s7_95-100aSOLUTION

s7_95-100aSOLUTION - ACM95a/100a Last updated November 25,...

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Unformatted text preview: ACM95a/100a Last updated November 25, 2009 Problem Set VII Solutions Prepared by Faisal Amlani 1. (20 pts.) Integrate exp(−z 2 ) around the boundary of the sector |z | < R, 0 < θ < π/4 and deduce that ∞ ∞ ∞ 1 2 sin(t2 )dt = cos(t2 )dt = √ e−t dt. 20 0 0 [Hint: On the curved part of the contour | exp(−z 2 )| = exp(−r 2 cos 2θ). Show analytically that cos 2θ ≥ 1 − π 4 θ, 0 ≤ θ ≤ . π 4 This makes it easy to estimate the integral by a calculation similar to that in the proof of Jordan’s lemma.] The first two integrals are known as Fresnel integrals. Even though the integrands don’t decrease with increasing t, the integrals are finite because of the increasingly rapid oscillation of the √ integrands as t → ∞. The latter integral is known from Calculus to equal π/2. SOLUTION : Figure 1 visualizes the two functions cos(2θ) ≥ 1 − 4 θ, π 0≤θ≤ π . 4 4 Figure 1: cos(2θ) and 1 − π θ To prove this inequality analytically, we can note that (1) the two functions are equal at the endpoints of the interval, (2) 1 − 4θ/π is linear and (3) cos(2θ) is concave down on the interval, i.e. π d2 cos(2θ) = −4 cos(2θ) ≤ 0 for 0 ≤ θ ≤ . 2 dθ 4 Let C be the positively oriented closed curve that bounds the domain 0 < r < R, 0 < θ < π/4. Then if CR denotes the quarter-circle arc of radius R, C1 the boundary θ = 0 and C2 the boundary θ = π/4, we have that e−z dz = C CR 2 e−z dz + C1 2 e−z dz + C2 2 e−z dz 2 =0 since the integrand is clearly analytic inside C so the left hand side vanishes by Cauchy’s integral formula. Along the arc, e CR −z 2 π /4 dz ≤ ≤ ≤ e−(R e 0 π /4 0 π /4 0 iθ )2 Ri eiθ dθ dθ dθ R e−R R e−R 2 cos(2θ ) 2 (1−4θ/π ) π −R2 (1−4θ/π) e 4R 2 π 2 1 − e−R = 4R → 0 as R → ∞. =R Along C1 , where z = t ei0 = t =⇒ dz = dt, e−z dz = C1 0 2 π /4 0 R e−t dt ∞ 0 2 = e−t dt as R → ∞. 2 Along C2 , where the contour stipulates a direction from R to 0, z = teiπ/4 = 1+i √ dt, 2 e−z dz = C2 2 1+i √t 2 =⇒ dz = = = = = = 1+i √ dt 2 R R 1+i 2 1 + i −√ t √ dt e2 − 2 0 “ ”2 ∞ 1+i − √ t 1+i 2 √ dt as R → ∞ e − 2 0 1 + i ∞ −it2 dt e −√ 20 1+i ∞ cos(t2 ) − i sin(t2 ) dt −√ 20 ∞ ∞ i 1 sin(t2 ) dt − √ cos(t2 ) dt + −√ 2 2 0 0 e √ − 1+i t2 2 2 2 2 0 ∞ 0 cos(t2 ) dt − ∞ 0 sin(t2 ) dt . Putting it all together, we see that as R → ∞, e−z dz C 2 = =⇒ CR e−z dz + C1 ∞ 0 e−z dz + C2 e−z dz 0 = 0+ e−t dt ∞ 0 2 ∞ 0 e −t2 =⇒ dt = 1 −√ 2 1 √ 2 ∞ 0 cos(t2 ) dt + 0 ∞ 0 ∞ i sin(t2 ) dt − √ 2 ∞ 0 ∞ 0 cos(t2 ) dt − ∞ 0 ∞ 0 sin(t2 ) dt . cos(t2 ) dt + i sin(t2 ) dt + √ 2 cos(t2 ) dt − sin(t2 ) dt . Since the integral is real, the imaginary part is zero which implies ∞ 0 cos(t2 ) dt = 0 ∞ sin(t2 ) dt. From this, the remaining real part shows that ∞ 0 cos(t2 ) dt = 0 ∞ 1 sin(t2 ) dt = √ 2 ∞ 0 e−t dt 2 2. (30 pts.) Evaluate the following integrals using contour integration: (a) (10 pts.) 0 ∞ dx (1 + x2 )2 ◦ , (b) (20 pts.) 0 ∞ dx . 