Chapter 5 定积分计算
Methods of Mathematical Physics (2012.03) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
定积分积算Chapter 5
留定理及其积用定积分、积分主积数——Abstracts:
、1留定理和留的求法数数(Residue theorem and methods of finding residues)
zf(z)f(z)z,1留的定积,积数是解析函数的孤立奇点~积在的留定积数00
1cz()Resf(z)=fzdz积~其中积积的积曲积且部无其奇点;内它00?cπ2i
Resf(z)Resf(z)积分沿正方向积行,~积积号或.z=z0?0c
(0
b>0)?0abcosx+
解[]
2?1:,?zz,,22,,22??2i::11()z11z1()其中 ==?=?,F(z)12?+()()??:,2aiz2bi2bizzzzzzz2212+++zzz1ab,,b2::
2222?+????aabaabzz=1. zz==, ,1212bb
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Methods of Mathematical Physics (2012.03) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
2xx,xxCAxxBA=+=?/,/积足积大定理的根两个[: ].AxBxC++=0121212
(z<1)F(z)z=z在积位积内有孤立奇点两个;二积奇点,~;一积奇z=0:1点,。因此~我积有
d1a2 Res(0)lim().FzFzzz==?+=()122 z 0d2izbib
1i22Res()lim().FzzzFzzzab=?=??=? ()()1112 zz 12ibb
12π 22IFFzzaab= +=? =??2iRes(0)Res()2i2.ππ[])11( 22ibb
三引理,个
R?z?aMarg(z?a)M(ε)>0~存在与无积的~使当积~ε>0[]
(z?a)f(z)?K<εf(z)dz?iK()θ?θ<εθ()?θ~所以;大模之比2121?CR
limf(z)dz=iK()θ?θ~积积积分相位了,~ 即;方向正向,RR/1=.21?CR??R
00积~存在与无积~使当积~ε>0[]
(z?a)f(z)?K<εf(z)dz?iK()θ?θ<εθ()?θ~所以~即2121?Cr
limf(z)dz=iK()θ?θ;方向正向,.21?Cr?0r
z??fzzz()0(0arg,Im0) π即引理引理,当积 3 (Jordan)(
imzlimf(z)edz=0(积常数m>0)C 积 ~其中是以原点积限制条件),?R??RCR
iθ积心~半积径的上半积周即,zR= 00
z=R>Mf(z)<εM(ε)>0的,~使当积~.
iθimRcosθzR= e0θπ据此~令~有()(Note||1e=)
zRiR=+cossinθθπimzmRmR??sinsinθθfzezfzezfzeR()d()d()d =θ CC0RR
ππ??mRmRsinsinθθ2<=ReRed2d.εθεθ 00
20?θ?sinθ由右积可积~当积~有0?θ?π/2.π
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Methods of Mathematical Physics (2012.03) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
2π?mRθππimzmR?2π积积~就积明了(m>0) fzezRee()d2d1.<=?<εθεε() C0Rmm
imzlimf(z)edz=0.???RCR
?f(x)dx,2?-
f(x)f(z)Imz>0)条件,由唯一定的解析函确数在上半平面;只有有(1)?
{},,,,bbbb=,限奇点个在积积上最多有有限积积积点个极kN12
{},,;aaaa=,jn12
zf(z)()z??0?argz?π在积积和上半平面内~当积~一致地积于零(2)
zf(z)?0积积积.
Nn fxxfbfa()d2iRes()iRes().=+ππ积积, kj - k1()1()==上半平面积积上j积明,;,先考积在积积上有奇点的情。没况1
C()z=R以的积积积一积~积充上半积周:?R~RR
Rfzzfxxfzz()d()d()d=+, CRC?R
N
=2iRes().fbπ,kkc=1()内
limzf(z)=0因积 ~根据大积弧z??
