chapter5定积分计算
Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
Chapter 5 定积分计算
Abstracts:留数定理及其应用——定积分、积分主值
一、留数定理和留数的求法(Residue theorem and residue calculations)
1(留数的定义:设是函数的孤立奇点(isolated singularity),即除过zf(z)0
点以外函数是解析的,则在的留数定义为zz,zf(z)f(z)00
1fzfzz,,其中为绕的闭曲线(积分沿正方向进Res()dzc,,00, ,cc,i2
行)且内部无其它奇点,记号为或. Resf(z)Resf(z)z,z00
(1)有限远孤立奇点的留数:在邻域(0,z,z,r)内(不含zf(z)00
,1,1其它奇点)的罗朗级数(Laurent series)展开的 次幂项的(z,z)0
1fzfzza,,系数称为在奇点的留数。即. Res()dzaf(z),,01,0,1 ,c,i2
此定义基于如下的事实:
,1()fzk,其中 . ,,,azdf(z),az,z,kk0k,1 ,c,,2()izzk,,,0
令函数沿以孤立奇点z为中心的一个圆周积分 cf(z)0
,,kk, ,,,,f(z)dz,az,zdz,az,zdz,k,k00,,,ccckk,,,,,,
2 (1),ik,,,kfzzia()d2,,而 所以 . zzz,,d,,,,10 , ,cc0 (1),k,,,
,1,,可见,级数中仅仅项对积分有贡献,积分后唯有a这个系az,z,1,10
数留下来,故名之为留数(residue).
R,z,,z,0(2)无穷远点的留数:f(z)在以为中心,环内(不含0
,1,1a其它奇点)的罗朗级数展开的次幂项(z,z)的系数的反号称为,10
1ffzza,,,,Res()df(z)在点的留数。即 (此定义直观)。 ,,,,1, c,i2
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
这是因为:对于无穷远点,以为展开中心、在区域R,z,,z,,
里展开的罗朗级数与以为中心、在区域展开的罗朗级R,z,,z,00
,k数有相同的形式: 换言之,以为中心、在区域z,0fzaz().,,0kk,,,
b展开罗朗级数亦可,其中任意(实际为的邻域)。Rzb,,,,z,,
(Chapter 1:无穷远点只有一个,其模,而幅角不定)。 ,,
同时注意到,对无穷远点的邻域来讲,的正方向为顺时针方向。因此, cR
,,,kkkfzzazzazzazzia()dddd2.,,,,,,,,,,1kkk, ,,,, cccRRR ,kkk,,,,,,,,,cR
clockwise clockwise clockwise counter clockwise
2( 留数定理:如果在区域D中有个孤立奇点,而除z,z,?,znf(z)12n了这些奇点外,是解析的,那么 f(z)
fzzfzzfzz()d()d()d,,,? ,,, ccc12n
,,,,,2Res()Res()Res()ifzfzfz?,,12n n ,2Res(),ifz,,k,1k
l其中c,c,?,c分别是围绕奇点z,z,?,z的小圆周(反方向, 与外界同12n12n
方向),再根据复连通域的柯西定理(Cauchy’s theorem),可以得到
nn
, fzzfzzifz()d()d2Res(),,,,,k ,,,,11kklck
l是区域D的外境界线,也可以是境界线之内的任一条闭合曲线,只要它包围n个孤立奇点并且沿确定(正)方向围绕奇点一圈。这就是留数定理。
l[留数定理]:如果函数在闭曲线所围的区域内,除具有有限个孤立f(z)
lzzkn,,(1,2,,)?奇点(isolated singularities)外是解析的,在上也是解析k
ll的,则f(z)沿的回路积分(逆时针方向)等于f(z)在内所有奇点的留
n
2,i数之和的倍,即 fzzifz()d2Res().,,,k ,l,1k
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
2( 留数定理的推论:若在闭复平面内(包括无穷远点)除有限个孤立奇点f(z)
外处处解析,则在全平面上全部留数之和为零(挖去所有奇点并且计及f(z)
n
无穷远点): Res()Res()Res()0.fzfzf,,,,,,k,,,()1kzk
说明:
* 留数定理
表
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明了解析函数沿闭曲线的积分与它的孤立奇点之间的关系,体现了解析点与奇点的内在联系。这是解析函数在不同点取值之间的相互关联这个性质的又一表现,即它是单(复)通域Cauchy定理的推广(变形)。
;** Laurent series的负幂次由有限内环内的奇异性引起,其积分方向为 r, rLaurent series的正幂次由有限内环以外(即外环R以外甚至直接至)的奇r,异性引起,其积分方向为. , R
*** 可以是函数的奇点亦可以不是奇点,只要存在它就是无,a,f(z)z,,,1穷远点的留数 Res().f,
3(留数的求法(Residue calculations)
(定义是定义,定理是定理,计算留数是另一回事)。
(1) 罗朗级数法: 一般地,对于本性奇点,例如中含指数函数、三角函 f(z)
z数()等,虽然极点是高阶的,罗朗级数展开有无穷多项,但 e,sinz,?
是我们仅仅需要与相关的项即可,这样往往比较简单。 a,1
lim()fzbb,,(2) 可去奇点:若是的可去奇点(),有限, 则 f(z)Res()0.fb,zb,
0注意:即使点是的可去奇点,其留数也不一定为,除非在f(z)f(z),
0一切有限远点的留数之和为. 例如,, f(z),1,1/z
Resf(,),,1, Resf(0),1.
b(3) 高阶极点(Multiple pole,high order pole):若是f(z)的m(m,1)阶极点,
,k即,,, f(z),az,b (a,0),k,mk,,m
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
,1m1dm则 . Res()lim,,fb,z,b,,,,fz,1m,zb,,1!dm,z
z,b [证明]: 如果是的阶极点,则在这点的邻域内罗朗级数是 mf(z)
,aak1,m,m,, f(z),,,?,az,b (,,a,0),k,mmm,1,,,,z,bz,bk,,m
mm,1???zbfzaazbazb,,,,,,,,,(), ,,,,,,,,,,mm11
mm,,11,ddmkm,,,,,zbfzazb,,,(),,,,,kmm,,11,,,,zzddkm,, ,k,1?kmkmkazb,,,,,,,,,,,,,, 12.,kk,,1
z,bk,,1取极限后右端只留下项,即. 所以 ,,m,1!a,1
,1m1dm. Res()lim,,fb,a,z,b,,,,fz,1,1m,zb,,1!m,dz
,,,,Resf(b),lim,,z,bfzm,1b(4) 单阶极点(Simple pole): 当时,为单极点. z,b
P(z)b特别地,如果可以写成的形式,其中和均在点解f(z)Pz()Qz()Q(z)
zb,析,而且为的一阶零点,即那么 Qz()Qz()0,,QzPz'()0,()0,,,
,,P(z)P(z)P(b),,,,,,Resf(b)lim,,zbfzlimzblim,,,,,,. ,,z,bz,bz,bQ(z)Q(b),Q(z)Q'(b),,
zb,
1z,bfbfzz,Res()d (5)根据定义:,其中为绕一圈的闭曲线且c,, ,c,i2
其内部无其它奇点,积分沿正(沿奇点的反)方向进行。 ,c
5. 例题(Examples)
1
zez,0f(z),z,1Example 1. 求函数在,,点的留数。 z,,1,z
[解] z,,0,1,分别是本性奇点,一阶奇点和一阶零点。
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
法一(Expand to the Laurent series):
z,0z,0? 是的本性奇点,因此,将在的邻域作罗朗级数展开 f(z)f(z)
11111,,23fzzzz()11,,,,,,,,,??,,,,23zzz2!3!,,
111,,,,,,,,???1,,,2!3!z,,
,111zn. Resf(0),1,,,?,e,1?ezz,,[and1],2!3!n!0n,
1,nz? 设并且 (其余的虽然复杂,但是我们用不到),则 ce,cecz,,(1),0nnn,0
1,ze1n ,,,,fzcz()(1).,n,,11zzn,0
故 Res(1).face,,,,,,10
? (全复平面留数之和为零)。 Res(0)Res(1)Res()0fff,,,,
Res()1.f,,
法二(Formula):
z,1? 是的一阶极点,因此 f(z)
1,,ze,, Res(1)lim1.fze,,,,,,,z1,,1z,,,,,
? 将在的邻域作罗朗级数展开 f(z)z,,
111/zfze(),,zz11/,
111111111,,,,,,,,,,,,,,,,11?? ,,,,,,2323zzzzzzz2!3!,,,,,,
1,,,,??(1),z
Resf(,),,(,1),1.