1 + x3 [Hint for (b): use the boundary of a 120 circular sector of large radius as integration contour] SOLUTION : (a) We have that by evenness, ∞ 0 dx 1 = 2 )2 2 (1 + x ∞ −∞ dx . (1 + x2 )2 Let C be the positively oriented closed semi-circle in the upper half plane. Then if CR denotes the arc of radius R, C1 the boundary on the x-axis, we have that dz (1 + z 2 )2 = = = = = = =⇒ = dz dz 2+ 2 22 CR (1 + z ) C1 (1 + z ) 1 ,z = i 2πi Res (1 + z 2 )2 1 2πi Res ,z = i (z − i)2 (z + i)2 1 d 2πi lim z →i dz (z + i)2 −2 2πi lim z →i (z + i)3 π 2 dz + (1 + z 2 )2 dz (1 + z 2 )2 C π 2 CR C1 by the residue theorem. Clearly, dz (1 + z 2 )2 ≤ = = ≤ = → dz 22 CR (1 + z ) π 2 1 |Ri eiθ |dθ (1 + (R eiθ )2 0 π R dθ (1 + R2 e2iθ )2 0 π R dθ (R2 − 1)2 0 R π2 (R − 1)2 0 as R → ∞. dz = 0. (1 + z 2 )2 CR One could also use Result 1 from Homework 6 to show the above. So we have as R → ∞, CR We also have that from the direction of the contour along the x-axis, dz (1 + z 2 )2 =− → dx dx (1 + x2 )2 R ∞ dx dx 22 −∞ (1 + x ) −R C1 as R → ∞. Putting the results together gives as R → ∞, π 2 = = = ∞ 0 dx (1 + x2 )2 =⇒ = dz dz + 2 )2 22 CR (1 + z C1 (1 + z ) ∞ dx 0+ dx 22 −∞ (1 + x ) ∞ dx 2 (1 + x2 )2 0 π . 4 (b) Let C be the closed boundary of 0 < r < R, 0 < θ < 2π/3, positively oriented. Factoring, we see that 1 1 = , 3+1 iπ/3 )(z + 1)(z − e−iπ/3 ) z (z − e which implies a simple pole at eiπ/3 in the interior of C for R > 1. Then by the residue theorem, we have that z3 dz = 2πi Res +1 1 , z = eiπ/3 +1 1 (z − eiπ/3 ) 3 = 2πi lim iπ/3 z +1 z→ e 1 = 2πi lim (z + 1)(z − e−iπ/3 ) z → eiπ/3 2πi = . iπ/3 + 1)( eiπ/3 − e−iπ/3 ) (e z3 C Let CR be the circular arc portion of the contour. Then, we can split the contour C making note of the directions into the arc plus the two lines as dz 3+1 z R = 0 C dx + 3+1 x 0 CR R dz − 3+1 z R 0 e2iπ/3 dx x3 + 1 dz +1 = (1 + e−iπ/3 ) → (1 + e−iπ/3 ) ∞ 0 x3 dx + +1 dx 3+1 x CR z3 as R → ∞ since CR z3 Thus we have that as R → ∞, C dz 2πR 1 ≤ → 0 as R → ∞. +1 3 R3 − 1 = =⇒ = =⇒ = = = (1 + e (1 + e −iπ/3 dz 3+1 z ) 0 ∞ x3 ∞ dx +1 2πi iπ/3 + 1)( eiπ/3 − e−iπ/3 ) (e ∞ 0 −iπ/3 ) 0 x3 dx +1 x3 dx +1 2πi (1 + + eiπ/3 )( eiπ/3 − e−iπ/3 ) 2πi √ √1 √ 1 (3 + i 3) 2 (3 − i 3)(i 3) 2 e−iπ/3 )(1 2π √. 33 3. (30 pts.) Show that (a) (15 pts.) 0 ∞ π3 (ln x)2 dx = 1 + x2 8 , (b) (15 pts.) 0 ∞ ln x dx = 0. 