《复变函数论 Theory of Complex Variable Functions》课程授课教案（教材讲义，复分析 Complex Analysis）Chapter Ⅶ Applications of Residues

-245 -Chapter VIIApplications of ResiduesWeturn nowto someimportant applications of thetheory of residues, which wasdeveloped in the preceding chapter. The applications include evaluation of certaintypes of definite and improper integrals occurring in real analysis and appliedmathematics.Considerable attention is also given to a method, based on residues,forlocatingzerosoffunctions.s7.1.Evaluationof Improper IntegralsIn calculus, the improper integral of a function f(x)over the semi-infiniteinterval [0,oo) is defined by means ofthe equationJ° f(x)dx= lim J。f(x)dx.(7.1.1)When the limit on the right exists, the improper integral is said to converge to thatlimit. If f(x) is continuous for all x, its improper integral over the infiniteinterval (-oo, oo) is defined by writing[ f(x)dx = Jlim |f(x)dx+ limf(x)dx;(7.1.2)and when both of the limits here exist, integral (7.1.2) converges to their sum.Another value that is assigned to integral (7.1.2) is often useful. Namely, theCauchy principal value (P.V.) of integral (7.1.2) is the numberPV J" f(x)dx = lim J-Rf(x)dx,(7.1.3)provided this single limit existsIf integral (7.1.2) converges, its Cauchy principal value (7.1.3) exists; and thatvalue is thenumbertowhich integral (7.1.2)converges.This isbecause[,f(x)dx= J%(x)dx+J,f(x)dxand the limit asR-co of each of the integrals on theright exists when integral(7.1.2)converges.It is not, however, always truethat integral (7.1.2) convergeswhen its Cauchy principal valueexists,as the following example shows.Example.Observethat

- 245 - Chapter Ⅶ Applications of Residues We turn now to some important applications of the theory of residues, which was developed in the preceding chapter. The applications include evaluation of certain types of definite and improper integrals occurring in real analysis and applied mathematics. Considerable attention is also given to a method, based on residues, for locating zeros of functions. §7.1. Evaluation of Improper Integrals In calculus, the improper integral of a function over the semi-infinite interval xf )( ∞),0[ is defined by means of the equation . (7.1.1) ∫ ∫ ∞ ∞→ = 0 0 )(lim)( R R dxxf dxxf When the limit on the right exists, the improper integral is said to converge to that limit. If xf )( is continuous for all x , its improper integral over the infinite interval ∞−∞ ),( is defined by writing ∫∫∫ ∞ − − ∞ ∞→ ∞→ = + 0 1 0 2 1 2 )(lim)(lim)( R R R R dxxf dxxf dxxf ; (7.1.2) and when both of the limits here exist, integral (7.1.2) converges to their sum. Another value that is assigned to integral (7.1.2) is often useful. Namely, the Cauchy principal value (P.V.) of integral (7.1.2) is the number ∫ ∫ ∞ − − ∞ ∞→ = R R R dxxfVP )(lim)(. dxxf , (7.1.3) provided this single limit exists. If integral (7.1.2) converges, its Cauchy principal value (7.1.3) exists; and that value is the number to which integral (7.1.2) converges. This is because ∫∫∫ − − = + R R R R dxxfdxxfdxxf 0 0 )()()( and the limit as of each of the integrals on the right exists when integral (7.1.2) converges. It is not, however, always true that integral (7.1.2) converges when its Cauchy principal value exists, as the following example shows. R ∞→ Example. Observe that