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
zeExample 2. 求函数在点的留数。 f(z),z,15,,z,1
[解一]
234zz,1,,zzz111,,,,,,,,,eeee, fzz()11,,,,,,,,,,?,,,,5552!3!4!zzz111,,,,,,,,,,,,,
ee Res(1).f,,4!24
[解二] 是的五阶极点,因此 z,1f(z)
514,zz,,1d1deee5 Res(1)lim1lim.fz,,,,,,,,5514,zz,,1151!d4!d24zz,,,z1,,,,,,,
1zi,(X)Example 3. 求函数在点的留数。 f(z),32,,z,1
2zi,, [解] 是的三阶极点,因此 f(z)zzizi,,,,1()(),
,,22,,1d11d13i3,,Res()lim.fizi,,,,, ,,,,3322zi,2,,2!d2d16zzzi,,,,,z1,,,,,zi,,,
sin2z(X)Example 4. 求函数的留数。 fz(),3z,1,,
z,,1[解]是的三阶极点,是本性奇点,因此 f(z)z,,
21dfzz,,,,,Res(1)limsin22sin2|2sin2. z,,12z,,1z2!d
Res()2sin2.f,,,
1f(z),(X)Example5. 求函数z,n,在(n为整数)的留数。 sinz
z,n,[解一]是的单极点,因此 f(z)
zn,,',,1n,, Res()limlim1.fnzn,,,,,,,,,,,,,,,,,znznsinsin'zz,,,,
11nfn,,,,,Res()1.[解二] ,,zzsin'cos,,,,zn,,zn
4zi,Example 6. Find the residue of ,, at . sinz/,,1,z
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
[解一]zi,是的单极点,因此 f(z)
sinsinsinzzi,,Res()limlimfizi,,,,,,42,,zzi,,i14,zi1z,,zi,,,,,, ,,11,,eeee1,,,sinh(1).(2)(4)84ii
sinsinsin1zzi [解二] ,,,,Res()sinh(1).fi33,,,444zi4zi,,1z,,zi,
imzem,0Example 7. 求函数(实常数)的留数。 ()fz,21,z
[解]z,,i是的一阶极点,是本性奇点(高振荡),因此 f(z)z,,
imzmimzm,eeeeRes();Res();,,,,, fifizizi,,,,,,22ziizii
i,mmRes()[Res()Res()]()sinh.,,,,,,,,, ffifieeim2
22(1)z,(X) Example 8. 求函数()的留数。 ,,,,,,1,fz(),2,,zzz()(),,
z,0[解]是的一阶极点,是二阶极点,解析,因此 z,,,,f(z)f()1,,
22222(1)(1)(1/)z,,,,,, Res();f,,,,,,,,z,,22zz()(),,,,,,,,,
22(1)z, ,,,Res();f,,,z,,2zz(),,
2z,0aza:是在附近Taylor展式中的一次项的系数 Res(0)fzfz()11
2222,,(1)(1)11zz,,2zfz(),,,,,,,,,,,()()zzzz,,,,,,,
2222,,,,,,,,(1)11(1)1111zzzz,,,,,,,,,,,,,(1)(1).???z,,,,,,,,22,,,,,,,,,,,,,,,,,,,,,,
,,11122,, (放心取项z,0). ?,,,,Res(0);f(1)z,,,22,,,,,,,
Res()[Res()Res()Res(0)]().ffff,,,,,,,,,,,,
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU 二、留数定理在定积分计算中的应用
(Residue theorem’s application to the definite integral calculations)
A. 基本思路与方法:
目标:为了计算实函数在整个实轴上或实轴上某一线段If(x)
b上的积分,可分为以下三个步骤(方法): f(x)dx,a
1(将看作复变函数当的特殊情况,即将延z,xf(x)f(z)f(x)
拓至复平面 (和之间有千丝万缕的关系)。 Fz()f(z)Fz()
2(将实积分路径改变为复平面闭曲线,其内部为D——或做变
bMa换,将ab变为闭曲线,或补一段使其成为闭曲线。
3( 在D上对或应用留数定理计算闭路上的积分,这样就把实轴上Fz()f(z)
的积分转化为计算在D内奇点的留数以及了。即 f(z)dzf(x)f(z),bMa
b fxxfzzfzzifz()d()d()d2Res().,,,,,j,,, labMaj
Key:()d?(fzz,see the Lemma 1,2 and 3 below) ,bMa
B. 基本类型:
2,,,Rsinx,cosxdx1. ,其中是的有理式。 R(sinx,cosx)sinx,cosx,0
ix0~2,z,e方法:作单位园变换,,即将的直线路径变为单位圆,,x,0,2,
,,11zzzz,,1z,1周路径。并且 sin, cos, ddxxxz,,,,则 22iiz
,,112,,,zzzz,,1RxxxRzsin,cosd,d,. ,,,,,, 01z,22iiz,,
,,11,,1zzzz,,FzR(),,令,那么 ,,izi22,,
2,RxxxiFzsin,cosd2Res(),,. ,,,,0,z1
,2I,sinxdxExample 1. . ,0
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
,1cos(2),x,[解一] .(简单~) dI,x,,022
[解二]
2,122,,,1111(1)zzz,,2Ixxzz,,,,sinddd,,3,,, z,,,1z1,2228iiziz,, 22展开,,,1(1)1z,,,,,,,,,,,,(2)Res=(2)(2).ii,,,,,,3882izi,,,,,,z,0
22,sinx Example 2. . I,dx (a,b,0),0a,bcosx
[解]
2,1,,zz,2222,,zz11,,,,,,2i111,,,其中 Fz(),,,,,,122azz,izbibizzzzz22,,,,,,,,2212zzz1,,ab,,,b2,,
2222,,,,,,aabaab zz,,, ,zz,1.1212bb
2AxBxC,,,0[韦大定理: 的两个根满足]. xx,xxCAxxBA,,,,/,/121212
z,0(z,1)在单位圆内有两个孤立奇点:(二阶奇点),(一阶奇点)。z,zF(z)1
(1)z,(一阶奇点)在单位圆外。因此,我们有 zz,2
计算d1a2,,,,,,,Res(0)lim().FzFzzz ,,122,,z,0d2zbiib
计算1i22,,,,,,,,,Res()lim().FzzzFzzzab ,,,,1112,,2zz,12bib
12,,,22 ,,,,,,,,,IiFFzizaab2Res(0)Res()22.,,,,,,11,,22bib,,
三个引理:
R,z,a,,引理1(大圆弧引理):如果在区域D:,,,arg(z,a),,f(z)12
K上连续,且当zz(D),,,时,(z,a)f(z)一致地趋于,简记为()(),zafzK,,
RCa,,,则,,,其中是以为圆心,为半径,夹角为的limf(z)dz,iK,,,2121R,C,,RR
zaRza,,,,,,arg().,,圆弧, 12
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
zd证明:因为,所以 ,,,i,,,21,CRz,a
K,,()d()d,,,,,,fzziKfzz,,21,,,,CCRR,za,,
dz ()(),,,zafzK,,,CRza,
dz()().,,,zafzK,CRza,由于当,时,一致地趋于,这意味着任K,,arg(z,a),,z,,(z,a)f(z)12
,,0给,存在[与无关,D内各向同性的],使当zaRM,,,arg(z,a)M(,),0
i,,,,,f(z)dz,iK,,,,,,,,时,,所以(大(z,a)f(z),K,,zaRe,,,2121,CR
RR/1,模之比,仅仅积分相位即可),即 (方向正向)。 ,,limf(z)dz,iK,,,21,CR,,R
引理2(小圆弧引理):若函数在区域D:0,z,a,r,f(z)
k上连续,且当 时,一致地趋于,,,arg(z,a),,zza(D),,(z,a)f(z)12
则,其中是以为圆心,为半径,夹角为的Ca,,,lim()dfzzik,,,,r,,r2121,Cr,0r
zarza,,,,,,arg().,,圆弧, 12
zd,,,i,,,证明: 因为, 21,Crz,a
,i,zre,nnin,,1(1),{(,)ddIrzzire,,,,n,,cr0 n,1rin(1),,,,,,,,,[1],,1,2,eN?,,,,,21 n,1
r,0Iri(,);,,,IrNn[(0,),2/(1)](0,),,,,,,,,,小圆弧引理; n,0,1
Ir(,2)0;,,IrNn[(0,),2/(1)](,0),,,,,,,,, r,,大圆弧引理。n,-1n,-2
亦即,高阶极点的一段圆弧的积分发散,而整个圆弧的回路积分为零;或者说,
虽然复平面是各向同性的,但是积分值与相对位相差相关} 所以
k,,fzzikfzz()d()d,,,,,,,,21,,,,CCrrza,,, dzdz=()()()().zafzkzafzk,,,,,,,,,CCrrzaza,,
k,,arg(z,a),,z,a(z,a)f(z)由于当,时,一致地趋于,这意味着任给12
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU ,,0,存在[与无关,D内各向同性的],使当时,z,a,r,,arg(z,a),(,),0
i,fzzik()d,,,,,,,,,,所以(小模之()()zafzk,,,,zare,,,,,,,2121,Cr
r/r,1比,仅仅积分相位即可),即
(方向为正向)。 lim()dfzzik,,,,,,21,C,r0r
引理3 (Jordan引理):当时, (此z,,(0,argz,,,即Imz,0)f(z),0
