s4.5.ExamplesThe purpose of this section is to provide examples of the definition in Sec.4.4 of path integralsand to illustrate various properties that were mentioned there. We defer development of theconcept of antiderivatives of the integrands f(=) in path integrals until Sec.4.7.Example1.Letusfindthevalueofthe integralI =[zdz(4.5.1)when C is the right-hand half (Fig. 4-7)0Z=2ei022ofthe circle ==2,from z =-2i to z =2i.According to definition (4.4.2), Sec.4.4,I=[2eig(2ei0)'d0 = 2e-82iei8d0=4id0=4mJIy21B41+iC0210xFig. 4-7Fig. 4-8Note that when a point zis on the circle==2,itfollows that zz=4,or z=4/zHencetheresult I=4元icanalsobewritten[4-[ d-[ =m (4.5.2)Jc44JcExample 2. Let f(=)= y-x-i3x(z= x+iy), C, denote the path OAB shown inFig. 4-8, and C, denote the segment OB of the line y= x.Find the integerals of f alongCh and C,.Solution. From the formula (4.4.6), we getJe, F(2)dz= Jo,(2)dz+JanJ()dz(4.5.3)Since the leg OA may be represented parametrically as z=0+iy(O≤ y≤1), by definition(4.4.2) we compute thatJo,5(=)dz =J'yidy=if,d =号On the leg AB: z = x+ i(0≤ x≤1), we have[an J(=)dz=J(1-x-13x)-1dx = J'(1- x)dx-3if'x2 dx=In view of equation (4.5.3), we now see thatJe f(a)dz = 1-i(4.5.4)1
§4.5. Examples The purpose of this section is to provide examples of the definition in Sec. 4.4 of path integrals and to illustrate various properties that were mentioned there. We defer development of the concept of antiderivatives of the integrands zf )( in path integrals until Sec. 4.7. Example 1. Let us find the value of the integral ∫ = C dzzI (4.5.1) when C is the right-hand half (Fig. 4-7) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ≤≤−= 22 2 π θ iθ π ez of the circle z = 2|| , from −= 2iz to = 2iz . According to definition (4.4.2), Sec.4.4, ∫− = ′ 2/ 2/ )2(2 π π θθ deeI θ ii ∫− ∫− − = = = 2/ 2/ 2/ 2/ 422 4 π π π π θθ θ πθ ididiee ii . Note that when a point z is on the circle z = 2|| , it follows that zz = 4 , or = /4 zz . Hence the result = 4πiI can also be written Fig. 4-7 Fig. 4-8 ∫∫ ∫ === CC C idzzdzz z dz π 4 1 4 1 . (4.5.2) Example 2. Let , denote the path OAB shown in Fig. 4-8, and denote the segment of the line )( )(3 2 +=−−= iyxzxixyzf C1 C2 OB y = x . Find the integerals of along and . f C1 C2 Solution. From the formula (4.4.6), we get ∫∫∫ = + 1 )()()( C OA AB dzzfdzzfdzzf . (4.5.3) Since the leg OA may be represented parametrically as = + ≤ yiyz ≤ )10(0 , by definition (4.4.2) we compute that ∫ ∫∫ === OA i ydyiyidydzzf 1 0 1 0 2 )( . On the leg xixzAB ≤≤+= )10(: , we have ∫∫ ∫ ∫ −−=⋅−−= −= AB idxxidxxdxxixdzzf 1 0 1 0 1 0 2 2 2 1 3)1(1)31()( . In view of equation (4.5.3), we now see that ∫ − = 1 2 1 )( C i dzzf . (4.5.4)
Since C, has a parametric representation z= x+ix(0≤x≤1), we see thatJe, J(=)dz = J'l-i3x(1+ i)]dx = 3(1-i)]°xdx =1-i.(4.5.5)Evidently, then, the integrals of f along the two paths C, and C, have different values eventhough those paths have the same initial and the same final points.Observe how it follows that the integral of f over the simple closed path C, -C2, i.e.,OABO, has the nonzero valueJ, (a)d-Je (=)d =-1+i2Example 3. We begin here by letting C denote an arbitrary smooth arcz=z(t) (a≤t≤b)from a fixed point =, to a fixed point z, (Fig. 4-9).PCc可xFig. 4-9In order to evaluate the integral(t)(t)dt=dwe note that, according to Exercise 1(b), Sec. 4.2,d [:(0 =≥()(0).dt2It follows from theformula (4.2.4)that1=[2(0P_ [=(b)]}° -[=(a)]°22But z(b) = z2 and z(a) = z1; and so1=(2 -2)/2This shows that the value of I depends only on the end points of C, and is not independent ofthearcthatistaken.Thus,wemaywrited-3-(4.5.6)2(Compare Example 2, where the value of an integral from one fixed point to another depended onthe path that was taken.)Expression (4.5.6) is also valid when C is a path that is not necessarily smooth since a pathconsists of a finite number of smooth arcs C(k =1,2,...,n),joined end to end. More precisely,suppose that each C extends from zkto zk+1.Then by (4.5.6)we have