- 246 -t/eDV=lim0=0xdx=limxdx=1lim(7.1.4)R- 2RR→0On theotherhand, xa = lm 'xad + m xdxR→03/3+ lim = lim Ri-→0 2Re-→ 2l0I-RR2R?+ lim = - lim (7.1.5)R-→ 2R2→002and since these last two limits do not exist, we find that the improper integral (7.1.5)fails to exist.But suppose that y= f(x)(-oo<x<oo) is an even function, that is,f(-x)=f(x) for all x. The symmetry of the graph of y= f(x) withrespectto the y axis enables us to write1rRJ(x)dx=f(x)dx,and we see that integral (7.1.1) converges to one half the Cauchy principal value(7.1.3) when that value exists. Moreover, since integral (7.1.1) converges and since[,f(x)dx=f(x)dx,integral (7.1.2) converges to twice the value of integral (7.1.1). We have thus shownThefollowingresult.Theorem 7.1.1 when f(x)(-oo < x < oo) is even and the Cauchy principalvalue (7.1.3) exists, both of the integrals (7.1.1) and (7.1.2) converge andPV [ f(x)dx= [f(x)dx =2]f(x)dx.(7.1.6)We now describe a method involving residues, to be illustrated in the nextsection, that is often used to evaluate improper integrals of even rational functionsf(x)=p(x)/g(x),wheref(-x) is equal tof(x)and wherep(x) andq(x) are polynomials with real coefficients and no factors in common.We agreethat g()has no real zeros but has at least one zero above the real axis.The method begins withthe identification of all of the distinct zeros of thepolynomial q()thatlieabovethereal axis.Theyare,of course,finitein number(see Sec.4.14)and may be labeled z,z2...,=n,where n is less than or equal

- 246 - ∫ ∫ ∞ − − ∞ ∞→ − ∞→ ∞→ = = == R R R R R R R x xdxVP xdx 00lim 2 . lim lim 2 . (7.1.4) On the other hand, ∫ ∫ ∫ ∞→ ∞ − − ∞ ∞→ = + 2 1 1 2 0 0 lim lim R R R R xdx xdx xdx 2 2 1 1 0 2 0 2 2 lim 2 lim R R R R x x ∞→ − ∞→ = + 2 lim 2 lim 2 2 2 1 1 2 RR R ∞→ R ∞→ +−= ; (7.1.5) and since these last two limits do not exist, we find that the improper integral (7.1.5) fails to exist. But suppose that )(( −∞= < xxfy < ∞) is an even function, that is, =− xfxf )()( for all x . The symmetry of the graph of with respect to the axis enables us to write = xfy )( y ∫ ∫− = R R R dxxfdxxf 0 )( 2 1 )( , and we see that integral (7.1.1) converges to one half the Cauchy principal value (7.1.3) when that value exists. Moreover, since integral (7.1.1) converges and since ∫ ∫ − = 0 0 1 1 )()( R R dxxfdxxf , integral (7.1.2) converges to twice the value of integral (7.1.1). We have thus shown The following result. Theorem 7.1.1 when )(( −∞ < xxf < ∞) is even and the Cauchy principal value (7.1.3) exists, both of the integrals (7.1.1) and (7.1.2) converge and ∫∫ ∫ ∞ ∞− ∞ ∞− ∞ = = 0 )(2)()(. dxxfdxxfdxxfVP . (7.1.6) We now describe a method involving residues, to be illustrated in the next section, that is often used to evaluate improper integrals of even rational functions = xqxpxf )(/)()( , where −xf )( is equal to and where and are polynomials with real coefficients and no factors in common. We agree that has no real zeros but has at least one zero above the real axis. xf )( xp )( xq )( zq )( The method begins with the identification of all of the distinct zeros of the polynomial that lie above the real axis. They are, of course, finite in number (see Sec. 4.14) and may be labeled , where is less than or equal zq )( n , zzz 21 K n

247tothedegreeof g(z).Wethen integratethequotientf(2)= P(a)(7.1.7)q(=)around thepositively oriented boundary of the semicircularregion shown in Fig.7-1.That simpleclosed contour consists of the segment of the real axis from z = -R to z = R and the top halfof the circlez=R,described counterclockwise and denoted by C.It is understood that thepositive number R is large enough that the points ,2,...,n all lie inside the closed pathyIC2,2.2x-RoRFig. 7-1TheCauchyresiduetheoremtellsusthat f(x)dx + Jc. f()dz = 2niZRes f(2)k=l =That is,(x)dx = 2niZRes f(=)-Jf(z)dz(7.1.8)Now we arrive at the following conclusion.Theorem 7.1.2. Let a function f be given by (7.1.7) and the points Z,-2....,=n havethedescribedpropertyabove.Ifthelimitlim /c f(=)dz = 0,PthenP.V. " (x)dx = 2niZ Res f(=).(7.1.9)kecIf f(x), in addition, is even, then (x)dx = 2元iZ Res f(a)(7.1.10)k=l sandf(x)dx = niZRes f(2).(7.1.11)l