imz限制条件为一致地趋于0,仅仅存在解析部分), 则 limf(z)edz,0,C,,RR
,其中是以原点为圆心,半径为的上半R(实常数m,0)CR
i,圆周,即 zR,,,e(0).,,
证明:由于当,时,一致地趋于0,z,,0,argz,,f(z)
,,0这意味着任给,存在(与无关的,D内各向同性argz
的)z,R,Mf(z),,,使当时,.据此,令M(,),0
i,imRcos,zR,,,e0,,,有 [Note:1). 2). 请检查||1.e,,,
是否满足条件,否则积分不存在或者重新补积分回路]
zRiR,,cossin,,,imzmRmR,,sinsin,,fzezfzezfzeR()d()d()d,,,,,,CC0RR ,,,,mRmRsinsin,,2,,ReRed2d.,,,,,,00
20,,,,/20,,,sin,由右图可见,当时,有. ,
2,,mR,,,imzmR,2,fzezRee()d2d1.,,,, ,,,,,,,,C0Rmm
imzm,0() 这样,就证明了. limf(z)edz,0,C,,RR
,f(x)dx2( ,-,
条件:(1)由f(x)唯一确定的函数f(z)除在上半平面(Imz,0)只有有限个奇
{},,,bbbb,?点以及在实轴上最多有有限个单阶极点kN12
{},,aaaa,?,以外是解析函数; jn12
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
(2) 在实轴和上半平面内,当时,一致地趋,,0,argz,,z,,zf(z)
于零,简记为. zf(z),0
Nn,结论: fxxifbifa()d2Res()Res().,,,,,,kj,-,上半平面内复平面实轴上1()1(),,kj
证明:(1)先考虑在实轴上没有奇点的情况。
以,R~R的实轴为一边,补充上半圆周: ,,Cz,RR
Rfzzfxxfzz()d()d()d,, ,,,CRC,R N
,,2Res().ifb,kkc,1()内
limzf(z),0因为 ,根据大圆弧引理, z,,
N,R,,. 于是,取极限就得到 fxxifb()d2Res().,,limf(z)dz,0,k,,-C,R,,R上半平面k1(),(2)实轴上存在一阶极点的情况。
设在实轴上有单极点,我们 z,cf(z)
所取路径必须绕过点。以点为圆心, cc
充分小为半径的半圆周,如图所示 Crr
将点挖去,则 c
crR,fzzfxxfzzfxxfzz()d()d()d()d()d,,,, ,,,,,CRCcrC,,rR N又
,=2Res().ifb,kk1(c)内,
当取两个极限时,我们有, R,,,r,0
c,rR,,,limf(x)dx,f(x)dx,f(x)dx, ,,,,,,Rc,r,,R,,,,
? limzf(z),0,由引理1,. limf(z)dz,0,z,,C,,RR
Res()lim()().fzzcfz,,z,c由于是单阶极点,所以 由引理2(注意积zc,zc,
0,,分方向),我们得到 (相位差,幅角变化). lim()dRes ()fzzifz,,,,zc,C0r,r
N,fxxifbifc()d2Res()Res().,,,,因此, ,k,-,上半平面内1(),k
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
如果实轴上存在有限个单阶极点,则
Nn, fxxifbifa()d2Res()Res().,,,,,,kj,-,上半平面内复平面实轴上1()1(),,kj
自证:如果将实轴上单极点包含在回路内,则
但在回路内多了lim()d(2)Res (),fzzifz,,,,,,zc,Cr,0r
结果相同。 lim()d2Res ().fzzifz,, ,zc,C0r,r
,,dx1dxExample 1. 计算积分 I,,. 22,,0,,x,12x,1
,/2,x,tan,[解一] 令,则(简单~) I,,d.,,02
Rddd1zxz [解二] ,,,,,,,,,, 2Res(i)2ifi222 ,,,,CRCR,,,zxzi1112
1R,,zi,,z(在下半平面)。取极限,因为lim,0,根据引理1,所以2z,,z,1
,zd1,,0I,f(x)dx,.. 于是得到, 2,,C,,Rz,122
,dx(X) Example 2. 计算积分 , 其中为正整数。 I,nn1,,2,,,,x,1
zi,n,1[解] 唯一孤立奇点 (阶极点, Im0)z,
Rdddzxz,,,,,2Res()ifi,,,111nnn ,,,,CRC222R,,,zxz111,,,,,,
n,,1d1(2)!n,,,2,i,,,,,1n22nnnzn!d2(!),zi,,,,,,,zi
nn,,d1(1)(1)(2)(2)1(1)(2)(2)1(2)!nnnnnnn,,,,,?? .,,,,,,12122,nnnnnd(2)2222!ziiinzi,,,,,,,,zi
1R,,取极限z,因为连续且,所以,根据引理1, fz()lim,0n,12z,,,,z,1
,zdx(2n)!d. 因此 I. ,0,,,n,1n12n2,,,22C,,R2(n!),,z,,x1,1,
,dxI, Example 3. 计算积分 . 2,,,,,xx,1,,x,1
(如图严格相互抵消,因而有值)
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
1[解] f(z),2,,zz,1,,z,1
z,i除在上半平面有奇点外,
z,0在实轴上还有两个单阶极点
和. 取积分路线如图,有 z,,1
又,,fzzfzzifi()d()d=2Res(),,,,,,,, ,,,,,,,,,ClClClC11223R,, 1i,,,,,,21.ii,,,iiii(1)()2,,
R,,取三个极限,, r,0,r,012
zd1因为 ,由引理1,. zlim,,0,022,Cz,,R,,zz,,zzz,,z,1,1(,1),1
1又因为 fz,,,Res(0)lim1,2z,0zzz,,11,,,,
11 所以,由引理2 (小圆弧引fz,,,,,,Res(1)lim(1).2z,,12zzz,,11,,,,
dzi,理),,,,,,,,ifRes(1)(),,2C12,,zzz(1)1,,
dz. 因此, ,,,,,,,,isfiRe(0)(),2C2,,zzz(1)1,,
,,Ifzzfzzlim()d[limlimlimlim]()d,,,,,,,,,,,,,, ,,lllCCCCRRRrr,,,,,,,,0012312R,,12rrrr,0,0,,1212
,,,ii(1)0().ii,,,,,,,,,,222
[附] 柯西积分主值(Cauchy principal value):
bf(x)dx,,通常的定积分(Riemann积分)有两个基本假设:积分区间a,b是,a
,,a,b有界的,同时函数f(x)在上也是有界的。反之,如果积分区间无界或f(x)有奇异性,则这种积分属于反常积分,这时积分就有一般值和主值之分。例如,按广义积分定义,双边无界积分
,xR02, f(x)dx,limf(x)dx,limf(x)dx,,,,,,,,,,RxRR1012
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
R,只当这两个积分存在时,才存在。考察极限,显然这是limf(x)dxf(x)dx,,,R,,R,,
前者的特殊情形。当此极限存在时,前者可能不存在,但前者存在时,后者必存在,并且两者相等。故通常称前者为一般值,把后者称为积分主值(按柯西的定
,,R义)。 Pfxxfxx()dlim()d.,,,R,R,,,,
同理,当在上有不连续点[在点无界且仅是单极点]时,,,a,bcccf(x)f(x)
bcb,,1其一般值定义为 而其主值定义为 fxxfxxfxx()dlim()dlim()d,,,,,,00aac,,,,,,212
bcb,,,,Pfxxfxxfxx()dlim()d()d.,,上面例3在三个地方均取了积分主值。 ,,,,,aac,0,,,,,
,imx3( (是非零实常数,Fourier Transform) f(x)edxm,-,
m,0条件:设,
(1) 由唯一确定的函数除在上半平面(只有有限个f(x)f(z)Imz,0)
(孤立)奇点并且在实轴上最多有有限个单阶极点以外是解析的;
(2) 在实轴和上半平面内,当时,. z,,f(z),0
z,,,,fxx()d(上面计算时要求, 利用Jordan lemma, zfz()0,,,,
z,,,imxf(x)edx现在计算Fourier Transform 时, 要求) fz()0.,,-,
,imximzimz,,,,fxexifzeifze()d2Res()Res().,,,,结论: ,,,,,,,-,上半平面内复平面实轴上
证明:(1)先考虑在实轴上没有奇点的情况。
,R~R,,Cz,R以的实轴为一边,补充上半圆: R
Rimzimximzfzezfxexfzez()d()d()d,, ,,,,CRCR 又imz,,,=2Res().ifze,,, 内C
limf(z),0因为 ,根据引理3(Jordan lemma) z,,
imzR,, 于是,取极限就得到lim()d0.fzez,,,,CRR
,imzimz,,fxexifze()d2Res().,, ,,,,-,上半平面内
(2)实轴上存在一阶极点的情况。
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
设在实轴上有单阶极点,我们所取路径必须绕过点,以点为圆z,cccf(z)
心,充分小为半径的半圆周,如图所示,则 Crr
crR,imzimzimzimzimzfzezfzezfzezfzezfzez()d()d()d()d()d,,,, ,,,,,CRcrCC,,Rr 又imz,,,=2Res().ifze,,,内C
当取两个极限时,我们有 R,,,r,0
crR,,imximximx,, lim()d()d()d.fxexfxexfxex,,,,,,,,,,,RcrR,,,,0r,
imzlimf(z),0因为,,根据引理3(Jordan lemma) lim()d0.fzez,,z,,,,CRR
imzimz,,Res()=lim()().fzezcfze,由于是的单阶极点,所以 z,cf(z),,,,zczc
由引理2(注意积分方向),我们得到
imzimz,,, lim()dRes()fzezifze,,,,,,C,0,rzcr
,imximzimc,,,,因此 fxexifzeifce()d2Res()Res().,,,,,,,,,,-,上半平面内
