Solution Set IV for 18.075–Fall 2004

SOLUTION SET IV FOR 18.075–FALL 2004 10. Functions of a Complex Variable 10.12. Residues. . In the following, I use the notation Resz=z0 f (z) ≡ Res(z0 ) ≡ Res[f (z), z0 ], where Res is the residue of f (z) at (the isolated singularity) z 0 . 82. Evaluate the integral dz 2−1 z C when C is the curve sketched in Figure 10.21. I Solution. z 21−1 has two simple poles. One is at z = 1, the other is at z = −1. It’s easy to check that Res[ z 21−1 , 1] = 12 , and Res[ z 21−1 , −1] = − 12 . The pole at z = 1 is encircled in the counterclockwise (positive) sense, while the pole at z = −1 is encircled in the clockwise sense. Hence, I 1 1 dz = 2πiRes[ 2 , 1] − 2πiRes[ 2 , −1] = πi − (−πi) = 2πi. 2−1 z z − 1 z −1 C 88 Determine the residue of each of the following functions at each singularity: 1 1 1 π (a) e z , (b) e z2 , (c) cos z−π , (d) (1 + z 2 )e z . Solution. (a) We have 1 ez = ∞ X z −n n=0 n! = 1 + z −1 + ∞ X z −n n=2 n! . 1 So e z has an essential singularity at z = 0, and 1 Res[e z , 0] = 1. (b) We have 1 e z2 = ∞ X z −2n n=0 n! = 1 + z −2 + ∞ X z −2n n=2 Date: October 9, 2002. 1 n! . 2 SOLUTION SET IV FOR 18.075–FALL 2004 1 So e z2 has an essential singularity at z = 0, and 1 Res[e z2 , 0] = 0. (c) We have cos (Note that the coefficient of ∞ X π 2n π (−1)n = . 2n (2n)! z−π (z − π) n=0 1 z−π (d) We have 1 (1 + z 2 )e z π is 0.) So cos z−π has an essential singularity at z = π, and π , π] = 0. Res[cos z−π = (1 + z 2 ) ∞ X z −n n=0 n! = (1 + z 2 )(1 + z −1 + ∞ z −2 z −3 X z −n + + ) 2 6 n! n=4 = 3 7 −1 + z + [higher powers of z −1 ]. 2 6 1 So (1 + z 2 )e z has an essential singularity at z = 0, and 1 Res[(1 + z 2 )e z , 0] = 7 . 6 10.13. 10.13 Evaluation of Real Definite Integrals. . 90. Use calculus to evaluate the following integrals: R 2πresidue dθ (a) 0 A+B sin θ = √A2π (A > |B|), 2 2 R 2π −B dθ R 2π dθ (b) 0 a2 +sin2 θ = 0 a2 +cos2 θ = a√2π (a > 0), a2 +1 π R R π2 (c) 0 sin4 θ dθ = 02 cos4 θ dθ = 3π 16 , R 2π sin2 θ π (d) 0 5+4 cos θ dθ = 4 . Solution. (a) First make the substitution: z = e iθ , dz = ieiθ dθ. Now the complex z describes the unit circle C 1 in the positive sense as θ varies from 0 to 2π. So, as was discussed in class, the integral becomes Z 2π I dz dθ iz = z 2 −1 A + B sin θ 0 C1 A + B 2iz I 2dz = 2 C1 Bz + 2iAz − B SOLUTION SET IV FOR 18.075–FALL 2004 3 The poles of the integrand are simple and occur when Bz 2 + 2iA z − B = 0, which in turn gives √ −iA ± B 2 − A2 z± = B Furthermore, A2 − B 2 A2 = −1. z+ z− = − 2 + B B2 Therefore, z+ is a (simple) pole inside the unit circle. Now using the known formula