s6_95-100aSOLUTION

s6_95-100aSOLUTION - ACM95a/100a Last updated November 18,...

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Unformatted text preview: ACM95a/100a Last updated November 18, 2009 Problem Set VI Solutions Prepared by Faisal Amlani 1. (20 pts.) Use contour integration to evaluate the integrals 2π (a) (10 pts.) 0 dθ , 2 + sin(θ) π (b) (10 pts.) −π cos(nθ) dθ (|a| < 1). 1 − 2a cos(θ) + a2 For the second integral, assume n is a non-negative integer. SOLUTION : (a) Let C be the positively oriented unit circle about the origin. Defining a parametrization z = eiθ , dz = i eiθ dθ, θ ∈ (0, 2π ), we can write sin θ and the differential dθ in terms of z to get 2π 0 1 dθ = 2 + sin θ C 1 dz . 2 + (z − 1/z )/(2i) iz Then by the residue theorem, 2π 0 1 dθ = 2 + sin θ = C C = C 1 dz 2 + (z − 1/z )/(2i) iz 2 dz z 2 + 4iz − 1 2 √ √ dz z+ 2+ 3 i z+ 2− 3 i z+ 2+ √ 2 3i √ z+ 2− 3 i z+ 2− = 2πi Res 2 = 2πi √ 2 3i 2π =√. 3 (b) If a = 0, we have √ 3 i , z = −2 + √ 3i π cos(nθ ) dθ = −π 0 2π for n = 1, 2, ... for n = 0. If |a| < 1, a = 0 then sin(nθ ) 1 − 2a cos θ + a2 is an even function and hence its integral vanishes over the domain. So we can add it to the integral to get π −π cos(nθ ) dθ = 1 − 2a cos θ + a2 = π −π π −π cos(nθ ) dθ + 1 − 2a cos θ + a2 einθ dθ 1 − 2a cos θ + a2 π −π sin(nθ ) dθ 1 − 2a cos θ + a2 Let C be the positively oriented unit circle about the origin. Defining a parametrization z = eiθ , dz = i eiθ dθ, θ ∈ (−π . . . π ), zn dz . 1 − a(z + 1/z ) + a2 iz we can write the integrand and the differential dθ in terms of z to get π −π einθ dθ = 1 − 2a cos θ + a2 C Then by the residue theorem, π −π einθ dθ = 1 − 2a cos θ + a2 zn dz 1 − a(z + 1/z ) + a2 iz C zn dz = −i 2 2 C −az + (1 + a )z − a i zn = dz a C z 2 − (a + 1/a)z + 1 zn i dz = a C (z − a)(z − 1/a) zn i = 2πi Res ,z = a a (z − a)(z − 1/a) 2π an =− a a − 1/a 2πan = 1 − a2 Summarizing the two cases we get π −π cos(nθ ) dθ = 1 − 2a cos θ + a2 2π 2πan 1−a2 for a = 0, n = 0 otherwise 2. (30 pts.) Use contour integration to evaluate the integrals ∞ (a) (b) (10 pts.) −∞ ∞ dx , 1 + x4 x2 dx , (1 + x2 )2 (10 pts.) −∞ ∞ (c) (10 pts.) −∞ cos(x) dx. (1 + x2 ) SOLUTION : To aid with the rest of this assignment, we state the following result: Result 1 Let f (z ) be analytic except for isolated singularities, with only first order poles on the real axis. Let CR be the semi-circle from R to −R in the upper half plane. If R→∞ lim R max |f (z )| z ∈CR =0 n then − ∞ m f (x) dx = 2πi k =1 Res (f (z ), zk ) + πi k =1 Res (f (z ), xk ) −∞ where z1 , . . . zm are the singularities of f (z ) in the upper half plane and x1 , . . . , xn are the first order poles on the real axis. Now let CR be the semi-circle from R to −R in the lower half plane. If R→∞ lim R max |f (z )| z ∈CR =0 n then − ∞ −∞ m f (x) dx = −2πi k =1 Res (f (z ), zk ) − πi Res (f (z ), xk ) k =1 where z1 , . . . zm are the singularities of f (z ) in the lower half plane and x1 , . . . , xn are the