如果实轴上存在有限个单阶极点,则
,imximzimz,,,,fxexifzeifze()d2Res()Res().,,,, ,,,,,,,-,上半平面内复平面实轴上
,,imxfxmxxfxex()cosdRe()d;,变形: ,,--,,
,,imxfxmxxfxex()sindIm()d., Why? ,,--,,
imzimxmymymymy,,,||||m,0思考:的情况,满足各lemma条件)。 (||||eeeee,,,,下半平面内
,,,,cossinmxmx例1:,,, IxIxmd,d(0).cs22,,,,11xx,,,,
,,imxz,,e1fz()0.,,,,,,解: 满足Jordan lemma条件 IIiIxfzd().cs22,,,11xz,,
imzm,eeimzRes[()]||.,,efz1)mzi,,0:是一阶极点, zizi,,,2zii
1,,mmIieeII,,,,,,2,0 (odd function). csi2
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
1,,,imim()2)是一阶极点, 结果相同。 Iiee,,,,,2.mzi,,,0:,i2
,xmxsina,0 Example 2. 计算积分 (m是非零实常数,). ,Ixds22,,,,xa
imx,xem,0 [解一] 先设,且先考虑积分. I,xd22,,,x,a
imzimzimzRzezezedddzzz,,222222 ,,,,CRCR,,,zazaza
imzma,又,,zee,ma=2Res2.,,,,,,,iiie,,22za,2,,zai,
zR,,取极限,因为lim,0,根据引理3 22z,,za,
imzimz,zeze,ma, 于是得到 dz,0,,d.,Izie2222,,C,,Rz,a,za
,ma IIeII,,,,Im,Re0.,sc
,,xmxsinxmxsinma,m,0,,,,,,如果,则 Ixxedd.2222,,,,,,,,xaxa
m,0[解二] 仅考虑的情况。
,imx,xe考虑积分 'd.,Ix22,,,,xa
取积分路径如左图所示,因此
,,,imzimzimz,Rzezezeddd,,zzz222222 ,,,CRCR,,,zazaza
,,imzma又,,zee,ma,,,=2Res2.,,,,iiie,,22,2za,,zai,,
imz,zzeR,,lim,0dz,0取极限,因为,根据引理3:. 2222,Cz,,R'za,z,a
,,imzimz,,,zeze,ma,ma'dd.,,,,,,于是得到 IIe,,,Im'., Izzies2222,,,,,,,zaza
,sinxI,dx Example 3. 计算积分 . 1,0x
1z,0f(z), [解] 在实轴上有单阶极点, 取积z
17
Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU 分闭曲线如左图所示。考虑积分
ixizixixiziz又,,,Reeeeee ,d.ddddd=0.,,,,Ixzxxzz, ,,,,,,,,CRCC,R,xzxxzz
1取两个极限,前两项即为积分; 因为,根据Ilim,0R,,,,,0z,,z
izizeez,0Jordan lemma ; 又因为z(是实轴上的单阶dz,0lim,,1,C,0zRzz
ize极点),据lemma 2,. dz,i,1,(,,),,i,,C,z
1,I,i,因此. 所以,, IIIm.122
,,cosx副产品(严格相互抵消) ,,dRe0.xI,x,,
推论:
,,sinmxsinmx,dx,d(mx), (m,0),,00xmx2
,,sinmxsinmx,dx,,dx,, (m,0). ,,00xx2
2,sinx,,Example 4. . I,dx,,2,0x,,
2,,1sin11cos2xx,,,[解] Ixx,,dd.,,2,,2,,,,24xx,,
2cos212sinxx,,为和差化积,反之为积化和差(即倍频分析)。
2iz1,ez,0f(z),令,存在单阶极点,取积分闭曲线如上图所示,2z
iziz22,,R11,,ee,,0dd.,,,,,zz ,,,,,22,,,CRCC,R,,,zz
ix2,,e1I,dx取两个极限R,,,,,0,前两项即为. 因为2,,,x
2iz1,eizixyy2222,,z,,(Imz,0), z,,0[, ||||1(0)eeey,,,,2z
izixyyyy22222222,,,,|1||1|(1cos2)(sin2)1(),,,,,,,,,eeexexOe],
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
iz2iziz22,,1,e11ee,,根据引理1,. 又因为,dz,0Reslim2,,,,zi,,2,22C,z0Rzzz,,,z0
iz21,e据lemma 2,. dz,i,(,2i),(,,),,2,2,C,z
ix2ix2,,111,e,1,e因此. 所以. dx,I,2,Ix,,,,Red2,22,,2,,,,x442x
,,sin2x副产品: ,,d2Im0.xI2,x,,
ix2,12,,eiz其实 ,,Ixi,,,d|2.,z0,2,,xz
3,sinx,,(X) Example 5. ,Ixd.,,3,0x,,
22[解]?sin3sincos2cossin2sin(12sin)2sin(1sin),xxxxxxxxx,,,,,,
13 ?,,,sin(sin33sin).xxx4
33ixix,,,,,,1sin111sin33sinxeexx,,,,Ixxx,,,,ddd. ,,3,,,,,33,,,,,,2228xxix,,,,
33ixixizize,3esin3x,3sinxe,3ez,0f(z),因为, 但是由于是 的三阶极Im,333zxx
点(一般来说,如果积分路线上有被积函数的高阶极点,则此积分是发散的),需
3izize,3e,2z,0F(z),要重新构造 为使成为单阶极点,只要改取即可,Fz().3z
sin33sinxx,Im().Fx,此时仍有 取积分闭曲线如上图所示, 3x
iziziziz33,,Reeee,,,,3232,,0dd.,,,,,zz ,,,,,33,,,CRCC,R,,,zz
ixix3,e,e,32dx取两个极限R,,,,,0,前两项即为. 因为 3,,,x
3izize,3e,2z,,(Imz,0), z,,0. 3z
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
iziz3e,3e,2根据引理1: . 又因为dz,03,CRz
iziziziziziz333eeeeee,,,,,,323232Reslimlim,,,z332zz,,00zzzz,0 iziz3ieie3393,,,,,,,|3.,z022z
iziz3e,3e,2据lemma 2,. dz,i,(,3),(,,),3,i3,C,z
ixix3,e,3e,2因此 . dx,,3,i3,,,z
ixix3,13213ee,,,所以 Ix,,,,,,,Imd(3).,3,3,,888x
(X) 4(多值函数积分的两种类型
,,,1)xQ(x)dx(,是非整数) (1,0
条件:(1) 由所唯一确定的在全平面上的单值解析函数仅有有Q(x)Q(z)
限个奇点,其中在正实轴上最多有有限个单极点;
,z,0(2) 当和时,一致地趋于零。 z,,zQ(z)
解法:(1)先考虑在正实轴上没有奇点的情况。
,,1z由于是多值函数,它的支点是
z,0和,我们沿正实轴作割z,,
线,并考虑单值分支,0,argz,2,
因此
规定
关于下班后关闭电源的规定党章中关于入党时间的规定公务员考核规定下载规定办法文件下载宁波关于闷顶的规定
上岸,,l有,则在argz,01
,,下岸l有. 取闭积分路argz,2,2
径C如左图所示,因此
,,,-1-1-1,,,,zQ(z)dz,,,,zQ(z)dz,2,iReszQ(z),,,,,,ClClC,,1R2,,,0,argz,2,
R,,,,,0当取极限时,
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
,-1-1,,在上,, largz,0??zQ(z)dz,xQ(x)dx1,,l01
在上, largz,2,?2
00,-1ii,-122(1)-1,,,,,zQzzxeQxxexQxx()d()d()d,,,,,,,l,,2 , ?,0ii2-12-1,,,,,,,,,exQxxexQxx()d()d.,,,0
,z,0当和时,一致地趋于零,即 z,,zQ(z)?
,,1,,1,,,,limz,zQ(z),0limz,zQ(z),0,,由引理1,2, zz,,,0
,-1,-1, . limzQ(z)dz,0limzQ(z)dz,0?,,CC,R,,0,,R
,,,,,,,-1i2-1-1,,xQ(x)dx,exQ(x)dx,2iReszQ(z),因此 ,,,000,argz,2,
,2,i,,-1-1,,,xQxxzQz()dRes(),i2,,,,,01,ez0arg2,,,
i,,(1),,2ie,-1,,,,Res()zQz,ii,,,,,,,ee,z0arg2,,, -1,i,,,,,,Res()zeQz,,,,,sin,,0arg2z,,,,,
-1,,,,,,Res().zQz,,,,,sinarg()z,,,,,,,,(2)正实轴上存在一阶极点的情况。 设在正实轴上有单极点 Q(z)
,所取积分路径如图所示,则 z,c
取积分闭曲线如左图
,,,,,,,,,lClC12,R,,-1-1,,zQzzzQzz()d()d, ,C,,,,, ,,,,,,lClC'34,,,,
-1,,,,2Res()izQz,,,,,,0arg2z,
当取极限R,,,,,0,0,,时,
c-1-1,,largz,0在上,, ??zQ(z)dz,xQ(x)dx1,,0l1
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
,,-1,-1在上,, largz,0??zQ(z)dz,xQ(x)dx2,,lc2
在上,l,largz,2,?34
c,,-1ii,-12,2,,,-1, ?,,zQ(z)dz,xeQ(x)dx,,exQ(x)dx,,,lc,3
0c-1,-122-1,,,,,ii ?,,zQ(z)dz,xeQ(x)dx,,exQ(x)dx,,,0lc4
,z,0 当和时,一致地趋于零,即 z,,zQ(z)?
