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Unformatted text preview: ACM 95a/100a Problem Set 8 Solutions Prepared by: Zhiyi Li 11/30/2007 Total: 132 points Include grading section: 2 points Problem 1 (2 × 30 points) Evaluate the following improper integrals using appropriately indented contours a) P Z ∞∞ x 2 x 4 1 dx b) Z ∞ x 1 / 3 ( x + 8)( x + 1) dx Solution 1 a) If we consider the contour given in Figure 1, which takes into account the poles of the integrand at z = ± 1 , ± i , we see I ≡ P Z ∞∞ x 2 x 4 1 dx = lim ² → , R →∞ Z Γ 1 +Γ 2 +Γ 3 + C 1 + C 2 z 2 z 4 1 dz. If we also consider Z C R z 2 z 4 1 dz, we can apply the ML theorem as R → ∞ : Z C R z 2 z 4 1 dz ≤ πR R 2 R 4 1 → as R → ∞ . This means we can write I = lim R →∞ , ² → I Γ 1 +Γ 2 +Γ 3 + C 1 + C 2 + C R z 2 z 4 1 dz, and apply the residue theorem. The pole at z = i will contribute 2 πi Res( f ( z ) , i ) to the integral, while the poles which are enclosed as ² → 0 will contribute πi (Res( f ( z ) , 1) + Res( f ( z ) , 1)), so we have I = 2 πi Res z 2 z 4 1 , i + πi Res z 2 z 4 1 , 1 + Res z 2 z 4 1 , 1 . Each of these points is a simple pole, so we know Res z 2 z 4 1 , i = z 2 ( z 1)( z + 1)( z + i ) z = i = i 4 , Res z 2 z 4 1 , 1 = z 2 ( z i )( z + 1)( z + i ) z =1 = 1 4 , Figure 1: The contour and poles for problem 1a). and Res z 2 z 4 1 , 1 = z 2 ( z i )( z 1)( z + i ) z = 1 = 1 4 . So, we find I = π 2 . b) We want to define z 1 / 3 using a branch cut, but we have to choose the location of the cut carefully so it doesn’t intersect the contour we use for this integral. In the problems we have done so far, we typically chose a contour which travels along the positive real axis, around a semicircle at infinity, and back to the origin along the negative real axis, permitting a cut anywhere in the lower half of the plane. However, there is no way to parameterize any straight contour in the complex plane to reproduce this integral unless the contour is the positive real axis. Thus, we must integrate “out” along the positive real axis, and “back” to the origin along the same axis. These integrals will cancel each other unless there’s a discontinuity at the axis so that the values of the function z 1 / 3 when integrating from 0 to ∞ and then from ∞ to 0 differ. To achieve this, consider the “pacman” contour shown in Figure 2. A branch cut in this location can be achieved using the definition z 1 / 3 = r 1 / 3 e iθ/ 3 , θ ∈ (0 , 2 π ) ....
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 Fall '06
 NilesA.Pierce
 lim, Konrad Zuse, Methods of contour integration, Pole

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