# Course hero - 19.5 Exercises Answers to selected...

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This preview shows page 1 out of 7 pages. Unformatted text preview: 19.5 Exercises Answers to selected odd-numbered problems begin on page ANS-000. In Problems 1–6, use a Laurent series to find the indicated residue. 2 1. f (z) ϭ ; Res ( f (z), 1) 1z 2 121z 1 42 1 2. f (z) ϭ 3 ; Res ( f (z), 0) z 11 2 z2 3 4z 2 6 3. f (z) ϭ ; Res ( f (z), 0) z12 2 z2 2 4. f (z) ϭ (z ϩ 3)2 sin ; Res ( f (z), Ϫ3) z13 2 5. f (z) ϭ e22>z ; Res ( f (z), 0) e2z 6. f (z) ϭ ; Res ( f (z), 2) 1z 2 22 2 In Problems 7–16, use (1), (2), or (4) to find the residue at each pole of the given function. z 4z 1 8 7. f (z) ϭ 2 8. f (z) ϭ 2z 2 1 z 1 16 1 1 9. f (z) ϭ 4 10. f (z) ϭ 2 z 1 z3 2 2z2 1z 2 2z 1 22 2 5z2 2 4z 1 3 11. f (z) ϭ 1z 1 121z 1 221z 1 32 2z 2 1 12. f (z) ϭ 1z 2 12 4 1z 1 32 cos z ez 13. f (z) ϭ 2 14. f (z) ϭ z e 21 z 1z 2 p2 3 1 15. f (z) ϭ sec z 16. f (z) ϭ z sin z In Problems 17–20, use Cauchy’s residue theorem, where appropriate, to evaluate the given integral along the indicated contours. 1 17. dz 1z 2 121z 1 22 2 C C 18. (b) |z| ϭ 3 2 (c) |z| ϭ 3 z11 dz z 1z 2 2i2 C C ᭢ 2 (a) |z| ϭ 1 z3e21>z dz 2 ᭢ C C (a) |z| ϭ 5 20. (b) |z ϩ i| ϭ 2 (c) |z Ϫ 3| ϭ 1 (b) |z Ϫ 2i| ϭ 3 (c) |z| ϭ 5 1 dz C z sin z C ᭢ (a) |z Ϫ 2i| ϭ 1 In Problems 21–32, use Cauchy’s residue theorem to evaluate the given integral along the indicated contour. 1 21. dz, C: |z Ϫ 3i| ϭ 3 z2 1 4z 1 13 C C 1 22. dz, C: |z Ϫ 2| ϭ 3 2 z3 1z 2 12 4 C C z 23. dz, C: |z| ϭ 2 4 C z 21 C z dz, C is the ellipse 16x2 ϩ y2 ϭ 4 24. 2 C 1z 1 121z 1 12 C zez 25. dz, C: |z| ϭ 2 2 C z 21 C ez 26. dz, C: |z| ϭ 3 3 2 C z 1 2z C tan z 27. dz, C: |z Ϫ 1| ϭ 2 C z C cot pz 28. dz, C: |z| ϭ 1 2 2 C z C ᭢ ᭢ ᭢ ᭢ ᭢ ᭢ ᭢ ᭢ (b) |z Ϫ 2i| ϭ 1 (c) |z Ϫ 2i| ϭ 4 ᭢ the rectangle defined by x ϭ 1 , x ϭ p , 2 ᭢ C is the rectangle defined by x ϭ Ϫ2, x ϭ 1, ᭢ ᭢ (a) |z| ϭ 1 2 19. C: |z Ϫ 3| ϭ 1 cot pz dz, C is C C y ϭ Ϫ1, y ϭ 1 2z 2 1 30. dz, z2 1z3 1 12 C C y ϭ Ϫ1 , y ϭ 1 2 eiz 1 sin z 31. dz, 4 C 1z 2 p2 C cos z 32. 2 2 C 1z 2 12 1z 1 C 29. ᭢ 92 dz, C: |z Ϫ 1| ϭ 1 19.6 Evaluation of Real Integrals Introduction In this section we shall see how residue theory can be used to evaluate real integrals of the forms # 2p F1cos u, sin u2 du, (1) 0 # q 2q f 1x2 dx, (2) 19.6 Evaluation of Real Integrals 79665_CH19_FINAL.indd 843 843 11/2/09 3:11:10 PM # q 2q f 1x2 cos ax dx and # q 2q f 1x2 sin ax dx, (3) where F in (1) and f in (2) and (3) are rational functions. For the rational function f (x) ϭ p(x)/q(x) in (2) and (3), we will assume that the polynomials p and q have no common factors. Integrals of the Form e02p F (cos U, sin U ) dU The basic idea here is to convert an integral of form (1) into a complex integral where