,,1,,1,,,,limz,zQ(z),0limz,zQ(z),0,,由引理1,2, zz,,,0
,-1,-1 , , limzQ(z)dz,0limzQ(z)dz,0?,,CC,R,,0,,R
i0现在考察沿,的积分,因为割线将点分开为(上岸)CC'cc,ce,,,
i2,,,0和(下岸),所以对而言,当时,时,Cz,cc,ce,,,
,,,,,,111,,,, lim()Res()Res(),zczQzzQzzQz,,,,,,i0,,,,zczczce,,,,,
,,,,,,111,,,,lim()Res()Res()zczQzzQzzQz,,,,,,i2,,,,,zczczce,,,,, i21,,,,,,,ezQzRes().i0,,zce,
根据引理2,
,,-1,1,,limzQ(z)dz,,i,ReszQ(z)i0,z,ceC,0,,
i,,-12,,,1 ,,limzQ(z)dz,,i,eReszQ(z)i0,z,ceC,'0,,
,,,,,i2-1-1,,,1,exQ(x)dx,2iReszQ(z),,,,0,,z,0arg2因此
i2,,,,1,,,i,,,1,eReszQ(z)i0z,ce
i2,,ie,1,,,,2i,,-1-11,,,,,,,xQxxzQzzQz()dRes()Res(),,,i0ii22,,,,,,,,,0,zce11,,ee,,0arg2z,
-1,,,,i1,,,,,,,,Res()ctgRes()zeQzzQz,,,,,,i0,,,,,zce,,sin,,0arg2z,,,
-1,1,,,,,,,Qz().,,,Res()ctgReszQzz,,,i0,,,,,zce,,,sin,,,argz,,,,
Q(z)推广到在正实轴上有有限个单极点的情况,
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
i2,,ie,1,,,,2i,,-1-11,,,,,,,xQxxzQzzQz()dRes()Res(),,,,ii22,,,,,,,,,011,,ee,,,0arg2arg0zz,
-1,,,,i1,,,,,,,,Res()ctgRes()zeQzzQz,,,,,,,,,,,,,sin,,,0arg2arg0zz,,,
-1,1,,,,,,,Qz().,,,Res()ctgReszQzz,,,,,,,,,,,sinzarg0,,,,argz,,,,
思考:取其它单值分支,结果如何,
1,,,x0,,,1,Example 1. 计算积分 ()(欧拉积分) Idx,0,x1
[解]沿正实轴作割线,考虑单值分支,并规定上岸有 ,,l0,argz,2,1
,则在下岸有. ,,largz,0argz,2,2
,,-1-1zz,,ddzz,,,, ,,,,,,,ClClC,,12Rr11zz,,
,1,,,zi,1,,i,,,,,,,,,,,,,2Res22.iieie,,z,1,i,,,ze
当取极限时, 在上,lR,,,r,0argz,0?1
-1-1,,,zx,dzdx, 在上, largz,2,??2,,l01,,z1x1
,-100-1,1xii,-122(1),,,,,,zQzzxexex()ddd,,,,,l,,2,,xx11 , ?-1,,xi2,,,,exd.,0,x1
,-1,-1zzzzlim,,0lim,,0 ,,由引理1,2, ?z,z,,0zz,1,1
,-1,-1zzlimdz,0limd0.,z, ?,,CCR,,r,0rRz,1,1z
,,-1-1,,xx2i,,i,,dx,edx,,2,i,e因此 . ,,00x,1x,1
,,,-1i,xie,,2,,,d.x,,, ,ii,i2,,,,,,0ee,xe,,11sin,,
,2i
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
,,1,x0,,,2Example 2. 计算积分 (). Ix,d2,0x,1
1,z,0[解法一(全平面)令],当和时,显然满足. Q(z),z,,zQ(z),02z,1
3,,ii,i,222仅有单极点和(), 因此 z,ez,eQ(z)zie,,,,
,,,,,,11,11,,,,,,ii,,,,Izeze,,ResRes,,,,,,22,,3,,,,iizzsin11,,,,,,,,22,,zeze,,,,
,,ii,,,(1)(1),,,,22ee,,,,,,,.,,iisin22,,,,,,2sin,,2[解法二](上、下半平面)
,,1zC设为如图所示的上半平面,fz(),,则 2z,1
,,,,11zz,,dd. ,,,,zz ,,,,,22ClClC,,12rR,,11,,zz
定义路经则 larg0,z,2
iii0,, lzxelzxeCzRe:,:,:(0).,,,,,,,21R
,,,,11zzz,0dd0,,显然zz(虽然支点处 ,,22CCrR,,11zz
是多值函数)。因为
,,,,11,0xxi,,()dd,()ded,,,fzzxfzzx 所以,,,,22ll0,21,,11xx
,,,,,,11/2i,,,xiziei22()/2,,,,dRes|.x,,, i,/2,,22ii,ze,,,,,0xezei,,,,11112sin(/2),,,,
1,,,x0,,,1,IdxExample 3. 计算积分 (). ,0,1x
1z,1Qz(),,[解法一] 令为唯一单阶极点(在实轴上)。1,z
,,1,,zI,,,,,,,,,ctgRes|ctg. i0,,ze,,11,z,,
,,l0,argz,2,[解法二] 沿正实轴作割线,考虑单值分支,并规定上岸有1
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
,则在下岸有. ,,largz,0argz,2,2
,,1,,1zz,,. dz,,,,,,,dz,0,,,,,,,,,,,ClClClClC'1,2R3,4,,,1,z1,z当取极限时, R,,,,,0,,,,
11,,,,czx,在上,. dzdxlargz,0??1,,0l1,,1z1x
,,1,,1,zx在上,,. dzdxlargz,0??2,,lc2,,1z1x
在上,l,largz,2,?34
,1,c,,-1z11ii2,2,,,-1,,,,,dzxedxexdx, ?,,,lc,3,,,1z1x1x
1,,0c-1,z1122-1,,,,ii,,,,,dzxedxexdx. ?,,,0lc4,,,1z1x1x
,,1,,1,,,,zzlim,,0lim,,0zz ,,由引理1,2, ?,,,,,0,,zz1,z1,z,,,,
,,1,,1zzlimdz,0limdz,0 , . ?,,CCR,0,,,,R1,z1,z
i0现在考察沿C,C'的积分,因为割线将实轴上单极点分为 z,e,,,
i2,(上岸),(下岸): z,e,
,,,1,1,,zz,,zz, lim,,Res,,1,,,z,z,zz1,1,i0,,z,e
,,,1,1,,zzi2,,,1,,limRes,,z,z,,,e. ,,,z,z,11,z,zi2,,,z,e
根据引理2,
,,1zlimdz,i(,1)(0,,),i,.,C,0,,1,z
,1,zi2,,i2,,limdz,i,,,e,,,2,ie,,,. ,C'0,,,1,z
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
,1,,xi,,2i2,,因此 . ,,,,1,edx,,i,1,e,01,x
,,,,,iiee,,,,,1i2,,xie,,,1,2即 . xd,,,,,ctg,,,i,,i,,i,,2,0eexe,1,1,
i,2
,n(2)(). lnxQ(x)dxn,0,1,2,?,0
条件:(1) 由所唯一确定的在全平面上的单值解析函数仅有有限Q(x)Q(z)
个奇点,其中在正实轴上最多有有限个单极点;
z,0(2) 当和时,一致地趋于零。 z,,zQ(z)
解法:(1)先考虑在正实轴上没有奇点的情况。
n,1n,1考虑,由于lnzf(z),lnzQ(z)
z,0是多值函数,它的支点是和,z,,
我们沿正实轴作割线,并考虑单值分支
,因此规定上岸有,,l0,argz,2,1
,则在下岸有.取,,largz,0argz,2,2
闭积分路径C如左图所示,因此
n11,,n,,ln()dln()dzQzzzQzz,,,, ,,,,,,,ClClC12R,,, n,1,,,2Resln().izQz,,,,,0arg2,,z
当取极限时, R,,,,,0
,11nn,, 在上,. largz,0??lnzQ(z)dz,lnxQ(x)dx1,,0l1
在l上,argz,2,?2
0,1n,112,nni,,,?,,,,lnzQ(z)dz,lnxeQ(x)dx,,lnx,i2,Q(x)dx,,,0l,2
z,0z,, 当和时,一致地趋于零,可以证明, zQ(z)?
n,1n,1,,,,limz,lnzQ(z),0limz,lnzQ(z),0,,由引理1,2, z,0z,,
nn,1,1, . 因此limlnzQ(z)dz,0limlnzQ(z)dz,0,,CC,R,0,,,R
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
,,n,1n,1n,1,,,,.lnxQ(x)dx,lnx,i2Q(x)dx,2iReslnzQ(z),,,,,000,argz,2,
,n分离实部和虚部,可以得到的递推表达式。 lnxQ(x)dx,0
,,,n,0例如, : . ,,lnxQ(x)dx,lnx,i2,Q(x)dx,,i2,Q(x)dx,,,000
,n,1,,Q(x)dx,,ReslnzQ(z)所以 . ,,00,argz,2,
,,22ln()dln2()dxQxxxiQxx,,,,,,,00n,1: ,,2,,,,22ln()d2()d.ixQxxiQxx,,,,,,00
,,,1n,1,,所以 lnxQ(x)dxIm2,iReslnzQ(z),,,,,022,,,0,argz,2,,,
,,33ln()dln2()dxQxxxiQxx,,,,,,,00n,2: ,,,232,,,,,,32ln()d32ln()d2()d.ixQxxixQxxiQxx,,,,,,,,,,000
??
当在实轴上有单极点时,只要注意到奇点在上、下岸的不同Q(z)
取值,同类型(1)相似做法。
(2)正实轴上存在一阶极点的情况。
设在正实轴上有单极点, z,cf(z)
所取积分路径如图所示,则
当取极限时, R,,,,,0,,,,
在上, largz,0?1
c11,,nn, ?lnzQ(z)dz,lnxQ(x)dx,,0l1
,nn11,,在l上,, argz,0??lnzQ(z)dz,lnxQ(x)dx2,,lc2
l,l在上argz,2,, ?34
c,nn11,,n1,, ?,,,,lnzQ(z)dz,lnx,i2,Q(x)dx,,lnx,i2,Q(x)dx,,,lc,3
0c11,,nn1,n, ?,,,,lnzQ(z)dz,lnx,i2,Q(x)dx,,lnx,i2,Q(x)dx,,,0lc4
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
,z,0 当和时,一致地趋于零,可以证明 z,,zQ(z)?