g(z) 0) = hg(z Resz=z0 h(z) 0 (z ) , where z0 : simple zero of h(z), we get: 0 Res(z+ ) = = = Thus, by the residue theorem, Z 2π 0 2 2Bz+ + 2iA 1 √ −iA + i A2 − B 2 + iA 1 √ 2 i A − B2 dθ A + B sin θ = = (b) We manipulate the integrand as follows: a2 2πi √ i A2 − B 2 2π √ 2 A − B2 2 1 1 = . = 2 1−cos 2θ 2 2 2a + 1 − cos 2θ + sin θ a + 2 So, with the new variable ϕ = 2θ, Z 2π dθ 2 a + sin2 θ 0 2π 2dθ + 1 − cos 2θ 0 Z 4π dϕ = 2 + 1 − cos ϕ 2a 0 Z 2π dϕ = 2 2 2a + 1 − cos ϕ 0 Z 2π dϕ , = 2 2 2a + 1 − sin ϕ 0 = Z 2a2 where we shifted the integration variable by π/2 in the integral of the third line and used the periodicity of the integrand. By using the result of part (a) above with A = 2a 2 + 1 and B = −1, we get Z 2π π 2π dϕ = √ = p . 2 2 2 2a + 1 − sin ϕ a a2 + 1 (2a + 1) − 1 0 4 SOLUTION SET IV FOR 18.075–FALL 2004 Thus, Z 2π a2 0 dθ 2π = √ , 2 + sin θ a a2 + 1 and, hence, Z 2π 0 dθ 2 a + cos2 θ = Z = Z = Z 2π a2 0 3π 2 a2 dθ + sin2 θ a2 dθ + sin2 θ − π2 2π 0 dθ + sin2 (θ − π2 ) π √ . a a2 + 1 The integral of the third line ensues from the periodicity of the integrand. (c) By subtracting the given integrals, we get Z π Z π Z π 2 2 2 4 4 sin θ dθ − cos θ dθ = (sin4 θ − cos4 θ) dθ = 0 0 0 π 2 Z = (sin2 θ − cos2 θ)(sin2 θ + cos2 θ) dθ 0 π 2 Z = (sin2 θ − cos2 θ) dθ 0 π 2 = − Z = − sin 2θ π2 | 2 0 cos 2θ dθ 0 = 0. So, Z π 2 4 sin θ dθ = 0 Then, Z 2π 4 sin θ dθ = 0 = Z Z π 2 4 sin θ dθ + 0 π 2 Z = 2 = 4 Z π 2 π 2 0 sin4 θ dθ. cos4 θ dθ. 0 4 Z 3π 2 4 sin θ dθ + π Z 2π 3π 2 sin4 θ dθ π 3π ) + sin4 (θ + π) + sin4 (θ + )] dθ 2 2 (sin4 θ + cos4 θ) dθ 0 π 2 π 2 sin θ dθ + [sin4 θ + sin4 (θ + 0 Z π Z SOLUTION SET IV FOR 18.075–FALL 2004 5 This shows Z π 2 4 sin θ dθ = 0 Z π 2 1 cos θ dθ = 4 4 0 Z 2π sin4 θ dθ. 0 But Z 2π z 2 − 1 4 dz ) 2iz iz C1 I 2 (z − 1)4 1 dz 16i C1 z5 I 4 sin θ dθ = 0 = ( 2πi (z 2 − 1)4 Res[ , 0] 16i z5 2πi 6 16i 3π , 4 = = = where C1 is the unit circle with center at origin. The residue was found easily by noticing that (z 2 − 1)4 z 8 − 4z 6 + 6z 4 − 4z 2 + 1 = , z5 z5 by which the coefficient of z −1 is 6. So, Z π 2 4 sin θ dθ = 0 Z π 2 cos4 θ dθ = 0 3π . 16 (d) Again, by the usual replacement z = e iθ , Z 2π 0 (z 2 −1)2 dz (2iz)2 z 2 +1 iz + 4 2z (z 2 − 1)2 sin2 θ dθ = 5 + 4 cos θ I C1 5 = I C1 −2iz 2 (10z + 4z 2 + 4) dz The simple poles of this integrand occur when 4z 2 + 10z + 4 = 0, i.e., when z = − 21 or z = 2, while a double pole occurs at z = 0. Since z = 2 is not within the unit circle, we disregard it. 6 SOLUTION SET IV FOR 18.075–FALL 2004 1 Res(− ) = 2 lim z→− 21 (z + 21 ) · (z 2 − 1)2 −2iz 2 (2z + 1)(z + 2) 2 ( −3 4 ) −2i( 23 ) 3i = 16   