247 to the degree of zq )( . We then integrate the quotient )( )( )( zq zp zf = (7.1.7) around the positively oriented boundary of the semicircular region shown in Fig. 7-1. That simple closed contour consists of the segment of the real axis from z = −R to z = R and the top half of the circle , described counterclockwise and denoted by . It is understood that the positive number || = Rz CR R is large enough that the points all lie inside the closed path. n , zzz 21 K Fig. 7-1 The Cauchy residue theorem tells us that ∫ ∫ − ∑= = + = π R R C n k R z zk f x dx f z dz i f z 1 ( ) ( ) 2 Res ( ). That is, ∫− ∑ ∫ = = = π − R R n k z zk CR f x dx i f z f z dz 1 ( ) 2 Res ( ) ( ) . (7.1.8) Now we arrive at the following conclusion. Theorem 7.1.2. Let a function be given by (7.1.7) and the points have the described property above. If the limit f n z ,z , ,z 1 2 K ∫ = R→∞ CR lim f (z)dz 0 , then ∫ ∑ ∞ −∞ = = = π n k z z PV f x dx i f z k 1 . . ( ) 2 Res ( ) . (7.1.9) If f (x) , in addition, is even, then ∫ ∑ ∞ −∞ = = = π n k z z f x dx i f z k 1 ( ) 2 Res ( ) (7.1.10) and ∫ ∑ ∞ = = = π 0 1 ( ) Res ( ) n k z z f x dx i f z k . (7.1.11)

$7.2.ExamplesWeturn nowto an illustration of themethod in Sec.7.1for evaluating improper integrals.2dx,we start with the observation thatExample.In orderto evaluatetheintegralJo x6 +122has isolated singularities at the zeros of - +1, which are the sixththe function f(z)=2°+1roots of-1,and isanalytic everywhere else.Themethod in Sec.1.8for finding roots of complex（元2k元）numbers reveals that the sixth roots of -1 are Ci =expi(k = 0,1,2,...,5), and{66it is clear that none of them lies on the real axis. The first three roots, Co = ein/6,c, = iandC,=e15z/6lie in the upper half plane (Fig. 7-2) and the other three lie in the lower one.Fig. 7-2When R >1, the points C (k = O,1,2) lie in the interior of the semicircular region bounded bythe segment z=x(-R≤x≤ R) of the real axis and theupper half CR ofthe circle[z=Rfrom z=R to z=-R. Integrating f(-) counterclockwise around the boundary of thissemicircular region, we see thatJ,(x)dx + Jc f(z)dz=2i(B + B, + B,),(7.2.1)where B is the residue of f(z) at c,(k = 0,1,2)With the aid of Theorem 6.8.2 in Sec.6.8, we find that the points Ck are simple poles of fand that2-c1(k = 0,1,2)B,=Res6ck=Ck 2+16c6Thus(111)-元2 mi(B。 + B, + B,) = 2元l(6i6i6i)-3and equation (7.2.1)canbeput intheformJ (x)dx=-J. (=)dz,(7.2.2)3which is valid for all values of R greater that 1.Next,weshowthat thevalueof the integral on theright in equation (7.2.2)tendsto0 asRtends to oo. To do this, we observe that when I= |= R,[-2 H==R2