n,1n,1,,,,limz,lnzQ(z),0limz,lnzQ(z),0,. 由引理1,2, z,0z,,
nn,1,1 , . limlnzQ(z)dz,0limlnzQ(z)dz,0?,,CC,R,0,,,R
i0现在考察沿,的积分,因为割线将c点分开为(上岸)和CC'c,ce,,,
i2,,,0(下岸),所以对而言,当时,时,Cz,cc,ce,,,
nnn,,,111,,,,limln()Resln()Resln().zczQzzQzzQz,,,,,,i0,,,,zc,zczce,,,,
n,1nn,,11,, ,,limln()Resln()Resln2().zczQzzQzziQz,,,,,,,,,,,,i0,,zc,zc,,,zce,
根据引理2,
nn,,11,,limln()dResln(),zQzzizQz,,,i0,,,C,,0zce,,
n,1n,1,, limln()dResln2().zQzziziQz,,,,,,,,i0,,,C',0,zce,
因此,
,,n,1nn,,11,,ln()dln2()d2Resln()xQxxxiQxxizQz,,,,,,,,,,,,00,0arg2,,z n,1n,1,,,,,, Resln()Resln2().,,,izQzziQz,,,,i0,,i0,,,,zce,zce,,,
,1,IdxExample 1. 计算积分 . 3,0,x1
5,,iilnzi,33Q(z),[解一]令,它有三个单极点(),. 0,argz,2,z,e,z,e,z,e3z,1
因此
lnlnzz,,,,,,ddzz33 ,,,,,,,ClClC12R,,,,,z1z1
,,lnlnlnzzz,,,,,, ,,,,2ResResResi,,,,,,,,3335,,iiz1z1z1,,,i,,,,,,33,,,,zezeze,,
2,,2343i,,,,,,,2.i,,,,,39,,
取极限R,,,,,0,我们得到
,lnlnzx,dd,zx 33,,l01,,zx11
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
0,lnln2ln2zxixi,,,, ddd.zxx,,,333,,,l,02zxx,,,111
1
zzzlnlnz, z?lim,,lim,lim,lim,03333z,z,z,z,0000zzzz,1,12,12,1
2zz
1
zzzlnln1z, z?lim,,lim,lim,lim,lim,033332z,,z,,z,,z,,z,,zzzzz,1,12,12,16
2zz
lnzlnz (引理2), (引理1) ?limdz,0?limdz,033,,CC,,0R,,,Rz,1z,1
2,,lnln243xxii,,,因此, dd.xx,,,,,3300xx,,119
. I,23,9
1f(z),[解二] 令,去闭积分路径如图所示,3z,1
,i3再所围区域中仅包含单极点z,e.
1,, ,,fzzfzzi,()d()d2Res,,,,,3, ,,,,,,,,ClCli12R,,z,1,,3,ze
1i 2.,,,,2i3e3
2,i,113R,,,dzdxz,re取极限,,在上, l233,,l01,,z1x1
22,,ii0,,,11133zreexddd.,,,因此 ,,333,,,l,02,2zx,,11i,,,,3re,1,,
,,
11zlim,,0limdz,0又因为 ,由引理1,, 33,Cz,,,,RRz,1z,1
i23,2323,,ie,, I.i23,,19e
i,说明:如果f(x)满足,就可以考虑此算法,如f(x),f(xe)
,,11,,Ixd,Ixd., 41004100,,00,,x1x1
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
,dxExample 2. 计算积分 . I,2222,0,,,,x,alnx,,
22i,,i,x,0[解] 考虑到,当时,, ,,,,lnx,,,lnx,i,lnx,i,,ln,,,,xelnxe
1,i,因此,可选取,沿负实轴作割线,并取下岸,f(z),z,,e22,,z,alnz
,i,i,i02上岸. 有三个单极点, f(z)z,,ez,ae,z,e
,,fzzfzz()d()d,,,, ,,,,,,,ClClC12rR,,
,,,ii0i22,,,,2[Res()Res()Res()]ifaefaefe
,, 111,,,,,2i,2,,,,,,,,iiii1a,22222ln()2ln()aeaeaeae,,,,
,,11,2.i,,,,,,,2222ln41aaa,,,,,
取极限,我们得到 R,,,,,0
,,ii0,ee1,,z,,,ddd222222,,,,ii,0,l,,,,1z,az,ae,aelnlnln,,,,,,,,,, ,1,d,22,0,,,,,a,iln,,,
,i,,,e11z,,, dd,d,2222i,22,,,,00l2,,,,z,az,a,,,,e,a,i,,lnlnln,,,,,
221/()zza,?limlim0,z,,, 22zz,,00lnzzaz,ln,,
221/()zza,?limlim0z,,,, 22zz,,,,lnzzaz,ln,,
1?limdz,0 (引理2), 22,C,,0,,,z,alnz
1?limdz,0 (引理1), 因此 22,C,,RR,,z,alnz
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
,,11,,d,d2222,,00,,,,,,,,,a,,ln,i,,,a,,ln,i ,,,,,d11,,,,2i,,2i,,,,2222222,0,,,,,aln,,2alna,41a,,,,,,
将换成,即 ,x
,11. I,,2222alna,,41,a
5. 几个特殊积分(Taking special loops)
,,22(1)菲涅耳(Fresnel)积分,,. Ixx,sindIxx,cosdsc,,00
图 5.11 i,[解] 取大圆弧:则 zR,e,
22222i,zReRiR,,,cos2sin2.,,
2iz设,但不能取上半大圆周。这是因为f(z),e
,R,,,argz,,在左上半平面内,当时,2
2222izRiRR,sin2cos2|sin2|,,,不满足大圆||||,eeee,,,,
弧引理和Jordan引理所要求的条件,因而不能得
到 因此我们尝试如图5.11所示的fzz()d0.,,CR
积分路径(右上半平面部分),
22iizz,,0dd.,,,,ezez ,,,,,,ClClR,,12
2iRcos2,iRcos,R,,e当取极限时,和证明Jordan lemma相似,可以证明:e类似于,
22222izizRiRR,,sin2cos2sin2,,,R,, (). ezd0,||||0,eeee,,,,CR
,,22izix22在l上z,x, ??ezexxixxIiIddcossind.,,,,,,,1cs,,,l001
0,222iziyiy,,z,iy在l上, ??ezeiyieyiIiIddd.,,,,,,,,2cs,,,l,02
IiIiIiI,,,,0.II,因此, 从而. 它们的具体值呢, ,,,,cscscs
进一步将积分区域缩半,取图5.12所示闭合路径,我们有
22iziz,,0dd.,,,,ezez ''' ,,,,,,ClClR,,12
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
R,,当取极限时, 同样可以证明
2izR,, (). ezd0,,CR
2iz在上, z,xezIiId.,,l??1cs,l1
,i'4, z,rel:?2
2i,402ire,,izi,4ezeredd,,,',,,l2 (自证)。 ?,2,,iri44,,图 5.12 ,,,,eered,02
,,,222,i,4因此 从而 IiIei,,,,II,,..,,cscs4244
,RRRRR2222222,,,,,,,2()axayaxyaraR注意: Iexeyexyerre,,,,,dddddd(1).,,,,,,,4a000000
a,R,,:. 因此 I,2
(2) Gaussian Fourier Integral(高斯函数的富氏变换)
,,221axax,,I(b),ecosbxdx,ecosbxdx ,,0,,2
(实数 ab,,0,0).
2222,,,azaRiaRcos2sin2,,,az,ibz[解]设,这是解析函数,则 eee, f(z),e
,,3,,在发散(). 取如图5.13所示积分路径(弧状行不通,,?cos20,,,,,R44
hh试矩形状,正实数待定,二次配方可同时解决待定和问题)。因此
22--azibzazibz,,,,0dd.ezez,,,,, ,,,,,,,Cllll1234,,
2-az,ibzN,,当时,edz,0和 ,l2
2-azibz,ezd0., 证明如下: ,l4
,,lz,N,iy0,y,h在上,, 2
图 5.13
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
22h22-aN-yby,,,-azibz-azibz,,ezezeyddd,,,,,ll022 上式第一个等式用2222h--aN-haN-h,,,,,,,eyhed0. ,0
-aiNyibN2,22到了第二个不等式用到了maximum at最后一步||1,e,ayahby,,,,0,
2-az,ibzhN,.N,,用到了 同样可以证明: (). edz,0,l4
,22-z-aibzaxibx,,正是需要求的积分或与其相关(还带了副产品)。 ezexJbdd(),,,,l,,1
z,x,ih关键在于能否计算或用待求的表示。在上,,因此 lIb()3,l3
2,,,2222-i-2axhibxihaxibahx,,,,,,,,,,,-zaibzahbhahbh,,, ezexeexeJbahddd(2).,,,,,,,,,l,,,3
bb,2ah,0只要令,即取,上式就易积分。又由h,Jbah(2),2a
22bb,,,,,222,,,-ax--axibx,,axibx4a4aexd,,则可得edx,,e,0,即 exed.,, ,,,,,,,,,aaa