d 2 1 Res(0) = (z f (z)) (2 − 1)! dz z=0    (z 2 − 1)2 d = dz −2i(4z 2 + 10z + 4) z=0   −4i(2z 2 + 5z + 2)(4z 3 − 4z) − (z 4 − 2z 2 + 1)(−16iz − 20i) = (−4i)2 (2z 2 + 5z + 2)2 z=0 −5i = 16 Alternatively, you may expand the integrand in z (considering |z| “small”) and find the coefficient of z −1 . (Try it for practice!) Thus,   Z 2π sin2 θ 3i 5i dθ = 2πi − 5 + 4 cos θ 16 16 0 π = 4 = 91. Use calculus to evaluate the following integrals: R ∞residue dx (a) −∞ (x+b)2 +a2 = πa (a > 0), R∞ π (b) 0 (x2 +a2dx 2 2 = 2ab(a+b) (a > 0, b > 0), R ∞ dx )(x +bπ ) (c) 0 x4 +4a4 = 8a3 (a > 0), R∞ dx π (d) 0 (x2 +a 2 )2 = 4a3 (a > 0). Solution. (a) The degree of the denominator is 2 greater than the degree of the numerator and the function is finite for all real values of x. Thus, we can employ the strategy given in class by closing the original path with a large semicircle in the upper half plane (or lower half plane). By shifting the integration variable by −b, we get Z ∞ −∞ dx (x + b)2 + a2 = Z ∞ −∞ dx = 2 x + a2 I C1 F (z)dz = 2πi X Res(zk ) k where the points zk are the poles of F (z) = 1 z 2 +a2 in the upper half-plane. SOLUTION SET IV FOR 18.075–FALL 2004 7 p The (simple) poles occur when z 2 + a2 = 0, that is when z = ±i (a2 ) = ±ia (since a > 0). So there is one (simple) pole in the upper half-plane, namely, at z 1 = ia. 1 1 = . Res(z1 ) = 2z1 2ia Thus, Z ∞ dx 1 π = 2πi · = . 2 + a2 (x + b) 2ia a −∞ 1 (b) (z 2 +a2 )(z 2 +b2 ) has two singularities on the upper half plane. One of these is at z = ai, the other is at z = bi; both of them are simple poles. Note that the denominator of 1 is of degree 4. Accordingly, (x2 +a2 )(x2 +b2 ) Z ∞ dx 1 1 = 2πi (Res[ 2 , ai] + Res[ 2 , bi]) 2 2 2 2 2 2 2 2 (z + a )(z + b ) (z + a )(z 2 + b2 ) −∞ (x + a )(x + b ) 1 1 + ) = 2πi ( 2ai(b2 − a2 ) 2bi(a2 − b2 ) π . = ab(a + b) Since 1 (x2 +a2 )(x2 +b2 ) Z ∞ 0 is even, we get 1 dx = 2 2 2 2 (x + a )(x + b ) 2 Z ∞ −∞ dx (x2 + a2 )(x2 + b2 ) = π . 2ab(a + b) √ πi has two singularities on the upper half plane. One of these is at z = 2e 4 a, √ 3πi the other is at z = 2e 4 a. Both of them are simple poles. Note that the degree of the 1 denominator of x4 +4a 4 is 4. Accordingly, Z ∞ √ πi √ 3πi 1 1 dx 4 a] + Res[ = 2πi (Res[ , 2e , 2e 4 a]) 4 4 z 4 + 4a4 z 4 + 4a4 −∞ x + 4a (c) 1 z 4 +4a4 πi Since 1 x4 +4a4 3πi e− 4 e− 4 = 2πi ( √ ) − √ 8 2a3 i 8 2a3 i √ √ √ √ 2 2 2 2 π √ = [( − i) − (− − i)] 3 2 2 2 2 4 2a π = . 4a3 is even, we have Z ∞ Z ∞ 1 π dx dx = = . 