1MTH302 Chapter 1The Residue Theorem and Improper IntegralsPoles of a functionExampleDetermine the poles ofz2(z2 ? 2z ? 2)eizSolution Let ( ) 2 2( 2 2)izef zz z z?? ?The poles of f z ( ) are obtained by solving z2(z2 ? 2z ? 2) ? 0One pole is z = 0 of order 2.Solving z2 ? 2z ? 2 ? 0 , we get2? 2 ? 4 ?8z ? or z = -1 + i and 1 i.Both -1 + i and 1 i are simple poles.Thus z = 0, -1 + i, 1 i are the three poles of orders 2, 1, 1, respectively.ExampleFind the poles of242 5( )16zf zz???SolutionThe poles of f (z) are obtained by solving the equation z4 + 16 = 0 or z4 = -16Hence 41z ? 2(?1) ? 2?cos(2n ?1)? ? isin(2n ?1)??14??????????4(2 1)sin4(2 1)2 cos n ? i n ?n = 0, 1, 2, 3 (by Demoivres Theorem). If n = 0, ?? ????? ? ?? ????? ?2 2124sin4z 2 cos? i ? i2If n = 1, ?? ????? ? ? ?? ????? ?2 21243sin43z 2 ? i ? iIf n = 2, ?? ????? ? ? ?? ????? ?2 21245sin45z 2 ? i ? iIf n = 3, ?? ????? ? ?? ????? ?2 21247sin47z 2 ? i ? iThus the poles of162 542? ?zzare given by z ? 2 ?i, 2 ?i . Each pole has order 1.Evaluation of the residue of f at its poleIf z0 is a simple pole of f, we can write??(??) = ??-1??-??0+ ??0 + ??1(?? ??0) + ?? ? ? ?01 0 limz za z z f z ??? ?If z0 is a pole of order m, then? ? ? ? ? ? 011 0 111 !mmmz zda z z f zm dz?? ??? ? ? ?? ? ?TheoremLet two functions p and q be analytic at a point z0. If p(zo) ? 0, q(zo) = 0, and q'(zo) ? 0,then z0 is a simple pole of the quotient p(z)/q(z) and000( ) ( )Re ( , )( ) ( )p z p zs zq z q z??3Residue TheoremIf f (z) is analytic inside and on a closed curve C except at a finite number of poles z1, z2 zninside C at which the residues are R1, R2 Rn respectively, then? ? ? ?cf (z)dz 2?i(R1 R2 Rn)? 2 residues ?i?ExampleEvaluate( 2)( 4)zCedz? z z ? ?wheni). C is the circle |z| = 5ii). C is the circle |z| = 3iii). C is the circle |z| = 1Solutioni). The integrand(z ? 2)(z ? 4)ezhas two simple poles at z = 2 and z = 4 inside |z| = 5 (since thedenominator (z 2) (z 4) = 0 gives z = 2 and z = 4).Residue at z = 2 is( 4)lim( 2)( 4)( 2)lim2 2??? ??? ? zez zz ezzzz 2 4 22 2????e eResidue at z = 4 isCC iziz1. .z2.zn542 2lim( 2)( 4)( 4)lim44 4ezez zz e zzzz???? ??? ?Sum of residues =2 2 22 4 4 2e e e ? e? ?Hence( 2)( 4)zCedz? z z ? ? = 2?i (sum of residues)2( )24 2e e? ?i ? = ?i (e4 e2)ii). Let |z| = 3, i.e. C is the circle with centre origin and radius 3 unitsThere is only one pole z = 2 inside |z| = 3and the corresponding residue at z = 2 is2 2e ?.Hence 2222( 2)( 4)ie iez ze dzcz??? ???? ?? if C is |z| = 3.iii). Let |z| = 1 or C is the circle with centre at the origin and radius 1 unit.There is no pole inside the circle |z| = 1. Since both the poles z =2 and z =4 are outside thecircle.Hence by Cauchy integral theorem 0( 2)( 4)zCe dzz z?? ?? . 0 3 4 05Cauchy principal value of an integralLet f (x) be a continuous function for all x in (-?, ?).The improper integral of f on (-?, ?) is defined by21 21( ) lim ( ) lim ( ) , is any fixed real number,c RR RR cf x dx f x dx f x dx c???? ????? ? ? ? ? , (1) if both limits exist.The Cauchy principal value of the improper integral is defined by. . ( ) lim ( )RRRPV f x dx f x dx????? ?? ? ? , (2)if the limit exists.Note that if (1) exists, then ( ) . . ( )R Rf x dx P V f x dx??? ?? ? ? , but the converse may not be true.For example, PV f x dx . ( ) 0???? ? but the improper integral f x dx ( )???? does not exist.Cauchy criterion for convergence of improper integralTheoremLet f (x) be a continuous function for all x in (-?, ?). Then the improper integral f x dx ( )???? isconvergent if and only if for every ? > 0, there exists a number N > 0 (depending on ?) suchthat for all s and t > N, we have| ( ) |t s? f x dx ??6ExampleLet f be defined on (-?, ?) bysinif 0( )1 0xxf x xx?? ?? ?? ? ?. Then f is continuous on (-?, ?).ClaimThe improper integral0f x dx ( )??is convergent.ProofFor t > s > 0, we use integration by parts to obtain22sin cos coscos cos costt t ss st sx x xdx dxx x xt s xdxt s x?? ?? ?? ? ? ? ? ?? ?? ??Hence2sin 1 1 1