§7.2. Examples We turn now to an illustration of the method in Sec. 7.1 for evaluating improper integrals. Example. In order to evaluate the integral ∫ ∞ 0 +6 2 1 dx x x , we start with the observation that the function 1 )( 6 2 + = z z zf has isolated singularities at the zeros of 1 6 z + , which are the sixth roots of , and is analytic everywhere else. The method in Sec. 1.8 for finding roots of complex numbers reveals that the sixth roots of −1 −1 are ),5,2,1,0( 6 2 6 exp ⎥ = K ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π + π = k k ick and it is clear that none of them lies on the real axis. The first three roots, icec and i 1 == 6/ 0 , π 6/5 2 i π = ec lie in the upper half plane (Fig. 7-2) and the other three lie in the lower one. Fig. 7-2 When R > 1, the points lie in the interior of the semicircular region bounded by the segment c (k = 0,1,2) k z = x(−R ≤ x ≤ R) of the real axis and the upper half of the circle from CR | z |= R z = R to z = −R . Integrating counterclockwise around the boundary of this semicircular region, we see that f (z) ∫ ∫ − + = π + + R R CR f (x)dx f (z)dz 2 i(B B B ) 0 1 2 , (7.2.1) where is the residue of at Bk f (z) c (k = 0,1,2) k . With the aid of Theorem 6.8.2 in Sec.6.8, we find that the points are simple poles of and that k c f ( 0,1,2) 6 1 1 6 Res 5 3 2 6 2 = = = + = = k c c c z z B k k k z c k k . Thus 6 3 1 6 1 6 1 2 ( 0 1 2 ) 2 π π π ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ + + = − + i i i i B B B i ; and equation (7.2.1) can be put in the form ∫− ∫ − π = R R CR f x dx f (z)dz 3 ( ) , (7.2.2) which is valid for all values of R greater that 1. Next, we show that the value of the integral on the right in equation (7.2.2) tends to 0 as R tends to ∞. To do this, we observe that when | z |= R , 2 2 2 | z |=| z | = R

and[=6 +12z /6 -1/= R6 -1.So, if ≥ is any pointon Cr.R?122-1 f(z) /=≤MRwhereMR=[≥6 +1]R6-1and this means thatc. ()d|≤MR=→0,(7.2.3)R°-1as R tends to co.Thus, lim Jf(z)dz=0. It now follows from equation (7.2.2)thatR→JCx21dr=元Rlim 3R-0J-R x°+1that isx-元P.V.["-dx=3J-0 x° +1Since the integrand here is even, we know from equations (7.1.6) in Sec.7.1 and Theorem 7.1.2thatx2dxr=元o(7.2.4)Jo x+16

and 1|1||||1| 6 6 6 Rzz −=−≥+ . So, if z is any point on , CR M R z z zf ≤ + = |1| || |)(| 6 2 where 1 6 2 − = R R M R ; and this means that RMdzzf R CR π≤ ∫ )( 0 1 6 3 → − = R πR , (7.2.3) as R tends to ∞. Thus, . It now follows from equation (7.2.2) that ∫ = R ∞→ CR dzzf 0)(lim ∞→ ∫− π = + R R R dx x x 31 lim 6 2 , that is ∫ ∞ ∞− π = + 31 . 6 2 dx x x VP . Since the integrand here is even, we know from equations (7.1.6) in Sec.7.1 and Theorem 7.1.2 that 0 61 6 2 π = + ∫ ∞ dx x x . (7.2.4)

s7.3.ImproperIntegralsFromFourierAnalysisResidue theorycanbe useful in evaluating convergent improper integrals of theform["f(x)sinaxdx or Jf(x)cosaxdx,(7.3.1)where a denotes a positive constant. As in Sec. 7.1, we assume that f(x)= p(x)/q(x),where p(x) and q(x) are polynomials with real coefficients and no factors in common.Alsoq(=) has no real zeros. Integrals of type (7.3.1) occur in the theory and application of the Fourierintegral.The method described in Sec.7.1 and used in Sec.7.2 cannot be applied directly here since(see Sec. 3.6)Isin az '= sin ax + sinh’ ayandI cos ax P= cos? ax + sinh’ ayMore precisely, sinceeay-e-aysinh ay2the moduli Isin az| and Icosaz increase like ey as y tends to infinity. The modificationillustrated in the example below is suggested by thefact that[f(x)cosaxdx +if"f(x)sinaxdx=frf(x)edx,togetherwiththefactthatthe modulusle Heia(+)Heei eyis bounded in the upper half planey ≥ 0.Example.Let us show that[.cos 3x d = 21(7.3.2)(x2+1)2eBecause the integrand is even, it is sufficient to show that the Cauchy principal value of theintegral exists and to find that value.We introduce the function1f(a)=(7.3.3)(22 + 1)2and observe that the product f(z)ei3= is analytic everywhere on and above the real axis exceptat the point z =i. The singularity z =i lies in the interior of the semicircular region whoseboundary consists of the segment - R≤ x≤ R of the real axis and the upper half C of thecircle I=-R(R>1) from ==R to z=-R. Integration of f(=)er3- around thatboundaryyields the equationei3xR(r +1 dx = 2元iB, - Jf(=)ei3=dz(7.3.4)whereB, = Res[f(=)ei3]Sinceer3f(=)ei3==_()whereΦ()(z-i)?(z+i)2the point z=i is evidently a pole oforder m=2 of f(z)ei3=; and