引理,
limf(z)dz=0于是~取限极就得到R??. ?C??RR
N fxxfb()d2iRes().=π,k - k1()=上半平面
;,积积上存在一积点的情。极况2
f(z)z=c积在积积上有积点极~我积
cc所取路必积积积径点~以点积积心~
rC充分小积半的半积周径~如积所示~r
积
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Methods of Mathematical Physics (2012.03) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
crR?fzzfxxfxxfzzfzz()d()d()d()d()d=+++, CRcrCC?+Rr
N
=2iRes().πfb,kk1(c)=内
R??,r?0当极取限积~我积有~
c?rR?,,limf(x)dx+f(x)dx=f(x)dx~???,,RcrR,,?+????
, limzf(z)=0limf(z)dz=0~由引理~1.?Cz????RR
lim(z?c)f(z)=Resf(z)z=c由于是积积点~所以极由引理;注意积. 2z=czc?
lim()diRes ()fzzfz=?π?π分方向,~我积得到 ;相位差~幅角积化 zc=Cr 0r
N fxxfbfc()d2iRes()iRes().=+ππ,因此~. ,k - k1()=上半平面
如果积积上存在有限积积积点~积个极
Nn fxxfbfa()d2iRes()iRes().=+ππ kj - k1()1()j==上半平面积积上
dx1dx??==I积算积分 Example 1. .22??0x+12x+1??
Rddd1zxz2iRes(i)2i.f=+= = =πππ解一[] 222, CRC?Rzxz+++1112i
zd1zlim=0=0取限极~因积~所以~根据引理~R??1.22?Cz??Rzz1+1+
1?πI=f(x)dx=.于是得到~?22??π/2πI==d.~积解二令积积,θx=tanθ[] () 02
dx?=In积算积分 其中积正整。数Example 2. , n1?2+()x+1??
Im0)z> 积点~极(zi=解唯一孤立奇点[] n+1
Rdddzxz2iRes(i)f=+= πnnn+++111, CRC?222Rzxz111+++()()()
n 1d1(2)!n2i,= = ππn+1nn22nzn!d2(!)zi+() zi=
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Methods of Mathematical Physics (2012.03) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
nn d1(1)(1)(2)(2)1(1)(2)(2)1(2)!?++++nnnnnnn,,===. n+1nnnn2122+d(2)2222!ziiinzi+() zi=
1zlim=0fz()取限极~因积积积且~所以~根据引理~R??1n+12z??()z1+
zd=0因此 . n+1?C2R()z1+
dx(2n)!?I==π.n12n2?2+2(n!)??()x1+
dx?=I 积算积分 Example 3. .2?x()x+1()x+1??
如积积格相互抵消~因而有积()
1=f(z)解[] 2z()z+1()z+1
除在上半平面有奇点z=i外~
在积积上积有积积积点两个极z=0
和取积分路积如积~有. z=?1
=+++++= fzzfzzif()d()d2Res(i)π, ClClClC 11223R
1iπ= =?+21.ii()π++iiii(1)()2
r?0,r?0取限极~~R??12
zd1zlim?=0=0因积 ~由引理~1.22?Cz??Rz()z()zzz()z11++(1)1++
1limRes(0)1zf ==又因积 ~ 2z 0zzz++11()()
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Methods of Mathematical Physics (2012.03) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
11lim(1)Res(1)z+?=f?=?所以~由引理小积弧引. 2(2z??1()1()12zzz++
dziπ= ? ?=ifRes(1)()π理~~) 2C12zzz(1)1++()
dz= ?=?isfiRe(0)()ππ因此~. 2C2zzz(1)1++()
Ifzzfzz=++=???lim()d[limlimlimlim]()d , lllCCCCRRRrr 0012312R 12rrrr,0,0 1212
iiπππ=??????=?(1)0().iiπ222
附柯西积分主积,[] (Cauchy principal value)
b[]a,bf(x)dx通常的定积分;积分,有基本假积,积分积两个区是Riemann?a
f(x)f(x)[]a,b有界的~同积函数在上是有界的。如果积分积无界或区无界~积积积
积分于反常积分~积积积分就有一般积和主积之分。例如~按积积分定积~积无界积分属广双?xR02f(x)dx=limf(x)dx+limf(x)dx~??????????RxRR1012
R?f(x)dx只积积积积分存在积~当两个才存在。考察限极limf(x)dx~积然积是?????RR
??前者的特殊情形。此限存在积~前者可能不存在~但前者存在积~后者必存在当极~
并两称称且者相等。故通常前者积一般积~把后者积积分主积;按柯西的定积,。+ RPfxxfxx()dlim()d.= ?RR ?