the contour C is the unit circle centered at the origin. This contour can be parameterized by z ϭ cos u ϩ i sin u ϭ eiu, 0 Յ u Յ 2p. Using eiu 1 e2iu eiu 2 e2iu , sin u 5 , 2 2i dz 5 ieiu du, cos u 5 we replace, in turn, du, cos u, and sin u by dz 1 1 , cos u 5 1z 1 z212, sin u 5 1z 2 z212. iz 2 2i du 5 (4) The integral in (1) then becomes 1 1 dz F a 1z 1 z212, 1z 2 z212 b , 2 2i iz C C ᭢ where C is |z| ϭ 1. ■ EXAMPLE 1 Evaluate # 2p 0 Solution A Real Trigonometric Integral 1 du. 12 1 cos u2 2 Using the substitutions in (4) and simplifying yield the contour integral 4 z dz. i C 1z2 1 4z 1 12 2 C With the aid of the quadratic formula we can write z z f 1z2 5 2 5 , 2 2 1z 1 4z 1 12 1z 2 z02 1z 2 z12 2 ᭢ where z0 ϭ Ϫ2 Ϫ 23 and z1 ϭ Ϫ2 ϩ 23. Since only z1 is inside the unit circle C, we have z dz 5 2pi Res 1 f 1z2, z12. 2 C 1z 1 4z 1 12 C Now z1 is a pole of order 2 and so from (2) of Section 19.5, ᭢ 2 Res 1 f 1z2, z12 5 lim zSz1 d d z 1z 2 z12 2 f 1z2 5 lim 2 dz zSz1 dz 1z 2 z02 5 lim 2 zSz1 z 1 z0 1 5 . 3 1z 2 z02 623 Hence, 4 1 z 4 4 dz 5 и 2pi Res 1 f 1z2, z12 5 и 2pi и i C 1z2 1 4z 1 12 2 i i C 623 ᭢ # and finally 2p 0 1 4p du 5 . 2 12 1 cos u2 323 q Integrals of the Form e2q f (x) dx When f is continuous on (Ϫq, q), recall from calculus q that the improper integral e2q f (x) dx is defined in terms of two distinct limits: # q 2q 844 79665_CH19_FINAL.indd 844 f 1x2 dx 5 lim # 0 rSq 2r R f 1x2 dx 1 lim # f 1x2 dx. (5) RSq 0 CHAPTER 19 Series and Residues 11/2/09 3:11:12 PM If both limits in (5) exist, the integral is said to be convergent; if one or both of the limits fail to q exist, the integral is divergent. In the event that we know (a priori) that an integral e2q f (x) dx converges, we can evaluate it by means of a single limiting process: # q 2q f 1x2 dx 5 lim # R f 1x2 dx. (6) RSq 2R It is important to note that the symmetric limit in (6) may exist even though the improper integral is q divergent. For example, the integral e2q x dx is divergent since limRSϱ e0R x dx ϭ limRSϱ 1 R2 ϭ q. 2 However, using (6), we obtain lim # R x dx 5 lim c RSq 2R RSq 12R2 2 R2 2 d 5 0. 2 2 (7) The limit in (6) is called the Cauchy principal value of the integral and is written P.V. # q 2q f 1x2 dx 5 lim # R RSq 2R f 1x2 dx. y q In (7) we have shown that P.V. e2q x dx ϭ 0. To summarize, when an integral of form (2) converges, its Cauchy principal value is the same as the value of the integral. If the integral diverges, it may still possess a Cauchy principal value. q To evaluate an integral e2q f (x) dx, where f (x) ϭ P(x)/Q(x) is continuous on (Ϫq, q), by residue theory we replace x by the complex variable z and integrate the complex function f over a closed contour C that consists of the interval [ϪR, R] on the real axis and a semicircle