2b,,211,-,axibx4aIbexe,,所以 ()Red.,,,a22
* 仍为Gauss型,它是物理学第一大积分~ Ib()
实空间的Gauss分布在动量空间仍为Gauss分布。
,,2ax,** 副产品: ebxxsind0.,,,,
2bahbh,b,2ah,0h,JbeJbah()(2).,,取,即,上式就得到 *** 关系式: 2a
2b,2111,,ah4aIbJbJe,,,()Re()e(0). a222
'i,**** 如果取第I象限内直线积分 其中固定 ,,l:zrer,,,,,,(0,0/4),,,3
'C则按上例积分,大圆弧积分应该为零,但是的积分仍然求不出来: lR3
2ibbii22,,,-()aere,,,,2222ii,,-z-z-aibzaibziaeriberi,,,,,24aaezezeereerdddd,,,,',,,,00ll13 22bb,,,,11,,b,,why not44aa=1Erfi?ee,,,,,,,,win what' ay,,22aa2a,,,,
Erfi[]Erf[]/:zizi, imaginary error function虚误差函数是纯虚数,问题出在
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
2,,b,21,b,,,ax4aebxxesindErfi., (半无界破坏了空间的对称性~) ,,,2a2a,,0
complementary error function. Erfc[]1Erf[]:zz,,
''***** 有用积分公式[]: (cos)sin,(sin)cosbxxbxbxxbx,,,bb
,ncos2,4,6...bx22,,,,d()/4nnaxba,, Ibxexen()d,,,,,,,n,sin1,3,5...bx4dab,,,0
1dx(3) I,.,22,111,,xx,,
[解] Chapter 3 思考题:支点是什么类型的奇点, (Key:特别类,因为要割线,并非孤立奇点,但是挖走就行,并且围绕支点转一圈幅角有变化)
1,这是一个多值函数,它有设f(z),22,,1,zz,1
z,,1,1,1两个支点,沿实轴从到作割线,并规定在割线的上岸取和,由于在割线下arg(z,1),0arg(z,1),,
A'岸上的点(如点)可以看作是上岸上的相应点(A)逆
z,,1时针绕一周而达到,所以在割线的下岸有
而不变[教材上是这样选取arg(1)2,z,,,arg(z,1),,
z,1A'的]。当然也可以将看作是A顺时针绕一周而到达,这样不变,而[我们在这里是arg(z,1),0arg(z,1),,,
z,,i如此选取的]。取这样的单值分支后,有两个单极点。 f(z)
取积分曲线如上图,因此,
,,fzfzififi(z)d(z)d2Res()Res().,,,,,,,,, ,, ,,,,,,,,ClClCC'12rrR,,
当取两个极限R,,,r,0时,
dz1lim0.,z,由引理1, ?lim,,0,22C22R,,z,,R,,11zz,,zz1,,1,,
z,,1虽然是支点,我们也可以求它们的留数:
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
z11,1,即留数为零。 ,,z?lim,1,,lim,,0222z,1z,1zz1,,1,,zz1,,1
dz由引理2得, lim0.,,22Cr,0r11,,zz,,
111z,,即留数为零。 ?lim1lim0z,,,,,,,222zz,,,,1111,,zz11,,zz,,
zd由引理2得,. lim,0,22C',r0r,,zz1,,1
在上有和 larg(z,1),0arg(z,1),,?1
11ddidzxx, ,,,?,,,ii,222022l,,1111111111,,,,,,,zzxxexexx,,,,,,
,i,,i,注意这里用到了如下事实:虽然 但是 ei,,.e,1,
在上有:和,larg(z,1),0arg(z,1),,,?2
,11ddidzxx,,,,?,,,ii,l222022,11,21111111,,,,,,,zzxxexexx,,,,,,
111, fziRes(i)lim.,,,,,,2,,3z,iii22zzz111,,,,,,,,,44iieie,,211,,
111,fziRes(i)lim.,,,,,,,2,,3z,,i,,ii22zzz111,,,,,,,,,44iieie,,,,,,,211
1dx,,,因此 I. ,,2212,,11xx,,
2,,1dyyydd,xy,,, [解二]:令非线性变换: x,,,,21/223/223/22(1)(1)(1),,,yyy1,y,,
221,,yy22yy,,,1.则 积分上下限为 解得y,,1/2. 1,,,x,,,21,y
,1/2y,2cos,1/2,ddy,,,,,I.在以上变换下,此积分简化为 ,,2,1/2022,12y
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
bf(x),(4) . ,Idx(, 0)axb,,,,0,a,,,xxi0
这是理论物理中常用的一个积分,它的被积函数刚好在实轴的上方有一个单
,阶极点:,或下方有一个单极点: 因为正数 我们zxi,,,zxi,,,.,,0,00
可以等价地认为这极点是在实轴上,但积分路径是从的下方 [但是对应z,x0
2,,,,,,逆时针方向,相位差为] 或上方 [但是对应, 顺时zxi,,,zxi,,,00
0,,,,,针方向,相位差为] 沿小半圆绕过这个极点。不可以随意改变小半Cr
,圆的开口方向,否则当时绕不过这个极点了。设是解析的,而且r,0Cf(z)r
,我们有: ,,fx,00
bxrb,,,fxfxfzfx()()()()0 dlimddd.xxzx,,,,,,,,,aaCxr,0r,r0,,xxixxzzxx,,,,0000,,
xrbb,,,fxfxfx()()()0P limddd.xxx,,,,,,,axra,0r,0xxxxxx,,,000,,
fzf(z)()因为 ,,,,,,,所以 ,,. z,z,,fz,fxlimdx,,i,fxlim0000,Cz,z,r0r0z,zz,z00
bbfxfx()(),P,,0故 (). ,,,ddxxifx,,0,,aa,,,xxixx,00
物理上,时间顺序要求zxi,,,,即推迟(超前)效应:物理源首先发出信号,0
,0后有响应接收信号(先有响应接收到了信号,时间顺序后有产生它的源信号)。 还可以用符号形式将此式中被积函数除之外的部分记为: f(x)
11,,,0,,,P,, (). ixx,,0,,,xxixx,00
x这个式子的左边为Green函数,当x,为能谱时,其实部为准粒子能量,虚部0
为准粒子寿命的倒数(See Chapter 14). 源分布函数乘以Green函数,其积分(求和)亦即场叠加是个复变函数,其实部为主值积分,虚部与位相有关。
,xx,See the next chapter for the delta function:. ,,0
Home work: 5.1: (1), (4); 5.4: (1), (4); 5.5: (2), (4), (7); 5.7: (1), (3).
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
复习
i,zxiyre,,,0. 复变函数理论: ,, ,,,,,,,,2 (0,1,2,)nn…
, :CRC,,,fzuiv(),,,, 定理:Cauchy,,,
解析函数的唯一性(一致性)定理.
,,,,,,inn,,,22,,,,,,i22,,,,ieie,,, ,
,,1,1. ,2n,,,in21,,,2,,,iiee,,,, 1…
2/nikn,zze,,,1: ,zkn,0,2. 时,在实轴上; kk
zkn,,,,,1,2,,1knnn,,,,,,,1,2,,21时,在上半平面;k
z时,在下半平面。 k
ikn,,1/2/,,2nzze,,,1: ,zkn,,,,,0,1,2,,1 时,在上半kk
knnnn,,,,,,,,1,2,,21平面,时,在下半平面. 这是因zk
iii,,,,2n2n2n为在上半平面…,亦在上半平面;而在ze,ze,,ze,,0n1n,
i,,2nkn,2下半平面…,亦在下半平面。特别的,当为整ze,21n,
i,2n数时,zze,复原于了。 02n
NAzdfzzd0,,,,2:iA3. 尤其是 fzzfzzdd,,,,,,,,, , , ,,l,,bllk,zb,1k
i,2,Ariefi,AzbiAln()2,,,d2.,,,iA, zbre,,,i,i,0re
1()df,,nf!()d,,()n,4. fz(), ,fz().,n1 , ,ll,2iz2i,,,,z,,,
n,,z1zn5. ,,,,; ; ,,zzez,1,,,,,,,z!n100n,,n
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
21n,,zn…. sin1,,,,,zz,,,,,21!,n,,n,0
1,C,nnn6. ,(为奇点). RCzb,,,,limlimminzfzCzb,,,,,,,nknknn,,,,Cn0,n,1
kC,17. Laurent展开式的核心及如何寻找,且知 ,zzk
k11m,,,,C,k,1d2,,,,zCi ,,,,1,1mm,1m ,lk01.m,,zz,,,,,k,
k8. Residue, Res.fzC,,,,1k
Reslimfbzbfz,,1阶极点:. ,,,,,,zb,
m1d,1mm,1阶极点:fbzbfz,,. Reslim,,,,,,m,zbmz!d
9. 基本积分类型:
三大引理:小/大圆弧引理和Jordan引理:
2aR,,,,,,,a0,,aRaRsinsin,,22,,,,,,,eeed2d2d.,,,,,,,000aCR
N,1) ,z在上半平面fxxifzd2Res,,,,,,,kk,,,1,k
lim0zfz,内,. ,,z,,
,,11,2,,i,,,1zzzz,,ze,,,,,RiRsin,cosd2Res,,,,,2) , ,,,,,,,,0izi22z,1,,k,,zz,k
in an unit circle.