4 4 4 4 x + 4a 2 −∞ x + 4a 8a3 0 (d) The degree of the denominator is greater than twice the degree of the numerator and the function is finite for all real values of x. We can once again employ the strategy given in class. Also, note that the integrand is even so that Z Z ∞ dx 1 ∞ dx = 2 2 2 2 (x + a ) 2 −∞ (x + a2 )2 0 8 SOLUTION SET IV FOR 18.075–FALL 2004 The poles occur when (z 2 + a2 )2 = 0, that is when z = ±ia. Thus, there is one pole in the upper half-plane, i.e., at z = ia, and it is a double pole.  d (z − ia)2 dz (z 2 + a2 )2 z=ia   1 d = dz (z + ia)2 z=ia   −2 = (z + ia)3 z=ia 1 = 4ia3 Res(ia) =  Therefore, Z 0 ∞ (x2 dx + a 2 )2 = = 1 1 · 2πi 3 2 4ia π 4a3 92. Use calculus to evaluate the following integrals: R ∞residue mx π −am (a) 0 xasin (a > 0, m > 0), 2 2 dx = 2 e R ∞ +xcos mx −bm −am (b) 0 (x2 +a2 )(x2 +b2 ) dx = 2(b2π−a2 ) ( e a − e b ) (a > 0, b > 0, m ≥ 0, b 6= a), R ∞ cos mx (c) −∞ (x+b)2 +a2 dx = πa e−am cos bm (a > 0, m ≥ 0), R∞ sin mx π −am sin bm (a > 0, m ≥ 0), 2 2 dx = − a e −∞ (x+b) R ∞ +a cos mx π −am (d) 0 (x2 +a2 )2 dx = 4a3 e (1 + am) (a > 0, m ≥ 0), R ∞ cos mx π −am (e) 0 x4 +4a4 dx = 8a3 e (cos am + sin am) (a > 0, m ≥ 0), R ∞ x3 sin mx π −am cos am (a > 0, m > 0). (f) 0 x4 +4a4 dx = 2 e Solution. (a) z a2 +z 2 Z ∞ −∞ has a simple pole in the upper half plane, which is at z = ai. Z ∞ Z ∞ x sin mx xemxi x cos mx dx + i dx = dx 2 2 2 2 a2 + x 2 −∞ a + x −∞ a + x = 2πi Res[ zemzi , ai] a2 + z 2 aie−am 2ai = πi e−am , = 2πi where we close the path in the upper half plane for the last integral involving e imz , since m > 0. The first integral in the first line is of course 0 because the integrand is odd. So, Z ∞ x sin mx dx = πe−am . 2 + x2 a −∞ SOLUTION SET IV FOR 18.075–FALL 2004 Note that (b) get x sin mx a2 +x2 1 z 2 +a2 9 is even, we have Z Z ∞ 1 ∞ x sin mx π −am x sin mx dx = dx = e . 2 2 2 2 a +x 2 −∞ a + x 2 0 has a simple pole on the upper half plane, which it at z = ai. Since m ≥ 0, we Z ∞ −∞ cos mx dx + i x2 + a 2 Z ∞ −∞ sin mx dx = x2 + a 2 Z ∞ −∞ emxi dx x2 + a 2 = 2πi Res[ = 2πi emxi , ai] z 2 + a2 e−am 2ai πe−am , a where we close the path in the upper half plane. The second integral in the first line vanishes because the integrand is odd. So, Z ∞ πe−am cos mx dx = . 2 2 a −∞ x + a = Since cos mx x2 +a2 Similarly, is even, we have Z ∞ Z ∞ 1 πe−am cos mx cos mx dx = dx = . x2 + a 2 2 −∞ x2 + a2 2a 0 Z ∞ 0 πe−bm cos mx dx = . x2 + b 2 2b So, Z ∞ 0 cos mx dx = 2 (x + a2 )(x2 + b2 ) = = = ∞ cos mx cos mx 1 ( 2 − 2 ) dx 2 2 −a x +a x + b2 0 Z ∞ Z ∞ 1 cos mx cos mx ( dx − dx) 2 2 2 2 b −a x +a x2 + b 2 0 0 πe−am πe−bm 1 ( − ) b2 − a 2 2a 2b π e−am e−bm ( − ). 