cccf(x)f(x)[]a,b同理~当在上有不积积点;在点无界且是积点,积~极
bcb?ε1fxxfxxfxx()dlim()dlim()d,=+其一般积定积积 而其主积定积积 aac+ 00εεε212
bcb?ε =+Pfxxfxxfxx()dlim()d()d. aac+ 0εε
?imxmf(x)edx, 是非零积常数(,Fourier Transform)3?-
?条件,积~m>0
f(x)f(z)Imz>0)由唯一定的解析函确数在上半平面;只有(1)
有限奇点~在积积上最多有有限积积积点~个个极
f(z)?0z?? 在积积和上半平面~内当积~(2) .
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Methods of Mathematical Physics (2012.03) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
+ z z 积即此,Jordan lemma ;积算,fxdx()zfz()0 fz()0. ?
imximzimz fxexfzefze()d2iRes()iRes().=+ππ积积, - 上半平面积积上
积明,;,先考积在积积上有奇点的情。没况1
C()z=R以的积积积一积~积充上半积周,?R~RR
Rimzimximzimz fzezfxexfzezfze()d()d()d2iRes().=+=π,, CRC?RC内
limf(z)=0因积 ~根据引理;, 3Jordan lemmaz??
imzlim()d0.fzez= 于是~取限极就得到R?? CR R
imzimz fxexfze()d2iRes().=π, - 上半平面
;,积积上存在一积点的情极况。2
f(z)z=c积在积积上有积积点极~
cc我积所取路必积积积径点~以
r点积积心~充分小积半的半积径
C周~如积所示~积r
crR?imzimzimzimzimzfzezfzezfzezfzezfzez()d()d()d()d()d=+++, CRcrCC?+Rr
imz =2iRes().fzeπ, C内
R??,r?0当极取限积~我积有
crR? imximximx +=lim()d()d()d.fxexfxexfxex ?+? RcrR r 0
imzlimf(z)=0lim()d0.fzez=因积~~根据引理;, 3Jordan lemma Cz??R R
f(z)z=c由于是的积积点~所以极
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Methods of Mathematical Physics (2012.03) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
imzimz lim()()Res().zcfzefze?= zc zc=由引理;注意积分方向,~我积得到2
imzimz[]limf(z)edz=?πiResf(z)e~?z=cC0r?r
imzimzimz fxexfzefce()d2iRes()iRes().=+ππ因此, - 上半平面
如果积积上存在有限积积积点~积个极
imzimzimz fxexfzefze()d2iRes()iRes().=+ππ -上半平面积积上
imx积化,fxmxxfxex()cosdRe()d;= --
imx fxmxxfxex()sindIm()d.=Why? --
imzimxmymy??思考,的情,况m<0(||||).eee==
+ cosmxIdxm(0).= 例,1c2 1+x?
+ imxe1IIiIdxfz,().=+==解,cs22 11++xz?
imzm?eeimzmzi>=0:,1是一积点~极Res[()]||.efz==zizi==+2zii
1??mm IieeII====2,0ππ(odd function).cs2i
1???imim()mzi<=?0:,2Iiee=?=2.ππ是一积点~极?i2
?xsinmx=Idx 积算积分 是非零积常~数a>0Example 2. (m).S22?+xa??
imx?xe= 解一先积~且先考积积分m>0[] .Idx22?+??xa
imzimzimzRzezezedddzzz=+222222, CRC?Rzazaza+++
imzma? zee?ma= = =2iRes2ii.eπππ 22za+2 zai=
zlim=0取限极~因积~根据引理R??3 22z??za+
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Methods of Mathematical Physics (2012.03) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
imzimz zeze?ma于是得到, dz=0Ize==di.π22?22 C? Rza+za+
?maIIeII====Im,Re0.πSC
xmxsinxmxsin?ma如果~积m<0Ixxe==?=?dd.π2222 ? ? xaxa++
的情。况解二积考积m<0[]
?imx xe考积积分Ix'd.=22 ? xa+
取积分路如左积所示~因此径
???imzimzimz?Rzezezedddzzz=+222222, CRCRzazaza+++
??imzma zee?ma= = =2iRes2ii.πππe 22za+2 zai=?
imz?zzelim=0取限极~因积~根据引理,R??3dz=02222?z??CzaR'+za+
.