CR of radius large enough to enclose all the poles of f (z) ϭ P(z)/Q(z) in the upper half-plane Re(z) Ͼ 0. See FIGURE 19.6.1. By Theorem 19.5.3 we have ᭢ C C f 1z2 dz 5 # f 1z2 dz 1 CR # R CR zn z2 z1 z3 –R z4 0 R x FIGURE 19.6.1 Closed contour C consists of a semicircle CR and the interval [ϪR, R] n 2R f 1x2 dx 5 2pi a Res 1 f 1z2, zk2, k51 where zk, k ϭ 1, 2, … , n, denotes poles in the upper half-plane. If we can show that the integral eCR f (z) dz S 0 as R S q, then we have P.V. # q 2q ■ EXAMPLE 2 f 1x2 dx 5 lim # R RSq 2R n f 1x2 dx 5 2pi a Res 1 f 1z2, zk2. (8) k51 Cauchy P.V. of an Improper Integral y Evaluate the Cauchy principal value of # q 2q CR 1 dx. 1x 2 1 12 1x 2 1 92 3i i Solution Let f (z) ϭ 1/(z ϩ 1)(z2 ϩ 9). Since 2 (z ϩ 1)(z ϩ 9) ϭ (z Ϫ i)(z ϩ i)(z Ϫ 3i)(z ϩ 3i), 2 2 we let C be the closed contour consisting of the interval [ϪR, R] on the x-axis and the semicircle CR of radius R Ͼ 3. As seen from FIGURE 19.6.2, 1 dz 5 1z 1 121z2 1 92 C C ᭢ 2 # R 2R 1 dx 1 1x 2 1 121x 2 1 92 # CR –R R x FIGURE 19.6.2 Closed contour C for Example 2 1 dz 1z2 1 121z2 1 92 5 I1 1 I2 and I1 ϩ I2 ϭ 2pi[Res ( f (z), i) ϩ Res ( f (z), 3i)]. At the simple poles z ϭ i and z ϭ 3i we find, respectively, Res 1 f 1z2, i2 5 1 1 and Res 1 f 1z2, 3i2 5 2 , 16i 48i 19.6 Evaluation of Real Integrals 79665_CH19_FINAL.indd 845 845 11/2/09 3:11:14 PM I1 1 I2 5 2pi c so that 1 1 p 2 d 5 . 16i 48i 12 (9) We now want to let R S q in (9). Before doing this, we note that on CR, |(z2 ϩ 1)(z2 ϩ 9)| ϭ |z2 ϩ 1||z2 ϩ 9| Ն ||z|2 Ϫ 1| ||z|2 Ϫ 9| ϭ (R2 Ϫ 1)(R2 Ϫ 9), and so from the ML-inequality of Section 18.1 we can write ZI2 Z 5 2 # CR 1 pR dz 2 # 2 . 2 1z 1 121z 1 92 1R 2 121R2 2 92 2 This last result shows that |I2| S 0 as R S q, and so we conclude that limRS q I2 ϭ 0. It follows from (9) that limRS q I1 ϭ p/12; in other words, lim # R RSq 2R 1 p dx 5 or P.V. 2 2 12 1x 1 121x 1 92 # q 2q 1 p dx 5 . 2 12 1x 1 121x 1 92 2 It is often tedious to have to show that the contour integral along CR approaches zero as R S q. Sufficient conditions under which this is always true are as follows: Theorem 19.6.1 Behavior of Integral as R S q Suppose f (z) ϭ P(z)/Q(z), where the degree of P(z) is n and the degree of Q(z) is m Ն n ϩ 2. If CR is a semicircular contour z ϭ Reiu, 0 Յ u Յ p, then eCR f (z) dz S 0 as R S q. In other words, the integral along CR approaches zero as R S q when the denominator of f is of a power at least 2 more than its numerator. The proof of this fact follows in the same manner as in Example 2. Notice in that example that the conditions stipulated in Theorem 19.6.1 are satisfied, since the degree of P(z) ϭ 1 is 0 and the degree of Q(z) ϭ (z2 ϩ 1)(z2 ϩ 9) is 4. ■ EXAMPLE 3 Cauchy P.V. of an Improper Integral Evaluate the Cauchy principal value