§7.3. Improper Integrals From Fourier Analysis Residue theory can be useful in evaluating convergent improper integrals of the form ∫ ∞ ∞− sin)( axdxxf or , (7.3.1) ∫ ∞ ∞− cos)( axdxxf where denotes a positive constant. As in Sec. 7.1, we assume that , where and are polynomials with real coefficients and no factors in common. Also, has no real zeros. Integrals of type (7.3.1) occur in the theory and application of the Fourier integral. a = xqxpxf )(/)()( xp )( xq )( zq )( The method described in Sec. 7.1 and used in Sec. 7.2 cannot be applied directly here since (see Sec. 3.6) az ax ay 22 2 sin|sin| += sinh and ax ax ay 2 2 2 cos|cos| += sinh . More precisely, since 2 sinh ay ay ee ay − − = , the moduli and increase like as tends to infinity. The modification illustrated in the example below is suggested by the fact that az |sin| az |cos| ay e y ∫ ∫∫ − −− + = R R R R R R iax cos)( sin)( )( dxexfaxdxxfiaxdxxf , together with the fact that the modulus iaz iyxia iaxay ay eeeee + − − |||||| === )( is bounded in the upper half plane y ≥ 0 . Example. Let us show that ∫ ∞ ∞− π = + 22 3 2 )1( 3cos e dx x x . (7.3.2) Because the integrand is even, it is sufficient to show that the Cauchy principal value of the integral exists and to find that value. We introduce the function 22 )1( 1 )( + = z zf (7.3.3) and observe that the product is analytic everywhere on and above the real axis except at the point zi ezf 3 )( z = i . The singularity z = i lies in the interior of the semicircular region whose boundary consists of the segment − ≤ ≤ RxR of the real axis and the upper half of the circle from CR RRz >= )1(|| z = R to z = −R . Integration of around that boundary yields the equation zi ezf 3 )( ∫ ∫ − −π= + R R C zi xi R dzezfiBdx x e 3 22 1 3 2 )( )1( , (7.3.4) where ])([Res 3 1 zi iz ezfB = = . Since 2 3 )( )( )( iz z ezf zi − = φ where 2 3 )( )( iz e z zi + φ = , the point = iz is evidently a pole of order m = 2 of ; and zi ezf 3 )(

1B, =(i)=ie3By equating the real parts on each side of equation (7.3.4), then, we find thats dt--Rel.(a)eidR(7.3.5)J-R (x2 + 1)?e3JFinally, we observe that when z is a point on Cr1If(=)KMRwhere MR=(R2-1)?and that le3"e-3y ≤1 for sucha point. Consequently,[ReJe, (2)e"-d|≤c (a)e=de≤ MrR-→0,(7.3.6)JCas R tends to oo and because of inequalities (7.3.6), we need only let R tend to co inequation (7.3.5) to arrive at the desired result (7.3.2)

1 3 1 )( ie = φ′ iB = . By equating the real parts on each side of equation (7.3.4), then, we find that ∫ ∫ − − π = + R R C zi R dzezf e dx x x 3 22 3 )(Re 2 )1( 3cos . (7.3.5) Finally, we observe that when z is a point on , CR ≤ Mzf R |)(| where 22 )1( 1 − = R M R and that 1|| for such a point. Consequently, 3 3 ≤= zi − y ee )(Re )( 0 3 ≤ 3 →≤ ∫∫ RMdzezfdzezf R C zi C zi R R π , (7.3.6) as R tends to ∞ and because of inequalities (7.3.6), we need only let R tend to in equation (7.3.5) to arrive at the desired result (7.3.2). ∞