1 + x2 [Hint: Obtain both integrals simultaneously by integrating some branch of (log z )2 /(1 + z 2 ) around a large semicircle indented at z = 0.] SOLUTION : Let C be the closed counterclockwise curve bounding the indented semicircle ǫ < r < R, 0 < θ < π , for ǫ > 0. If we choose the branch of the logarithm as a cut on the negative imaginary axis with angle range −π/2 < θ < 3π/2, we can compute the integral C (log z )2 dz = 2πi Res 1 + z2 (log z )2 ,z = i 1 + z2 (log z )2 = 2πi lim z →i z + i (iπ/2)2 = 2πi 2i π3 =− 4 by the residue theorem. Let CR be the semi-circular arc from R to −R in the upper half plane and Cǫ be the semi-circle from −ǫ to ǫ in the upper half plane. Then we can split the integral over the contour C as C (log z )2 dz = 1 + z2 CR (log z )2 dz + 1 + z2 Cǫ (log z )2 dz + 1 + z2 ∞ 0 (ln x)2 dx + 1 + x2 0 −∞ (ln |x| + πi)2 dx 1 + x2 But along CR , (log z )2 (log z )2 dz ≤ πR max z ∈CR 1 + z 2 1 + z2 (ln R)2 + 2π ln R + π 2 ≤ πR R2 − 1 → 0 as R → ∞ (log z )2 (log z )2 dz ≤ πǫ max z ∈Cǫ 1 + z 2 1 + z2 (ln ǫ)2 − 2π ln ǫ + π 2 ≤ πǫ 1 − ǫ2 → 0 as ǫ → 0. (log z )2 dz 1 + z2 (log z )2 dz + 1 + z2 ∞ 0 CR and along Cǫ , Cǫ Thus as ǫ → 0 and R → ∞ for the integral along C , = CR C Cǫ (log z )2 dz 1 + z2 0 −∞ + = π3 4 =⇒ = 0 0 ∞ ∞ 0 ∞ 0 ∞ (ln x)2 1 + x2 dx + 0 (ln |x| + πi)2 dx 1 + x2 (ln x)2 1 + x2 dx + −∞ 0 −∞ 0 (ln |x| + πi)2 dx 1 + x2 (ln |x| + πi)2 dx 1 + x2 − (ln x)2 dx + 1 + x2 (ln x)2 dx + = = ∞ 0 2 (ln x)2 dx + 2πi 1 + x2 ∞ 0 ln x dx 1 + x2 =⇒ = (ln x + πi)2 (−dx) 1 + x2 1 + x2 ∞ ∞ (ln x + πi)2 (ln x)2 dx + dx 1 + x2 1 + x2 0 ∞ π2 0 1 π3 dx − . 1 + x2 4 To evaluate the integral on the right hand side, we consider the usual semi-circular C0 contour by noting the evenness of the integrand. That is, since ∞ 0 1 1 dx = 2 1+x 2 ∞ −∞ 1 dx, 1 + x2 then dz 1 + z2 = → as R → ∞ since lim R max 1 1 + z2 dz + 2 CR 1 + z ∞ dx 2 −∞ 1 + x ≤ lim R R −R C0 dx 1 + x2 R→∞ z ∈CR R→∞ R2 But by the residue theorem, we have ∞ −∞ 1 −1 = 0. 1 dx = 2πi Res 1 + x2 1 ,z = i 1 + z2 1 = 2πi lim z →i z + i = π. Thus we have that ∞ 0 1 dx = 1 + x2 = 1 2 π . 2 ∞ −∞ 1 dx 1 + x2 Plugging this into our original expression of interest, we have that 2 0 ∞ (ln x)2 dx + 2πi 1 + x2 ∞ 0 ∞ ln x 1 π3 dx = π 2 dx − . 1 + x2 1 + x2 4 0 3 3 π π − = 2 4 π3 = . 