of # q 2q 1 dx. x4 1 1 Solution By inspection of the integrand, we see that the conditions given in Theorem 19.6.1 are satisfied. Moreover, we know from Example 3 of Section 19.5 that f has simple poles in the upper half-plane at z1 ϭ epi/4 and z2 ϭ e3pi/4. We also saw in that example that the residues at these poles are Res 1 f 1z2, z12 5 2 1 422 2 1 422 i and Res 1 f 1z2, z22 5 1 422 2 1 i. 422 Thus, by (8), # P.V. q 2q 1 p dx 5 2pi fRes 1 f 1z2, z12 1 Res 1 f 1z2, z22g 5 . x4 1 1 22 q q Integrals of the Forms e2q f (x) cos Ax dx or e2q f (x) sin Ax dx We encountered integrals of this type in Section 15.4 in the study of Fourier transforms. Accordingly, q q e2q f (x) cos ax dx and e2q f (x) sin ax dx, a Ͼ 0, are referred to as Fourier integrals. Fourier q integrals appear as the real and imaginary parts in the improper integral e2q f (x)eiax dx. Using iax Euler’s formula e ϭ cos ax ϩ i sin ax, we get # q 2q f 1x2eiax dx 5 # q 2q f 1x2 cos ax dx 1 i # q 2q f 1x2 sin ax dx (10) whenever both integrals on the right side converge. When f (x) ϭ P(x)/Q(x) is continuous on (Ϫq, q) we can evaluate both Fourier integrals at the same time by considering the integral ͐C f (z)eiaz dz, where a Ͼ 0 and the contour C again consists of the interval [ϪR, R] on the real axis and a semicircular contour CR with radius large enough to enclose the poles of f (z) in the upper half-plane. 846 79665_CH19_FINAL.indd 846 CHAPTER 19 Series and Residues 11/2/09 3:11:15 PM Before proceeding we give, without proof, sufficient conditions under which the contour integral along CR approaches zero as R S q: Theorem 19.6.2 Behavior of Integral as R S q Suppose f (z) ϭ P(z)/Q(z), where the degree of P(z) is n and the degree of Q(z) is m Ն n ϩ 1. If CR is a semicircular contour z ϭ Reiu, 0 Յ u Յ p, and a Ͼ 0, then eCR (P(z)/Q(z))eiaz dz S 0 as R S q. ■ EXAMPLE 4 Using Symmetry # q x sin x dx. x2 1 9 0 Solution First note that the limits of integration are not from Ϫq to q as required by the method. This can be rectified by observing that since the integrand is an even function of x, we can write Evaluate the Cauchy principal value of # q 0 # x sin x 1 dx 5 2 x2 1 9 q 2q x sin x dx. x2 1 9 (11) With a ϭ 1, we now form the contour integral z eiz dz, z2 1 9 C C ᭢ where C is the same contour shown in Figure 19.6.2. By Theorem 19.5.3, # CR z eiz dz 1 2 z 19 # R 2R x eix dx 5 2pi Res 1 f 1z2 eiz, 3i2, x 19 2 where f (z) ϭ z/(z2 ϩ 9). From (4) of Section 19.5, Res 1 f 1z2 eiz, 3i2 5 zeiz e23 2 5 . 2z z 5 3i 2 Hence, in view of Theorem 19.6.2 we conclude eCR f (z)eiz dz S 0 as R S q and so P.V. # q 2q x e23 p eix dx 5 2pi a b 5 3 i. 2 x 19 e 2 But by (10), # q 2q x eix dx 5 x2 1 9 # q 2q x cos x dx 1 i x2 1 9 # q 2q p x sin x dx 5 3 i. x2 1 9 e Equating real and imaginary parts in the last line gives the bonus result P.V. # q 2q x cos x dx 5 0 along with P.V. x2 1 9 # q 2q p x sin x dx 5 3 . 