.8µµµ�P (x), Q(x)�õ�ª§�§Q(x) = 0â�§�Q(x)�gê
'P (x)�gê��püg"-f(z) = P (z)Q(z)§K∫ +∞
−∞
P (x)
Q(x)dx = 2πi
n∑k=1
Res[f(z), zk]. (10)
Ù¥zk(k = 1, 2, · · · , n)�f(z)3þ�²¡þ��áÛ:"
3ê�Vg�O�
3ê3,½È©O� . . .
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yyy²²²µµµ�þ��±CR : |z| = RÚ¢�ã[−R,R]|¤�^4�C§Xe㤫"
x
y
CR
R−R
z1
z2
zn
�¿©��R§¦C¤��«��¹f(z)3þ�²¡þ�¤kÛ:"d3ê½n§� ∫ R
−R
f(z)dz +
∫CR
f(z)dz = 2πi
n∑k=1
Res[f(z), zk].
dÚn5.1§�
limR→+∞
∫CR
f(z)dz = 0.
l § ∫ +∞
−∞
P (x)
Q(x)dx = 2πi
n∑k=1
Res[f(z), zk].
3ê�Vg�O�
3ê3,½È©O� . . .
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~~~KKK5.17µµµO�È©I =∫ +∞−∞
x2−x+2x4+10x2+9dx.
)))µµµ¼êf(z) = z2−z+2z4+10z2+93þ�²¡Skü���4:z = iÚz = 3i§�
Res[f(z), i] = limz→i
(z − i) · z2 − z + 2
z4 + 10z2 + 9=
1− i
16i,
Res[f(z), 3i] = limz→3i
(z − 3i) · z2 − z + 2
z4 + 10z2 + 9=
7 + 3i
48i.
Ïd§
I =
∫ +∞
−∞
x2 − x + 2
x4 + 10x2 + 9dx = 2πi(
1− i
16i+
7 + 3i
48i) =
5
12π.
~~~KKK5.18µµµO�È©I =∫ +∞
01
(x2+1)n+1dx.
)))µµµ
I =
∫ +∞
0
1
(x2 + 1)n+1dx =1
2
∫ +∞
−∞
1
(x2 + 1)n+1dx
= πiRes[1
(z2 + 1)n+1 , i] = πi1
n!
dn
dzn(
1
z + i)n+1|z=i
=(2n− 1)!!
(2n)!!· π2.
3ê�Vg�O�
3ê3,½È©O� . . .
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2.3. OOO���∫ +∞−∞
P (x)Q(x)e
iλxdx...ÈÈÈ©©©
ÚÚÚnnn5.2µµµ�C��±|z| = R�þ��±§¼êf(z)3CþëY§�
limz→∞
f(z) = 0,
K
lim|z|=R→+∞
∫C
f(z)eiλzdz = 0 (λ > 0). (11)
½½½nnn5.8µµµ�P (x), Q(x)�õ�ª§�§Q(x) = 0â�§�Q(x)�gê
'P (x)�gê��p�g"-f(z) = P (z)Q(z)§K∫ +∞
−∞
P (x)
Q(x)eiλxdx = 2πi
n∑k=1
Res[f(z)eiλz, zk]. (12)
Ù¥zk(k = 1, 2, · · · , n)�f(z)3þ�²¡þ��áÛ:"yyy²²²µµµaqu½n5.7�y²§Ñ"d½n5.8��µ∫ +∞
−∞
P (x)
Q(x)cos λxdx = Re(2πi
n∑k=1
Res[f(z)eiλz, zk]),∫ +∞
−∞
P (x)
Q(x)sin λxdx = Im(2πi
n∑k=1
Res[f(z)eiλz, zk]).
3ê�Vg�O�
3ê3,½È©O� . . .
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~~~KKK5.19µµµO�È©I =∫ +∞−∞
x cos xx2−2x+10dx"
)))µµµ¼êf(z) = zeiz
z2−2z+103þ�²¡Sk����4:z = 1 + 3i§�
Res[f(z)eiz, 1 + 3i] =zeiz
(z2 − 2z + 10)′|z=1+3i =
(1 + 3i)e−3+i
6i.
l
I =
∫ +∞
−∞
x cos x
x2 − 2x + 10dx = Re(
∫ +∞
−∞
xeix
x2 − 2x + 10dx)
= 2πiRes[f(z)eiz, 1 + 3i] =π
3e3(cos 1− 3 sin 1).
~~~KKK5.20µµµO�È©I =∫ +∞
0x sin bx
(x2+a2)2dx, (a > 0, b > 0)"
)))µµµ
I =
∫ +∞
0
x sin bx
(x2 + a2)2dx =1
2Im(
∫ +∞
−∞
xeibx
(x2 + a2)2dx)
= Im(πiRes[zeibz
(z2 + a2)2 , ai])
=πb
4ae−ab.
3ê�Vg�O�
3ê3,½È©O� . . .
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