,Niaxiaz,,efxxiefzd2Res,,z3) ,在上图所示区域内,,,,,,k,,,1k,zz,k,,
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
lim,0fza,a,0(ifthen上半平面下半平面)。 ,,,z,,
,
P2ResRes.fxdxifzifx,,,,,,,,,,,,nn,上半平面实轴上-,
,,1,4) . xQxdx,,,0
,n,1n5) , fzzQz,ln.lnxQxdx,,,,,,,0
10. 四个特殊积分
,,2,221) IxdxIxdx,,,,cossin.cs,,004
,2ax,2) Iebxdx,cos,ab,0, ,,f,0
,1122,,Ixxdx11.,,,3) ,,i,,1,,
,1,2222,,Ixaxdxln.,,,,4) ,,,,a,,,0
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
1. 习题5.1(2):
k11,1122,,zz ,, ?,Res0fzIifz,2Res.,Iez,,d0?e,,,,,1,1,, ,2,z2kz!,,,0k
dz1’. 习题5.1(3) .2 ,zzsinz,1
21dz,,z,0解:是三阶极点,,求导两次,繁~ Res0limf,,,,,2z,02!dsinzz,,
111111111,3,,,. ,,Iii2,,,,,,,zz,,,,,,,z,233!3sin3!360zzzzzzsin3!360
2,d,2. , I.22,0,,12cos,,,
,1,dz11,,i,d,,解:令ze,(in an unit circle), ,,则() ,,cos,,z,,,iz,12z,,,
d/d12zizizi,. ,,,,,Ii2,2212, ,,,111/1/1,,,,,,,zzzzz,,,,,,,,,,,,,,z,,zz,,11
iax,exd,3. : Ia,,,,,,,xi
iaziazezdezd,,a,,,,,,,,,,,, Iaiei22(0);Ia0(0).,,,, , ,,,zizi,,
,dx4. ,PI.4,42,,,,xx14,,,,
111z,,1f,,,,Res1,在上平面;,在实轴上. ; zii,,2,,3z5420z,,1
1111ii,Res;Res2.fifi,,,, ,,,,33412152260ii
112ii,,,,,. Iii,,,,,,2,,4,,,,2020126015,,,,
nzq!dqn,,,,,1,0,1,2,,5. 试求,的导数形式表示. 由复I,,nnz,1 ,,r,,,2izeq
,nkI变函数理论可知,的幂级数表示为,试用已知的Ikq,,nnk1,
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU ,qk验证你所表示的,并显式表示出和. III,q,1021,q1k,
NN1,qk解:等比级数部分和. qq,,1,q,1k
,qk,,(递推)仍此式。 IqqIq?,?,,,,n1,2,3,,,Iqq,,,,,,nn,1,01,q1k,
,nk. Izkzz,,1,,,,,n1k,
z1zIz,,(正是上式,但导数不变)。 ,,1,,01,z11,,zz
,IzzIzn1,2,3.,,,,, ,,,,,,nn,1
,,122nn226. 类似于习题5.4(4),InN,,,,,,,,(sin,的sindsind()2n,,,0,4
周期为). ,
2n111d,,z,,i,ze,解:令,. IziC,,,2,21n,,,,, ,z,142iziz,,,,
2nmm111n,,kmkk,n下面考察其留数: ,, ?abCab,,CC?,,1,,,,,m,,,12nii42k0,,,
22112!nnnn,,,,,,,,,,,nnC,,2!!2!nn,其中,[Note ] ,,2n2nn,,,,11,,n!,,
nn,,,12!!21!!21!!nnn,,,,,,,,C11,,,n2nIiC,,,,2. ,22nn2n22nn42222!!22!!nnn,,2ii,,
nm,,07. 类似于习题5.5(4),设均为整数,计算
222mmm,,,xxz11,,,PPIxPxzddd. p上半,,, 222nnn,,0,,,12121xxz
k,i2nnz,,10ze,,解:由解出(kn,,,,,0,1,2,,21)均为单阶极点,其中上半k
n,1kn,,,,,1,2,,1knnn,,,,,,,1,2,,21平面有个(),而在下半平面.当
z,,1kn,0,knn,,,,,2,21,时,是实轴上的两个单阶极点。当,又返
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU 回到上半平面了。
11,11 Res fRes f,,,;1,,;1,,,,21m,21m,nznnzn,,2222z,1z,,1
21m,2m,ikz1kn,Res ze,,,:f,,k,21n,2nnz2k
21m,,i2mn,1n等比数列求和z11121,,em,1k,Res tan.fz,,,,,,,k21m,21n,i,,2222nznnink,1kn1,e
n,1112121mm,,,,,,,11 I,,,,,,2Res Res 1Res 1tantanifziffi,,,,,,,,,,,,,,,,pn,,22222ninnn,,k,1
ax,e8. 习题5.7(1)Ia,d. , 01,, x,,a,x,,,1e
aa,,11多值,,,yy,,x e,dRes .,,,,yIy,,a,01sin1sinyaya,,,,,,y,,1
azezf=e,,10:解:,在上半平面有无穷多个奇点(单阶极点)。z,,z,1e
ia2,fziefz,2=,不能取整个上半平面的积分路径,但, ,,,,
可以取回路如图
,,,,NN,2,ia2Reslim,if,ifxxefxxfzzddd,,,,,, ,,,,,,,,,,,,,,,NNN,,,,,,ll23,,,,
aNiy,,,azaNee2,e1,,aN,,,,,,,,,dd20, 1,01, fzdzdziyyeNa,,,,,,,,NzNiy,0lll222,1e,,11ee
ia,e2,,i2iaia,,,,122,,,,,,IeIiie有,. ,,aaiaia,,,i,,eeasine,
,lnxI,d.x9. 习题5.6(4) 2,02,1x,,
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
lnzz,0zi,,zi,C解:,是支点,是二阶极点,其中在内. f,z,,22,1z,,
i,,,i,dln12lnzzi,2 () ln0ln,,,ieRes fi,,,,,,,,,,2232d84zzizzizzi,,,,,,,,,,,,,zi,zi,
2,rR,,fzzfzzifiidd2Res,,,,,,,,, ,,,,,,,,, ,,,,,CRrCC,R,,rR24r,0
其中
22,lndlndzzRz,,2lnlnRRfzzRRRdd2 0;,,,,,,,,,,,,,,,,,,,,,2222,,,,CCC02222RRRz,1RR,,11,,,,z,1,,
22,,11,lndlndzzrz,,2lnlnrrfzzrrdd2 r0 0.,,,,,,,,,,,,,,,,,,,,2222,,,,CCC02222rrrz,1rrr,,,111,,,,,,,,rrRlnxi,,lnxi,, fzzxxddd,,,,,22,,,,,RRr22xx,,11,,,,
,,I,,,,24,,lndxx,,,,,,,故 有 2d,xii,,,22dx,,,002224,.,,xx11,,,,,2,024x,1,,,,
,,1,xI,,,d 01.x,10. 求积分 ,,,,01,x
,,1,,多值z,1,,,,,,,i0IRes解:e. ,,,,,,,,,sin1sinsinz,,,,,,,z,,,,arg,,,,,,,,
,t,ettI,dxe,t若,正是习题5.7(1). dd,xe,t,,,01,e
11. 解题指导书p52,2.3
,cossin,,,,cos,e,zsinsin2,,,,ezd,,I,d,,,,,,0 Hint: ,c,, ,i0,1z,54cos,,,ze2,,s,,
i,cossin,,,ie2,ee(),,IIiI,,,d.Consider cs,054cos,,,,,,
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
1i,,,,,,1ii,,In an unit circle ,,. ,,,,,eze,ez,coszzizzdd,,,,,,,2
iii,,,11zezee/2i,,z,ineeziziezied/d2,22 ,,,,,,,,Iie2.,,z2,out ,,,1,,zz11,,,21/2221/223zz,,,,,,52zz,,
,1,,cossin,,,,,221,,,,,, ,,Iexpcos,,c,,,321s,,,,,sinsin.,,,,2,,,
12. 习题3.12(p.55)
x1,,,z,2,,,1m2z,,Jxmx,,,,,eJxz,cossind.,,,,,,, ,mm,02,m,,,
i,证明: 令(for convenience, consider in an unit circle) ze,
1ii,,,zeei,,,,,, 由Laurent 级数展开定理得 2sinz
x1,,z,,,2z,,,211eiximsin,,,,,,Jxzeeidd,,mm,1 ,,022ii,,,z0,, 0,,,z
2,1,,,,,, dcossinsinsin.xmxm,,,,,,,,,,,,02,
aTT,ftTft,,fttfttdd,,For the periodical function , ,,,,,,,,,,0a
2,,sinsindsinsind0xmxm,,,,,,,,,,we have (add function). ,,,,,,,0,
zz13. 求在处的留数。 z,,1,2,f,,,2(1)2zz,,,,
,z解:Res 1lim(1)()|1;,,,,,zfz f,,z,12z,1(2),z
'dz,,2 Res zfz,,,,f2lim(2)()|1;,,z,,,2z,2,z1dz,,
Res Res Res fff,,,,,[12]0. ,,,,,,
2,1I,,,,d 01.x14. 求积分 ,,,,0,1cosx,
ix解:复平面单位园上 ze,,dd.zizx,
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
2,11d2dzz I,,d =.x,,,, zz,,0||1||11112,ixix,iiz2,,,,,,,,1()1()1eezzz,22z
令被积函数的分母为零,即由因式分解及求根公式可得 (zz,)()zz,,,
111122, 其中并且 z,,,,1,,,b,,,,|z,,,,44(1)0,ac|(11)1.,,,22,,,,
21412,,故 Izz,,,i(,2)|.,,,,zz,2izzz,,(zzz,)(),,,1,,,,,
,115. 求积分 I,d.x22,,,,,(4)(2)xx
1解:令则 fz(),,2zz,,(4)(2)
2,,,,,fzzifi()d2(2).,Res ,,,,,, CC,,2rR
由于
fzziffzzzfz()d(2),()d0[()|0],,,,,,Res z,,,, CCrR
所以
2(1),,,ii, Res Res Iifiif,,,,,2(2)(2)||.,,222ziz,,2(2)(2)48zizz,,,
,xkxcos216. 求积分为实数,并且 I,,,d(,,xkk,,,0,40).,,c2,,,,,xx,,
ikxxez解:因为 所以令 fz(),.fx(),Re,22,,,,,,,,zzxx
ikzfzez()d,fz()|,0,考查积分 回路如图所示。因为 由Jordan引理 z,,, C
ikz 令被积函数的分母为零, fzez()d0.,, CR
(zz,)()zz,即由因式分解及判别式 ,,
22 可得 ,,b,,,,440,ac,,
,,,4,,0:z,,,,1(1) 全是实数。 ,222,
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Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
ikzikz,,Iifzefze,,,[()()]Res Res ,,,
,,,,,[(coscos)(sinsin)].izkzzkzzkzzkz,,,,,,,,2,4,,
,,,4,,0:(2)zi,,,, 分别位于上、下半平面。 1,222,
ikz,zeikz,,,,2()2,,Res Iifzei,,,zz,,
12,,4,,k,141141,,,2,,,,,,[(cos1sin)(sin1cos)].ekkikk,,,,2222222,,4,,,
,xkxsin故 I,,I,Re.IdIm.xI,,sc2,,,,,,,,,xx
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