??imzimz ? zeze?ma?ma于是得到 Im'.IIe==?πIzze'ddi.==?=?πs2222 ? zaza++
?sinxI=dx 积算积分 Example 3. .1?0x
1f(z)= 解在积积上有积积点极取积z=0[] , z
分积曲积如左积所示。考积积分
ix eIx=d. ? x
izixixiziz?δReeeee zxxzzddddd0.=+++=, CRCC?δRδzxxzz
1R??,δ?0lim=0取限极~前积积积积分两即~因积~根据Jordan Iz??zizizeez~又因积;是积积上的积积点,~据极z=0lemma dz=0lim?=1?C?0Rzzz
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Methods of Mathematical Physics (2012.03) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
ize~因此I=iπlemma 2. . dz=i?1?(?π)=?iπ?Cδz
1πII==Im.所以 122
+ cosxdxI==Re0.副积品积格相互抵消() x?
推积,
??sinmxsinmxπdx=d(mx)= (m>0).??00xmx2
sinmxsinmx??πdx=?dx=? (m<0).??00xx2
2:,?sinx=Example 4. Idx.,,2?0x::
2 1sin11cos2xx? 解[] Ixx==dd.2 2 ? ? 24xx
2积和差化积~反之积积化和差~积倍积分析。即cos212sinxx=?
2iz1?e令~存在积积点极~取积分积曲积如上积所示~z=0f(z)=2z
iziz22R?δ11??ee 0dd.==+++zz22, CRCC?δRδ zz
2ix?1e?R??,δ?0取限极~前积积积两即因积. Idx?2?x??
2ize?1izixyy2222??~[||||1(0)eeey== ]zzz??(Im?0), ??02z
iz2?1e根据引理~又因积1. dz=02?CRz
i2zi2z,,??1e1elimz?=Res=?2i~据~lemma 222,,?z0zzz=0,,
iz2?1e.dz=i?(?2i)?(?π)=?2π2?Cδz
2ix2ix ??1e111?eπ因此所以. . dx=I=2πIx== =Red2π222? ? ??442xx
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Methods of Mathematical Physics (2012.03) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
+ sin2xdxI==2Re0.副积品,2 x?
ix2 12??eiz其积 Ixi===d|2.ππz=02 ? xz
3 sinx Example 5. Ix=d.3 0x
22解[]Qsin3sincos2cossin2sin(12sin)2sin(1sin),xxxxxxxxx=+=?+?
13?=?+sin(sin33sin).xxx4
33ixix? 1sin111sin33sinxeexx?? Ixxx===?ddd. 3 33 ? ? ? 2228xxix
33ixixizize?3esin3x?3sinxe?3e因积但是由于是 z=0, Imf(z)==333xxz
的三积点极一般积~如果积分路积上有被积函的高积点~积此积分是积散的来数极~()
3izize?3e+2Fz().需要重新造构 积使成积积积点~只要改取极即z=0F(z)=3z
sin33sinxx?Im().Fx=可~此积仍有3x
取积分积曲积如上积所示~
iziziziz33?δReeee?+?+3232 0dd.==+++zz33, CRCC?δRδ zz
3ixix?e3e2?+R??,δ?0取限极~前积积积两即因积. dx3?x??