$7.4.Jordan'sLemmaIn the evaluation of integrals of the type treated in Sec.7.3, it is sometimes necessaryto useJordan'slemma,which isstatedhereasatheorem.Theorem7.4.1(Jordan).Suppose that(i) a function f(z) is analytic at all points z in the upper half plane y≥0 that areexieriortoacircle==R;(ii) CR denotes a semicircle z= Re"(0≤0≤元), where R>R。 (Fig. 7-4);(ii) there is a constant MR such that If(=)≤ Mr, Vz eCr, and lim Mr = 0.Then,for every positive constant a,f(=)eldz= 0lim(7.4.1)yxRFig. 7-4Proof.The proof is based on a result that isknown as Jordan's inequality:fe-Rsinedeo0)(7.4.2)RTo verify this inequality, we first note from the graphs of the functions y=sin andy=20/元when0≤≤元/2(Fig.7-5)thatsin0≥20/元forallvaluesofinthatinterval.Consequently,if R>0Se-2ROI when 0≤0≤Rsing2and soe-2R0/nd0=元(1-e-R)a-Rsinede2RHence-Rsinod00)(7.4.3)2RBut this is just another form of inequality (7.4.2), since the graph of y = sin is symmetricwith respect to the vertical line =元/2on the interval 0≤≤元Vy=sing20元-02元／

§7.4. Jordan’s Lemma In the evaluation of integrals of the type treated in Sec. 7.3, it is sometimes necessary to use Jordan’s lemma, which is stated here as a theorem. Theorem 7.4.1(Jordan). Suppose that i)( a function zf )( is analytic at all points z in the upper half plane that are exterior to a circle ; y ≥ 0 0 || = Rz ii)( denotes a semicircle , where (Fig. 7-4); CR π≤θ≤= )0( iθ eRz > RR 0 iii)( there is a constant such that M R ≤ Mzf R |)(| ,∀ ∈Cz R , and = 0lim . ∞→ R R M Then, for every positive constant a , ∫ = ∞→ CR iaz R dzezf 0)(lim . (7.4.1) Fig. 7-4 Proof. The proof is based on a result that is known as Jordan’s inequality: ∫ π θ− > π 0 , Rsinθ R /2 πθ ee− − ≤ when 2 0 π θ ≤≤ ; and so ∫ ∫ π π θ− πθ− − − π ≤θ =θ 2/ 0 2/ 0 sin /2 )1( 2 R R R e R dede . Hence ∫ π θ− > π <θ 2/ 0 sin )0( 2 R R de R . (7.4.3) But this is just another form of inequality (7.4.2), since the graph of y = sinθ is symmetric with respect to the vertical line π=θ 2/ on the interval 0 ≤ θ ≤ π

Turning now to the verification of limit (7.4.1), we accept statements ()-(iii) in thetheorem and write[c.(z)e' dz = J,'f(Rel)exp(iaRe')iRedo,Since1f(Re)≤ Mr and [exp(iaRe")Ke-aRksnoand in view of Jordan's inequality (7.4.2),it follows that.(z)e d≤MRR,eaRsin deMR元Limit(7.4.1)isthenevident,sinceM→0 as R→co,TheproofiscompletedExample.Let us find the Cauchy principal value of the integralxsinxdxx2+2x+2As usual, the existence of the value in question will be established by our actuallyfinding it.We writeNZf(=) =22 + 22 + 2 ~ (2 - 2,)(≥ -2)where z, =-1+i.The point z, which lies above the x axis, is a simple pole ofthe functionf(z)ewith residueB=Zet(7.4.4)2/ - 2)Hence, when R > /2 and CR denotes the upper half of the positively oriented circlel - R,xe"dx=2元iB,-f()edzJ-RX2+2x+2=and this means thatRxsinxdx= Im(2元iB,)- Im Jc f(=)e dz .(7.4.5)Rx?+2x+2We note that, when z isa point on Cr,I f(=)< MR whereRMR=→0(R→8)(R- ~2)2By Theorem 7.4.1,[ImJc f(z)ed≤[c f(2)e'd2→0(R-),(7.4.6)Consequently, equation (7.4.5), together with expression (7.4.4) for the residue B,tells us thatPV."xsin xdr=Im(2元iB,)="(sin1+ cos1).(7.4.7)x2+2x+2