Since has a parametric representation C2 = + ≤ xixxz ≤ )10( , we see that ∫∫ ∫ −=−=+−= 2 1 0 1 0 2 2 1)1(3)]1(3[)( C idxxidxixidzzf . (4.5.5) Evidently, then, the integrals of along the two paths and have different values even though those paths have the same initial and the same final points. f C1 C2 Observe how it follows that the integral of over the simple closed path , i.e., , has the nonzero value f − CC 21 OABO ∫ ∫ − + − = 1 2 2 1 )()( C C i dzzfdzzf . Example 3. We begin here by letting C denote an arbitrary smooth arc = ≤ ≤ btatzz )()( from a fixed point to a fixed point (Fig. 4-9). 1 z 2 z Fig. 4-9 In order to evaluate the integral ∫ ∫ == ′ C b a )()( dttztzzdzI , we note that, according to Exercise , Sec. 4.2, b)(1 )()( 2 )]([ 2 tztz tz dt d = ′ . It follows from the formula (4.2.4) that 2 )]([)]([ 2 )]([ 2 2 2 azbztz I b a − = = . But and ; and so 2 )( = zbz 1 )( = zaz 2/)( 2 1 2 2 −= zzI . This shows that the value of I depends only on the end points of , and is not independent of the arc that is taken. Thus, we may write C ∫ − = 2 1 2 2 1 2 2 z z zz zdz . (4.5.6) (Compare Example 2, where the value of an integral from one fixed point to another depended on the path that was taken.) Expression (4.5.6) is also valid when is a path that is not necessarily smooth since a path consists of a finite number of smooth arcs C nkC ),2,1( k = K , joined end to end. More precisely, suppose that each extends from to . Then by (4.5.6) we have Ck k z k+1 z
2-2)d=-zdz:(4.5.7)22k=lz, being the initial point of C and zn+1 its final point.Itfollowsfromexpression(4.5.7)thattheintegral ofthefunctionf()=zaroundeachclosed path in the plane has value zero. (Once again, compare Example 2, where the value of theintegral of a given function around a certain closed path was not zero.)The question of predictingwhen an integral around a closed path has value zero will be discussed in Sec.4.7, 4.9, and 4.11Example 4. Let C denote the semicircular pathz = 3ei0(0 ≤ 0 ≤元)from the point z = 3 to the point z = -3(Fig. 4-10).Although the branch (Sec.3.3)f(=)=Jrer012 (r>0,0<0<2元)(4.5.8)of the multiple-valued function ≥1/2 is not defined at the initial point ≥ = 3 of the path C,theintegralI =J.f(z)dz(4.5.9)of that branchneverthelessexistsJx-303Fig. 4-10To see that this is so, we observe that0+iV3singJ[=(0)]= V3e1012 = V3 cos→/3(0-→0)22Hence f is continuous on C wheneverits value at ≥ =3 is defined as V3. Consequently,I = f'/3e0/23ie" d0 = 3/3if,'er30/2d0;and22,139/2e13012d0 =(1+i)3i3i0Finally, we obtain I = -2/3(1+i)