3iziz?+ee32.zzz??(Im?0), ??03z
iziz3?+e3e2根据引理又因积1: . dz=03?CRz
iziziziziziz333eeeeee?+?+?+323232limReslimz ==332zz00 zzzz=0
iziz3ieie3393??+===?|3.z=022z
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Methods of Mathematical Physics (2012.03) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
iziz3?+e3e2据~lemma 2. dz=i?(?3)?(?π)=3πi3?Cδz
3ixix?+?e3e2因此 .dx=?3πi3???z
ixix3 13213ee?+π所以 Ix=?=? ?=Imd(3).π33 ? 888x
,多积函积分的积积型数两*4
1?α?xQ(x)dx;,;积是非整,数1?0
Q(x)Q(z) 件,条由所唯一定的在全平面上的积积解析函确数积有有(1)
限奇点~其中在正积积上最多有有限积积点~个个极
αz??zQ(z)当和积~一致地积于零。z?0(2)
解法,;,先考积在正积积上有奇点的情。没况1
α?1由于是多积函~的支点数它z
是和~我积沿正积积作z=0z=?
0?argz?2π割积~考积积积分支并~
argz=0因此积定上岸()l有~积1
argz=2π在下岸()l有取积积分. 2
路径如左积所示~因此C
ααα-1-1-1,,[]zQ(z)dz=+++zQ(z)dz=2πiReszQ(z)?ClClC?????1R2δ00[]
1?iπ=f(z)z=ρe因此~可积取~沿积积积作割积~取下岸并~22()z+alnz
πiiπ?f(z)i0z=ρe上岸有三积积点个极~ . 2z=ae,z=e
fzzfzz()d()d=+++, ClClC 12rR
ππii?i022=++2[Res()Res()Res()]πifaefaefe
111 =++2iπ2ππππ iiii??1a+22222ln()2ln()aeaeaeae
11π2.i=? ?π 2222ln41aaa++ π
R??,δ?0取限极~我积得到
ππ?ii0ee1==?ρρzddd???222222+++?iπiπ0ρρρρzazaeaellnlnln1()()()()()
?1=dρ22?0+a()+i()lnρρπ
πi???e11==?zddρdρ222222iπ????00l2()z+az+ae()()+a()?i()lnlnlnρρρρπ
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Methods of Mathematical Physics (2012.03) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
221/()zza+Qlimlim0,z ==22zz 00lnzzaz+ln()
221/()zza+Qlimlim0z ==~22zz lnzzaz+ln()
1?limdz=0 ;引理,~222?Cδ?0δ()zalnz+
1?limdz=0 引理因此(1), 22?C??RR()zalnz+
??11ρρ?dd??2222ρρπρρπ+++?00alnialni()()()()
ρπ,,?d11ππ2i2i=?=???,,2222222?0()()ρalnρπ2alnaπ41a++++,,
ρx将积成~即
π11=?I.2222alna+π41+a
,4几个特殊积分(Taking special loops)
22;,1菲涅耳;,积分~~FresnelIxx=sindIxx=cosd.sc 00
积 5.11解[]
iiθθ22222zRzReRiR===+e,cos2sin2.θθ
2iz积~不能取上半积周~因积在f(z)=e
π0,b>0)I(b)=ecosbxdx=ecosbxdx . ??02??
2?az+ibz解积~积是解析函~数[]f(z)=e
222???azaRiaRcos2sin2θθ积 eee=
ππ3cos20,θ< R在积散(<<). θ44
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Methods of Mathematical Physics (2012.03) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
取如积所示积分路径正积数待定~弧行不通~二次配方可解积积状决。5.13(h)
因此
22--azibzazibz++ 0dd.==+++ezez, Cllll1234 2azibz-+edz=0当积~~N???l2
2-azibz+ezd0.= 积明如下, l4
z=N+iy()0?y?hl在上~2
积 5.13
22h22-aN-yby?()++-azibz-azibzezezeyddd = ll022
2222h--aN-haN-h()() = eyhed0. 0
-aiNyibN2+上式第一等式用到了个||1.e=
2azibz-+edz=0同积可以积明, N??().?l4
?22-azibz-axibx++edz=edx正是需要求的。??l1??
lIb()积积在于能否积算出或者用表示。在上~~因此z=x+ih3 l3
22? 22-i-2axhibxihaxibahx++++?()()()-zaibzahbh+?ezexeexddd.==? l ? 3
b 2π-axh=只要令~取即~又由~b?2ah=0exd= ? 2aa
22bb?? ?2ππ2-axibx+-axibx4a+4a积可得~ 即edx??e=0exed.= ? ? aa??