4 Equate the real and imaginary parts gives both integrals as (a) 0 ∞ (ln x)2 π3 dx = 1 + x2 8 ln x dx = 0 1 + x2 (b) 0 ∞ 4. (20 pts.) Evaluate x−α dx 1 + x4 0 for α real. What are the restrictions on real α for this integral to exist? ∞ SOLUTION : Existence: We want to consider where the integral ∞ 0 x−α dx 1 + x4 converges to a finite number. Since the integrand behaves like x−α near x = 0 we must have −α > −1 =⇒ α < 1. Since the integrand behaves like x−α−4 at infinity we must also have −α − 4 < −1 =⇒ α > −3. The integral converges for −3 < α < 1. Computation: The function z −α f (z ) = 1 + z4 √ clearly has first order poles at z = (±1 ± i)/ 2 and a branch point at z = 0. If we define a branch by placing a cut on the positive real axis with 0 < arg(z ) < 2π , an obvious contour for integration is given by the counterclockwise cut contour depicted in Figure 2. Figure 2: Possible Path of Integration for f (z ) = z −α 1+z 4 After demonstrating that the integrals along Cǫ and CR vanish we would need to calculate four residues. However, there are other–perhaps easier–contours. For example, Consider the value of the integrand on the line arg(z ) = θ, f (r eiθ ) = r −α e−iαθ . 1 + r 4 ei4θ If θ is a integer multiple of π/2 then the integrand is a constant multiple of f (x) = r −α . 1 + r4 Thus any of the contours in Figure 3 can be used with the main difference the number of residue calculations. Figure 3: Possible Paths of Integration for f (z ) = z −α 1+z 4 For this solution write-up, we will choose the first one (call it C ) depicted in Figure 3, but any one can be used. This places a branch cut on the negative real axis by choosing the branch −π < arg(z ) < π . Then we have from the value of the integrand on the positive imaginary axis z = x eiπ/2 , x−α e−iπα/2 (x eiπ/2 )−α = , 1 + (x eiπ/2 )4 1 + x4 the integral expression z −α dz = 1 + z4 → z −α dz + 1 + z4 x dx + 1 + x4 −α C CR ∞ 0 Cǫ 0 ∞ z −α dz + 1 + z4 x −α R 0 x−α dx + 1 + x4 0 R x−α e−iπα/2 iπ/2 e dx 1 + x4 e 1 + x4 −iπα/2 eiπ/2 dx as R → ∞ and ǫ → 0 by noting that from z = r eiθ we have |z −α | = r −α so that Cǫ z −α πǫ z −α dz ≤ max 1 + z4 2 z ∈Cǫ 1 + z 4 πǫ ǫ−α ≤ 2 1 − ǫ4 → 0 as ǫ → 0 z −α πR z −α dz ≤ max 1 + z4 2 z ∈CR 1 + z 4 πR R−α ≤ 2 R4 − 1 → 0 as R → ∞. and that CR By the residue theorem, we have that z −α dz = 2πi Res 1 + z4 z −α , eiπ/4 1 + z4 z −α (z − eiπ/4 ) = 2πi lim 1 + z4 z → eiπ/4 −αz −α (z − eiπ/4 ) + z −α = 2πi lim 4z 3 z → eiπ/4 e−iπα/4 = 2πi i3π/4 . 4e C Putting it altogether gives as R → ∞, ǫ → 0, 2πi e−iπα/4 4 ei3π/4 = = = ∞ 0 z −α dz 4 C 1+z ∞ x−α dx + 1 + x4 0 1 − eiπ(−α+1)/2 2πi x dx 1 + x4 −α =⇒ = = = = = = = x−α e−iπα/2 iπ/2 e dx 1 + x4 ∞ ∞ x−α dx 1 + x4 0 0 1 − eiπ(−α+1)/2 −1 πi iπ (−α+1)/2 1− e e−iπ −πi 1 − eiπ(−α+1)/2 1 −πi iπ (−α+1)/2 ) e−iπ (−α+1)/4 2(1 − e −πi 2( e−iπ(−α+1)/4 − eiπ(−α+1)/4 ) −2i π −iπ (−α+1)/4 − eiπ (−α+1)/4 ) 4 (e π csc 4 π (−α + 1) 4 . e−iπα/4 4 ei3π/4 e−iπα/4 2 ei3π/4 e−iπα/4 2 e−iπ/4 ...
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