2 x 19 e Finally, in view of (11) we obtain the value of the prescribed integral: # q 0 x sin x 1 dx 5 P.V. 2 x2 1 9 # q 2q x sin x p dx 5 3 . x2 1 9 2e Indented Contours The improper integrals of form (2) and (3) that we have considered up to this point were continuous on the interval (Ϫq, q). In other words, the complex function f (z) ϭ P(z)/Q(z) did not have poles on the real axis. In the event f has poles on the real axis, q we must modify the procedure used in Examples 2– 4. For example, to evaluate e2q f (x) dx by residues when f (z) has a pole at z ϭ c, where c is a real number, we use an indented contour 19.6 Evaluation of Real Integrals 79665_CH19_FINAL.indd 847 847 11/2/09 3:11:16 PM as illustrated in FIGURE 19.6.3. The symbol Cr denotes a semicircular contour centered at z ϭ c oriented in the positive direction. The next theorem is important to this discussion. y CR Theorem 19.6.3 –Cr –R c R x Behavior of Integral as r S 0 Suppose f has a simple pole z ϭ c on the real axis. If Cr is the contour defined by z ϭ c ϩ reiu, 0 Յ u Յ p, then lim # rS0 C r FIGURE 19.6.3 Indented contour f 1z2 dz 5 pi Res 1 f 1z2, c2. PROOF Since f has a simple pole at z ϭ c, its Laurent series is a21 f 1z2 5 1 g1z2, z2c where aϪ1 ϭ Res( f (z), c) and g is analytic at c. Using the Laurent series and the parameterization of Cr , we have # # f 1z2 dz 5 a21 p 0 Cr ireiu du 1 ir reiu p # g1c 1 re 2 e iu iu du 5 I1 1 I2. (12) 0 First, we see that I1 5 a21 # p 0 ireiu du 5 a21 reiu # p i du 5 pia21 5 pi Res 1 f 1z2, c2. 0 Next, g is analytic at c and so it is continuous at this point and bounded in a neighborhood of the point; that is, there exists an M Ͼ 0 for which |g(c ϩ reiu )| Յ M. Hence, ZI2 Z 5 2 ir # p g1c 1 reiu2 eiu du 2 # r 0 # p M du 5 prM. 0 It follows from this last inequality that limrS 0|I2| ϭ 0 and consequently limrS 0 I2 ϭ 0. By taking the limit of (12) as r S 0, we have proved the theorem. ■ EXAMPLE 5 Using an Indented Contour Evaluate the Cauchy principal value of y # q 2q CR sin x dx. x1x 2 2 2x 1 22 Solution Since the integral is of form (3), we consider the contour integral ͛C eiz dz /z(z2 Ϫ 2z ϩ 2). The function f (z) ϭ 1/z(z2 Ϫ 2z ϩ 2) has simple poles at z ϭ 0 and at z ϭ 1 ϩ i in the upper half-plane. The contour C shown in FIGURE 19.6.4 is indented at the origin. Adopting an obvious notation, we have ᭢ 1+i –Cr –R –r r R x ᭢ C C FIGURE 19.6.4 Indented contour C for Example 5 5 # # 1 CR 2r 2R 1 # 2Cr 1 # R 5 2pi Res 1 f 1z2 eiz, 1 1 i2, (13) r where e2Cr 5 2eCr . Taking the limits of (13) as R S q and as r S 0, we find from Theorems 19.6.2 and 19.6.3 that P.V. # q 2q eix dx 2 pi Res 1 f 1z2eiz, 02 5 2pi Res 1 f 1z2 eiz, 1 1 i2. x1x 2 2x 1 22 2 Now, 1 e21 1 i Res 1 f 1z2eiz, 02 5 and Res 1 f 1z2 eiz, 1 1 i2 5 2 11 1 i2. 