2(b2 − a2 ) a b Z b2 (c) The given integrals are evaluated by the standard prescription as follows. Z ∞ Z ∞ im(x−b) Z ∞ eimx e cos mx dx = Re dx = Re dx 2 2 2 2 2 2 −∞ (x + b) + a −∞ x + a −∞ (x + b) + a Z ∞ Z ∞ im(x−b) Z ∞ eimx e sin mx dx = Im dx = Im dx, 2 2 2 2 2 2 −∞ (x + b) + a −∞ x + a −∞ (x + b) + a 10 SOLUTION SET IV FOR 18.075–FALL 2004 where Z eimx eimz π dx = 2πi Res = e−ma . z=ia 2 2 2 2 x +a z +a a ∞ −∞ Hence, Z ∞  π  cos mx −imb π −ma = e−ma cos mb, dx = Re e e (x + b)2 + a2 a a −∞ ∞   sin mx π −imb π −ma dx = Im e e = − e−ma sin mb. 2 2 a a −∞ (x + b) + a R∞ imx (d) We can calculate the real part of −∞ (x2e+a2 )2 dx by noticing that our function is even. As in question 91(d), we have one pole in the upper half-plane. This is a double pole at z = ia. Z  (z − ia)2 eimz Res(ia) = (z 2 + a2 )2 z=ia    imz d e = dz (z + ia)2 z=ia   (z + ia)2 imeimz − eimz 2(z + ia) = (z + ia)4 z=ia  d dz  (2ia)2 ime−ma − e−ma 4ia (2ia)4   −ma ma + 1 = −ie 4a3 = Therefore, Z 0 ∞  Z ∞ 1 eimx dx Re 2 2 2 2 −∞ (x + a )    1 −ma ma + 1 = Re 2πi · −ie 2 4a3   ma + 1 = πe−ma 4a3 cos mx dx = (x2 + a2 )2 (e) Clearly, Z 0 ∞ cos mx 1 dx = Re 4 4 x + 4a 2 Z ∞ −∞ eimx dx. x4 + 4a4 We calculate the last integral involving e imz by closing the path in the upper half plane since m > 0. The poles of the integrand occur at z 4 + 4a4 = 0 and are all simple. The SOLUTION SET IV FOR 18.075–FALL 2004 11 √ √ poles that lie in the upper half plane are z 1 = eiπ/4 2a and z2 = e3iπ/4 2a. Accordingly,   Z ∞ eimz eimz eimx dx = 2πi Resz=z1 4 + Resz=z2 4 4 4 z + 4a4 z + 4a4 −∞ x + 4a  imz1  e eimz2 = 2πi + 4z13 4z23 2πi [−(cosma + i sin ma)(1 + i) + (cos ma − i sin ma)(1 − i)] = 16a3 π = (cos ma + sin ma) 4a3 It follows that ∞ cos mx π dx = 3 (cos ma + sin ma). 4 4 x + 4a 8a 0 R ∞ 3 eimx (f) Once again, we can calculate the imaginary part of 0 xx4 +4a 4 dx noting that our √ 4 4 function is even and m > 0. The poles occur when z + 4a = 0 ⇒ z = ± ±2ia2 , as in part (e) above. Thus the roots are: Z √ √ 1 1 2ia2 = a 2( √ + i √ ) = a + ia since a > 0. 2 2 = −a1 = −a − ia. p −2ia2 = ia − a. = = −a3 = −ia + a. a1 = a2 a3 a4 Thus there are two poles in the upper half-plane: a 1 and a3 (both simple poles). Res(a1 ) = lim (z − a1 )f (z) z→a1 (z − a − ia)z 3 eimz z→a1 z 4 + 4a2 z 3 eimz = lim z→a1 (z + a + ia)(z 2 + 2ia2 ) = lim = (a + ia)2 eim(a+ia) 8ia2 z 3 eimz z→a3 (z + ia − a)(z 2 − 2ia2 ) (ia − a)2 e−ima−ma = −8ia Res(a3 ) = lim 12 SOLUTION SET IV FOR 18.075–FALL 2004 Therefore, Z ∞ 3 x sin mx dx = x4 + 4a4 0 = = = = 1 Im 2 Z ∞ −∞ x3 eimx dx x4 + 4a4     1 (a + ia)2 eima−ma − (ia − a)2 e−ima−ma Im 2π 2 8a hπ i 1 Im i(eima · e−ma + e−ima · e−ma ) 2 2   π −ma eima + e−ima e 2 2 π −ma e cos ma 2