2b? 211π-axibx+4a所以 仍积型。 GaussIbexe()Red== ? a22
物理第一大积分,积空积的学分布在积量空积仍积分布(GaussGauss)
+ 2?axebxxsind0.=副积品, ?
ncos2,4,6...bx22 πd??()/4nnaxbaIbxexen()d,=== n sin1,3,5...bx4dab 0
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Methods of Mathematical Physics (2012.03) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
1dxI=.;, 322?1+?11xx()
1=f(z)~积是一解积[]22()1+zz?1
个数它两个多积函~有支点~z=?1沿积积从到作割积~积定在割积并?1+1
arg(z+1)=0的上岸取和
arg(z?1)=π~由于在割积下岸上的点;如点,可以看作是上岸上的A'
相积点;,逆积积积一周而到~所以达Az=?1
arg(z+1)=2π在割积的下岸有~而arg(z?1)=π不积教材上是积积积取的。然也当 [ ]可以将看作是积积积积一周而到~达AA'z=1arg(z+1)=0arg(z?1)=?π积积不积~而积里[
f(z)积积积取。取积积的积积分支后~有积积两个极]
点z=?i。取积分曲积如上积~因此~
=++++=+?fzfzififi(z)d(z)d2Res()Res().π[], ClClCC' 12rrR
R??,r?0当极取限积,
dz1zlim0.=,lim?=0~由引理~ 122CR 22z??R11+?zz()zz()11+?
z?111z(),lim?1?=lim?=0~由引理~2222z?z?11zz11++()zz11+?
dzlim0.= 22Cr 0r11+?zz()
111z+Qlim1lim0z+ = =()~由引理~222211zz ? 11+?zz11+?zz()
zdlim=0.?22C'?r0r()zz11+?
arg(z+1)=0arg(z?1)=πl在上有和,1
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Methods of Mathematical Physics (2012.03) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
1ddzx= 2220iiπl?1111111zzxxexe+?++?()()
l在上有?,21?idx=, ?12211+?xx()()
arg(z+1)=0arg(z?1)=?π和~?
?dzdxidx11==????,?iiπ?l222202211()()()()1+zz?11+xx+1ex?1e1+x1?x
1Res(i)limfzi=?()2z i111++?zzz()()()
11?==.3ππ22ii44211iieie+?()
1Res(i)limfzi?=+()2z ?i111++?zzz()()()
11?==.ππ322??ii44??+??211iieie()
xd1π=因此 .??2212()()xx11+?
yy==xx解二,令~或者令~积此积分可用以前积的方法解。学[]221?y1+y
bf(x)+=Idx(, 0)axb<< ε;, 4.?0a?,εxxi0
积是理积物理中常用的一积分~的被积函积好在积积的上方有以积积点个它数极
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Methods of Mathematical Physics (2012.03) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU
zxi=+?()zxi=?+()εε或下方有以积点极~因积正数~我积可以等价地积ε?000
z=xzxi=+?ε()积积点是在积积上~但积分路是极径从的下方积积或上方积积[][00
f(z)f()x?0zxi=?+()εC沿小半积积积积点。积个极是解析的~而且~我积有]0r0
?,,0bxrbf(x)f(x)f(z)f(x)=++dxlimdxdxdx ,,????0+?r0aaCxr,εrxxixxzzxx????0000,,
xrbb? fxfxfx()()()0limddd.xxx+=P axra+r 00xxxxxx???000
fz()()()()z?z?=fz=fxlim因积 ~000z?z0zz?0
f(z)limdx=?iπf()x所以 .0?C?r0rzz?0
bbfxfx()()ddxxifx= πP() ε?0).(0 aa??xxixx,ε00
f(x)积可以用符形式此式中被积函除号将数之外的部分积积
11ixx= ?Pπδ() (ε?0).0xxixx??,ε00
Home work: 5.1: (1), (4); 5.4: (1), (4); 5.5: (2), (4), (7); 5.7: (1), (3).
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