Turning now to the verification of limit (7.4.1), we accept statements in the theorem and write − iiii )()( ∫ ∫ π θ θθ = θ CR iaz i ii deiReiaReRfdzezf 0 )( )exp()( . Since R i ≤ MeRf θ |)(| and θ−θ ≤ sin |)exp(| i aR eeiaR and in view of Jordan’s inequality (7.4.2), it follows that . )( 0 sin a M deRMdzezf R aR R C iaz R π θ π θ 2 and denotes the upper half of the positively oriented circle , CR || = Rz ∫ ∫ −π= − ++ CR iz R R ix dzezfiB xx dxxe 2 )( 22 2 1 ; and this means that ∫ ∫ − −π= ++ R R C iz R iB dzezf xx xdxx )(Im)2Im( 22 sin 2 1 . (7.4.5) We note that, when z is a point on , CR ≤ Mzf R |)(| where )(0 )2( 2 ∞→→ − = R R R MR By Theorem 7.4.1, )(Im ≤ ∞→→ )(0)( ∫∫ Rdzezfdzezf R CR iz C iz . (7.4.6) Consequently, equation (7.4.5), together with expression (7.4.4) for the residue , tells us that B1 ∫ ∞ ∞− + π =π= ++ )1cos1(sin)2Im( 22 sin . 2 1 e iB xx xdxx VP . (7.4.7)

$7.5.IndentedPathsIn this and the following section, we illustrate the use of indented paths.We begin with animportant limit that will beused intheexampleinthis sectionTheorem 7.5.1.Suppose that(i) a function f(z) has a simple pole at a point z=X on the real axis, with a Laurentseries representation in a punctured disk 0z- Xo K R, (Fig. 7-7) and with residue Bo;(ii)Cp denotes the upper half of a circleIz-zo=p, where p<R, and theclochwise direction is taken.Thenimf(z)dz=-Bπi.(7.5.1)0yCPp--11x0Xo1PR,1Fig. 7-7Proof. From the assumption (i), the LictionfcanbewrittenasBof(z)= g(=) +-(0 4z-xkR2),z-Xowhere g(2)=a,(z-xo)"(Iz-xokR,).Thusn=0dzJc f(z)dz= J. g(=)dz + B J(7.5.2)Z-XoNow thefunction g() is continuous when z-z<R,,according to Corollary 5.8.1.Hence if we choose a number Po such that p< Po<R, (see Fig.7-7), it must be boundedon the closed disk I=-xPo,according to Sec.2.7.That is, there is a nonnegative constantM such thatlg()≤M whenever Iz-xoPo; and, since the length L of the path Cis L = πp, it follows thatg(2)dz≤ ML = MrpConsequently, g(2)dz = 0im(7.5.3)

§7.5. Indented Paths In this and the following section, we illustrate the use of indented paths. We begin with an important limit that will be used in the example in this section. Theorem 7.5.1. Suppose that i)( a function zf )( has a simple pole at a point 0 = xz on the real axis, with a Laurent series representation in a punctured disk 20 < − ||0 < Rxz (Fig. 7-7) and with residue ; B0 ii)( denotes the upper half of a circle Cρ − zz 0 || = ρ , where ρ < R2 and the clockwise direction is taken. Then ∫ ρ π−= →ρ C iBdzzf 0 0 )(lim . (7.5.1) Proof. From the assumption (i), the Laurent series of the function f can be written as Fig. 7-7 )()( )||0( 20 0 0 Rxz xz B zgzf <−< − += , where )|(|)()( . Thus 20 0 0 Rxzxzazg n n ∑ n −= <− ∞ = ∫ ∫ ∫ ρ ρ ρ − = + C C C x Bdzzgdzzf 0 0 z dz )()( . (7.5.2) Now the function zg )( is continuous when 20 − || < Rzz , according to Corollary 5.8.1. Hence if we choose a number ρ 0 such that ρ < ρ < R20 (see Fig. 7-7), it must be bounded on the closed disk 00 xz || ≤− ρ , according to Sec. 2.7. That is, there is a nonnegative constant M such that |)(| ≤ Mzg whenever 00 − xz || ≤ ρ ; and, since the length L of the path is Cρ L = πρ , it follows that πρ=≤ ∫ ρ MMLdzzg C )( . Consequently, ∫ ρ = →ρ C dzzg 0)(lim0 . (7.5.3)