first order poles on the real axis. (a) Consider ∞ −∞ x4 1 dx. +1 The z = {e , e plane. Since integrand z 41 is analytic on +1 iπ/4 i3π/4 i5π/4 i7π/4 ,e ,e lim the real axis and has isolated singularities at the points }. Let CR be the semi-circle of radius R in the upper half z4 1 +1 = lim R→∞ R→∞ R max z ∈CR R R4 we can apply Result 1 to get ∞ −∞ 1 −1 = 0, x4 1 dx = 2πi +1 Res z4 1 , eiπ/4 +1 + Res z4 1 , ei3π/4 +1 . The residues are Res 1 , eiπ/4 4+1 z z − eiπ/4 = lim 4 z → eiπ/4 z + 1 1 = lim 3 z → eiπ/4 4z 1 = e−i3π/4 4 −1 − i =√ 42 and Res 1 , ei3π/4 4+1 z = 1 4( ei3π/4 )3 1 = e−iπ/4 4 1−i = √, 42 Thus the integral from the residue theorem is ∞ −∞ x4 1 −1 − i 1 − i √ +√ dx = 2πi +1 42 42 π = √. 2 x2 dx. (x2 + 1)2 (b) Consider ∞ −∞ The integrand is analytic on the real axis and has second order poles at z = ±i. Since the integrand decays sufficiently fast at infinity, lim R max z ∈CR R→∞ z2 (z 2 + 1)2 = lim R→∞ R R2 (R2 − 1)2 = 0, we can apply Result 1 to get ∞ −∞ x2 dx = 2πi Res (x2 + 1)2 z2 ,z = i . (z 2 + 1)2 The residue is d z2 (z − i)2 2 z →i dz (z + 1)2 2 z d = lim z →i dz (z + i)2 (z + i)2 2z − z 2 2(z + i) = lim z →i (z + i)4 i =− . 4 Thus the integral from the residue theorem is simply Res = lim ∞ −∞ z2 ,z = i (z 2 + 1)2 x2 π dx = . 2 + 1)2 (x 2 ∞ −∞ (c) Consider cos(x) dx. (1 + x2 ) Since sin(x) 1 + x2 is an odd function, it vanishes on the domain. We can then add a multiple of it to our integral to get ∞ −∞ cos(x) dx = 1 + x2 = ∞ −∞ ∞ −∞ cos(x) dx + i 1 + x2 eix dx. 1 + x2 ∞ −∞ sin(x) dx 1 + x2 Since e /(1 + z ) is analytic except for simple poles at z = ±i, and the integrand decays sufficiently fast in the upper half plane, R→∞ iz 2 lim R max z ∈CR eiz 1 + z2 = lim R→∞ R we can apply Result 1 to get ∞ −∞ 1 R2 − 1 = 0, ∞ −∞ eix dx = 2πi Res 1 + x2 e−1 = 2πi 2i π = e =⇒ π cos(x) dx = . 2 1+x e eiz ,z = i (z − i)(z + i) 3. (20 pts.) For 0 < a < 1 show that ∞ (a) (b) (10 pts.) −∞ ∞ eax π dx = , ex + 1 sin(πa) cosh(ax) πa dx = π sec . cosh(x) 2 (10 pts.) −∞ [Hint: For the first integral integrate exp(az )/(exp(z ) + 1) around a rectangle with vertices at −R, R, R + 2πi and −R + 2πi. The second integral can be obtained similarly, or deduced from the first.] SOLUTION : (a) Consider eax dx. x −∞ e + 1 Let C be the positively-oriented rectangular contour with corners at ±R and ±R + 2πi. From the maximum modulus integral bound we see that the integrals on the vertical sides of the contour vanish as R → ∞: ∞ R+i2π R −R −R+i2π eaz eaR dz ≤ 2π R → 0 as R → ∞ ez + 1 e −1 e−aR eaz dz ≤ 2π → 0 as R → ∞. ez + 1 1 − e−R Thus in the limit R → ∞, the integral on the rectangular contour is the sum of the integrals along the top and bottom sides: eaz dz = ez + 1 ∞ −∞ C eax dx + ex + 1 ∞ −∞ −∞ ∞ ax ea(x+2πi) dx ex+2πi + 1 dx. = (1 − e−2aπi ) e ex +1 Since the only singularity of the integrand inside the contour is a first order pole at z = iπ , we can use the residue theorem to get eaz dz = 2πi Res ez + 1 eaz , iπ ez + 1 (z − πi) eaz = 2πi lim z →iπ ez + 1 a(z − πi) eaz + eaz = 2πi lim z →iπ ez aπi = −2πi e . C Equating the two results gives ∞ (1 − e−2aπi ) −∞ ∞ −∞ eax dx ex + 1 eax dx ex + 1 = =⇒ = = −2πi eaπi eaπi 2πi − e−aπi π sin(πa) (b) Consider ∞ −∞ cosh(ax) dx. cosh x From a change of variables x → 2x in the result of part (a), ∞ −∞ ∞ −∞ e2ax 2 dx e2x + 1 2e dx ex + e−x e dx cosh x (2a−1)x (2a−1)x = =⇒ = =⇒ = π sin(πa) π sin(πa) π . sin(πa) ∞ −∞ So if we let b = 2a − 1, we have that ∞ −∞ π π ebx dx = = cosh x sin(π (b + 1)/2) cos(πb/2) for − 1 < b < 1. Since cosine is even, we also have that ∞ −∞ e−bx π dx = cosh x cos(πb/2) for − 1 < b < 1. Thus ∞ −∞ cosh(ax) dx = cosh x = = = = eax + e−ax dx 2 cosh x −∞ 1 ∞ e−ax 1 ∞ eax dx + dx 2 −∞ cosh x 2 −∞ cosh x 1 π 1 π + 2 cos(πa/2) 2 cos(πa/2) π cos(πa/2) πa π sec . 2 ∞ 4. (30 pts.) Obtain the formulae ∞ (a) (b) (15 pts.) −∞ ∞ 1 − cos(x) dx = π , x2 sin(πx) dx = π . x(1 − x2 ) (15 pts.) 0 SOLUTION : (a) This is integrable since (1 − cos x)/x2 has a removable singularity at x = 0 and decays 1 like x2 at infinity. Since (sin x)/x2 is a odd function with a simple pole at x = 0, the principal value of its integral vanishes so we can subtract a multiple of it to get ∞ −∞ 1 − cos x dx = x2 ∞ 1 − cos x sin x dx − i − dx 2 2 x −∞ −∞ x ∞ 1 − cos x − i sin x dx =− x2 −∞ ∞ 1 − eix =− dx. x2 −∞ ∞ Let CR be the semi-circle of radius R in the upper half plane. Since R→∞ lim R max z ∈CR 1 − eiz z2 = lim R R→∞ 2 =0 R2 the integral along CR vanishes as R → ∞, CR 1 − eiz dz → 0 as R → ∞. z2 1 − eiz ,z = 0 z2 1 − eiz πi lim z →0 z −i eiz πi lim z →0 1 −i2 π π πi Res π. Then applying Result 1 gives ∞ − −∞ 1 − eix dx x2 = = = = = =⇒ = ∞ −∞ 1 − cos x dx x2 (b) Consider ∞ 0 Again, this is integrable since the integrand has removable singularities at the points x = 0, ±1. Then, since the integrand is also clearly even, we have that ∞ 0 sin(πx) dx. x(1 − x2 ) 1 sin(πx) dx = x(1 − x2 ) 2 ∞ −∞ sin(πx) dx. x(1 − x2 ) Since cos(πx) is an odd function with first order (removable) poles at x = 0, ±1, it’s x(1 − x2 ) integral vanishes over (−∞, ∞) so we can add it as follows: ∞ 0 1 ∞ sin(πx) sin(πx) dx = dx x(1 − x2 ) 2 −∞ x(1 − x2 ) ∞ i sin(πx) = dx 2i −∞ x(1 − x2 ) ∞ 1 ∞ cos(πx) sin(πx) i dx + − dx = 2i −∞ x(1 − x2 ) 2i −∞ x(1 − x2 ) 1 ∞ cos(πx) + i sin(πx) = − dx 2i −∞ x(1 − x2 ) eiπx 1∞ − dx = 2i −∞ x(1 − x2 ) i∞ eiπx =−− dx. 2 −∞ x(1 − x2 ) eiπz z (1 − z 2 ) 1 R (R 2 − 1) Let CR be the semi-circle of radius R in the upper half plane. Since lim R max z ∈CR R→∞ = lim R R→∞ =0 the integral along CR vanishes as R → ∞, i.e. CR eiπz dz → 0 as R → ∞, z (1 − z 2 ) so we can apply Result 1 to get eiπx i∞ −i −− dx = πi 2 −∞ x(1 − x2 ) 2 Res + Res eiz , z = 0 + Res z (1 − z 2 ) eiz ,z = 1 z (1 − z 2 ) π 2 π = 2 =π = ∞ 0 eiz , z = −1 z (1 − z 2 ) eiπz eiπz eiπz + lim − lim lim z →0 z (1 − z ) z →0 z (1 + z ) z →0 1 − z 2 −1 −1 + 1− 2 −2 sin(πx) dx = π. x(1 − x2 ) =⇒ ...
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