2 2| | ift ts sxdx dx t sx t s x s??? ?? ? ? ? ? ? ? ? ?? ? ? ?By the Cauchy criterion we have0f x dx ( )??convergent.As f(x) is an even function on (-?, ?), we also have0f x dx f x dx ( ) 2 ( ) .? ?? ?? ? ?Hence f x dx ( )???? is also convergent.7Theorem (Jordans Lemma) (Ref: Text, Lemma 3, page 332) Ifandisthequotientoftwopolynomials m ? 0 ( )( )p zq zdegreeof ( ) 1 degreeof ( ) q z p z ? ? , thenwhere is the positively lim 0, ( )( )RimzRp ze?? ? q z? ? ?R oriented upper half circle, centred at the origin and radius R.Theorem (Ref: Text Lemma 4, page 340)If f(z) has a simple pole at z = c. Let ?r be the contour given below.Then we have2 10lim ( ) ( )Res ( ( ), )rrf z dz i f z c ? ????? ? ?8Example (Ref: Text Examples 1 & 2, page 341-342)Show that sin xdxx????? ?Let OD = R and OC = ?Consider the contour F = semicircle DEF ? FA ? semicircle ABC ? CD Thenis analytic inside and on the contour F. Thus we have, by the Cauchy ize z0izFedzz? ? integral theorem.Thus we have0iz ix iz ix iz RF R ABC DEFe e e e edz dx dz dx dzz x z x z??? ?? ? ? ? ? ? ? ? ? ? (*)By the above two theorems, we know thatlim 0izRDEFedz?? z? ?and0lim ( 0)Res( ,0)iz izABCe edz i i? z z? ??? ? ? ? ??because 0 is a simple pole ofize zwith residue equals to 1 and ABC is an arc moving in aclockwise direction.Now taking limits on both side of (*), we getixedx ix????? ?CEF D?ABORMinus sign is used here because the path is clockwise9However, we also know thatcos sin cos sincos sine x i x x i x ixdx dx dx dxx x x xx xdx i dx ix x?? ? ? ??? ?? ?? ??? ??? ???? ? ?? ? ?? ? ? ?? ?Now equating the real part and imaginary part of the above integral, we getcosx dx 0x???? ?andsin sin x xi dx i dxx x? ?? ??? ??? ? ? ? ?The following theorems are also of interest. We omit details of their proofs.TheoremIf f is analytic in a closed and bounded region E, then f it is bounded in E. Hence | f (z) | ? Mwhere M is an upper bound of | f (z) | in E.TheoremIf f is analytic along a curve C of length L, then ( )C? f z dz ML ?where M is the upper bound of f on the curve C and L is the length of the curve C from A toB.ACB10TheoremIf f z ( ) MkR? then lim ( ) 0RRf z???? ? , where k > 1 and ?R is the counter clockwise upper halfcircle z R e ? ? ? i? , 0 ? ? .Further examplesExampleEvaluate ?? ??x2 ?1dxSolution We consider the integralthe curve ?R is the semicircle with centre at the origin and 21Rdzz?? ? radius R.(i) The poles of11( ) 2??zf z are obtained by solving z2 + 1 = 0 or z2 = -1. Hencez ? ?1 ? ?i are the two simple poles inside the circle |z| = R.Inside the semicircle ?R there is only one pole at z = i, the other being outside ?R. The residue at z =i foris given by 11( ) 2??zf z z i z i iz izz iz i z i 21( )( )( )lim) 1(lim2 ?? ???? ?? ?-R 0 R -R 0i ?Ri11(ii) Now ( ) ( ) ( ) 2 2 ( ) 1 12R RRC Rf z dz f z dz f z dz iR ii? ? ?? ?? ? ? ? ? ? ? ? where ?R is theupper semicircle and C R R R R ? ? ? ? [ , ].(iii) In the interval [-R, R], z = x+ iy becomes z = x since y = 0 on the x axis.On the semicircle ?R, z ? R ei? dz ? R ei?id? . Hence for R > 1, we have 21( ) ( )1R RRC Rf z dz f x dx dzz?? ?? ? ?? ? ? ?2 0 ? ?211 1R iiRR e iddxx R e? ?????? ?? ? ???? ?Thus2 2 0 ? ?21 1lim lim1 1 1R iR R iRR e iddx dxx x R e? ????? ???? ???? ? ?? ?? ? ? ? ? ?? ? ? ?? ? ?? ? ? ,since0 0 ? 1 ?2 2 1 0i iiiR e id e idR e R eR? ? ? ?? ?? ?? ?? ?? ?? ?? ? when R ? ?.Remark1 202??? ??xdxt.(TJs f+sfu:- I ri- d.rJ -o lt *( l- cCI a$= fT zdqltO3 cDSaXLt 9= ):q, lh* dg: zdq-(fl dg-)D Le+ z:qj0ttr,, , { ,,, 7z- 7 3-r d7L JC(orr)r1t2( t__ Tt@rl) dzLrLleN-ts-:a^I- 6=- ?,ffia,- -- 3- 2-7_-.- Jl- .*l 3f2lZ J1^d- V,zt is r.slJe Cqf) 4td{7zo1 qi:$.eari:ZI-2 -ttL-T_J Z- tt-2latG.--, ltris L-Fj f- t -7z.)r J?_!tar;4.i*Kd:c *- o4d* l*;141- a+ M=Xr^ {@ t*s si^gotar qf-[..,2,=tAZz:-a-/rL*+ be $^a Ct.sd Cbiibwz1t=glryCK(upFr sernic,r.le Fcn*rRd e/rq,a,,uJtj-rtftat^-R4or nR Zr rs insiJq -R.Gn zz is orrjg 4