2 4 Therefore, P.V. # q 2q 848 79665_CH19_FINAL.indd 848 1 e21 1 i eix 11 1 i2 b. dx 5 pi a b 1 2pi a2 2 4 x1x 2 2x 1 22 2 CHAPTER 19 Series and Residues 11/2/09 3:11:18 PM Using eϪ1ϩi ϭ eϪ1(cos 1 ϩ i sin 1), simplifying, and then equating real and imaginary parts, we get from the last equality P.V. # q # q 2q and P.V. 2q 19.6 p cos x dx 5 e21 1sin 1 1 cos 12 2 x1x 2 2 2x 1 22 sin x p dx 5 f1 1 e21 1sin 1 2 cos 12g. 2 x1x 2 2x 1 22 Exercises 2 Answers to selected odd-numbered problems begin on page ANS-000. In Problems 1–10, evaluate the given trigonometric integral. 1. 3. 5. 6. 8. 10. # # # # # # 2p # # In Problems 31 and 32, use an indented contour and residues to establish the given result. q sin x 31. P.V. dx 5 p x 2q q sin x dx 5 p11 2 e212 32. P.V. x1x 2 1 12 2q 33. Establish the general result 2p 1 1 du 2. du 1 1 0.5 sin u 10 2 6 cos u 0 0 2p 2p cos u 1 du du 4. 3 1 sin u 1 1 3 cos2 u 0 0 p 1 du [Hint: Let t ϭ 2p Ϫ u.] 0 2 2 cos u p 2p 1 sin2 u 7. du du 2 5 1 4 cos u 0 1 1 sin u 0 2p 2p cos2 u cos 2u du 9. du 3 2 sin u 5 2 4 cos u 0 0 2p 1 du cos u 1 2 sin u 1 3 0 # # # # # 0 # # # # # # # # # du ap 5 , a.1 1a 1 cos u2 2 1 2a2 2 12 3 and use this formula to verify the answer in Example 1. 34. Establish the general result In Problems 11–30, evaluate the Cauchy principal value of the given improper integral. q q 1 1 11. dx 12. dx 2 x 2 2x 1 2 x 2 2 6x 1 25 2q 2q q q 1 x2 13. dx 14. dx 2 2 2 2 2q 1x 1 42 2q 1x 1 12 q q 1 x 15. dx 16. dx 2 3 2 3 2q 1x 1 12 2q 1x 1 42 q q dx 2x 2 2 1 17. 18. dx 2 2 2 4 2 2q x 1 5x 1 4 2q 1x 1 12 1x 1 92 q 2 q x 11 1 19. dx 20. dx x4 1 1 x6 1 1 0 0 q q cos x cos 2x dx dx 21. 22. x2 1 1 x2 1 1 2q 2q q q cos x x sin x dx dx 23. 24. 2 2 2 2q x 1 1 0 1x 1 42 q q cos 3x sin x dx dx 25. 26. 2 2 2 0 1x 1 12 2q x 1 4x 1 5 q q cos 2x x sin x 27. 28. dx dx x4 1 1 x4 1 1 0 0 q cos x dx 29. 2 2 2q 1x 1 121x 1 92 q x sin x 30. dx 2 2 0 1x 1 121x 1 42 # # # # # # # # # # # p # 2p 0 sin2u 2p du 5 2 1a 2 2a2 2 b2 2, a . b . 0 a 1 b cos u b and use this formula to verify the answer to Problem 7. 35. Use the contour shown in FIGURE 19.6.5 to show that P.V. # q 2q eax p , 0 , a , 1. x dx 5 11e sin ap y 2pi pi C r –r FIGURE 19.6.5 Contour in Problem 35 36. The steady-state temperature u(x, y) in a semi-infinite plate is determined from 0 2u 0 2u 1 2 5 0, 0 , x , p, y . 0 2 0x 0y 2y u10, y2 5 0, u1p, y2 5 4 , y.0 y 14 u1x, 02 5 0, 0 , x , p. Use a Fourier transform and the residue method to show that u1x, y2 5 # q 0 19.6 Evaluation of Real Integrals 79665_CH19_FINAL.indd 849 x e2a sin a sinh ax sin ay da. sinh ap 849 11/2/09 3:11:19 PM ...
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