∫ ∑∫ ∑ = = + + − = − = = C n k C n k nkk k zzzz zdz zdz 1 1 2 1 2 1 22 1 2 2 , (4.5.7) 1 z being the initial point of C and its final point. n+1 z It follows from expression (4.5.7) that the integral of the function around each closed path in the plane has value zero. (Once again, compare Example 2, where the value of the integral of a given function around a certain closed path was not zero.) The question of predicting when an integral around a closed path has value zero will be discussed in Sec. 4.7, 4.9, and 4.11. )( = zzf Example 4. Let C denote the semicircular path πθ )0(3 θ = ≤≤ i ez from the point z = 3 to the point z = −3(Fig. 4-10). Although the branch (Sec.3.3) )( )20,0( 2/ πθ θ = rerzf <<> i (4.5.8) of the multiple-valued function is not defined at the initial point 2/1 z z = 3 of the path , the integral C ∫ = C )( dzzfI (4.5.9) of that branch nevertheless exists. To see that this is so, we observe that Fig. 4-10 )0(3 2 sin3 2 cos33)]([ 2/ → → + +== θ θ θ θ θ ezf i i Hence f is continuous on C whenever its value at z = 3 is defined as 3 . Consequently, ∫ ∫ = = π π θθ θ θ θ 0 0 2/ 2/3 3333 deidieeI ii i ; and )1( 3 2 3 2 0 2/3 0 2/3 i i e i dei i = +−= ∫ π θ π θ θ . Finally, we obtain I +−= i)1(32
S4.6.UpperBoundsforIntegralsIn this section, we will give an upper bound for the modulus of a path integral, wich is useful inestimation of an integral.Theroem 4.6.1 If a function f is contionuous on a countour C and M is a constantsuch that If(z)< M(Vz EC), thenf(z)d|≤ML(4.6.1)where L denotes thelengthof CProof. Let z = z(t)(a≤t≤b)be the parametric representation of C, Fisrt, weknowfrom definition (4.4.2), Sec. 4.4, and inequality (4.2.5) in Sec. 4.2 that()d-[=(2()d(=()dtSince I f(=) M(VzeC), we have[.J(2)d≤ M["1=(0)]dt = MLThis completes theproof.Note that since all of the paths of integration to be considered here are paths and theintegrands are piecewise continuous functions defined on those paths,a number M such as theone appearing in inequality (4.6.1) will always exist.y2iox2Fig. 4-11Example1.LetC bethe arc of the circleIz2from z=2 to z=2ithat lies inthe first quadrant (Fig. 4-11). Show thatz + 46元-(4.6.2)7-1Proof. Let z be any point on C. Then I=-2 and so4+4=6and7Thus, when z lies on C,wehave1z+41 - 62+423-1-23-1Writing M =6/7 and observing that L= π is the length of C, we may now use inequality(4.6.1) to obtain inequality (4.6.2). This completes the proof.Example 2. Suppose that CR is the semicircular pathz=Rem(0≤0≤元),and -1/2 denotes the branch
§4.6. Upper Bounds for Integrals In this section, we will give an upper bound for the modulus of a path integral, wich is useful in estimation of an integral. Theroem 4.6.1 If a function f is contionuous on a countour C and M is a constant such that ∈∀≤ CzMzf )(|)(| , then MLdzzf C ≤ ∫ )( , (4.6.1) where L denotes the length of C . Proof. Let be the parametric representation of . Fisrt, we know from definition (4.4.2), Sec. 4.4, and inequality (4.2.5) in Sec. 4.2 that ≤≤= btatzz )()( C ∫∫ ∫ = ′ ≤ ′ b a b C a |)(||)]([|)()]([)( dttztzfdttztzfdzzf . Since ≤ ∀ ∈CzMzf )(|)(| , we have MLdttzMdzzf b C a ≤ ′ = ∫∫ |)(|)( . This completes the proof. Note that since all of the paths of integration to be considered here are paths and the integrands are piecewise continuous functions defined on those paths, a number M such as the one appearing in inequality (4.6.1) will always exist. Fig. 4-11 Example 1. Let C be the arc of the circle z = 2|| from z = 2 to that lies in the first quadrant (Fig. 4-11). Show that = 2iz 7 6 1 4 3 π ≤ − + ∫C dz z z . (4.6.2) Proof. Let z be any point on C . Then z = 2|| and so ≤+ zz + = 64|||4| and 7|1||||1| . 3 3 zz =−≥− Thus, when z lies on C , we have 7 6 |1| |4| 1 4 3 3 ≤ − + = − + z z z z . Writing M = 7/6 and observing that L = π is the length of , we may now use inequality (4.6.1) to obtain inequality (4.6.2). This completes the proof. C Example 2. Suppose that is the semicircular path CR = π≤θ≤ )0( iθ eRz , and denotes the branch 2/1 z