ps7sol

# ps7sol - ACM 95a/100a Problem Set 7 Solutions Prepared by...

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ACM 95a/100a Problem Set 7 Solutions Prepared by: Zhiyi Li 11/20/2007 Total: 122 points Include grading section: 2 points Problem 1 (20 points) Evaluate the real integral Z π/ 2 0 sin 4 θ dθ Solution 1 We note that by symmetry Z π/ 2 0 sin 4 ( θ ) = 1 4 Z 2 π 0 sin 4 ( θ ) (1) Now, we can re-express the integral on the right hand side using the formula Z 2 π 0 f (cos( θ ) , sin( θ )) = I C f z + 1 z 2 , z - 1 z 2 i dz iz (2) where C ' z = re : r = 1 , θ = 0 to θ = 2 π . Thus, 1 4 Z 2 π 0 sin 4 ( θ ) = - i 4 I C ( z - 1 z ) 4 16 z dz = - i 4 2 πi Res ˆ ( z - 1 z ) 4 16 z , 0 ! = π 2 Res ˆ ( z - 1 z ) 4 16 z , 0 ! (3) since z = 0 is a pole which is enclosed positively by C . Since ( z - 1 z ) 4 16 z = 1 z 4 - 4 z 2 + 6 - 4 z 2 + z 4 16 z (4) the residue is 6 / 16 = 3 / 8. Therefore, Z π/ 2 0 sin 4 ( θ ) = π 2 3 8 = 3 π 16 (5) 1

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Problem 2 (20 points) Evaluate the improper integral Z 0 x 2 (1 + x 2 ) 2 dx, which arises in the study of Stokes flow as a sphere approaches a wall. Solution 2 x 2 (1+ x 2 ) 2 is even, so Z 0 x 2 (1 + x 2 ) 2 dx = 1 2 P Z -∞ x 2 (1 + x 2 ) 2 dx = lim R →∞ 1 2 Z Γ z 2 (1 + z 2 ) 2 dz (6) where Γ ≡ { z = x + iy : x = - R to x = R, y = 0 } . So, we consider I C z 2 (1 + z 2 ) 2 dz (7) where C Γ+ C R (see Figure 1) with C R ' z = re : r = R, θ = 0 to θ = π . Now, using the ML bound, lim R →∞ fl fl fl fl fl Z C R z 2 (1 + z 2 ) 2 dz fl fl fl fl fl lim R →∞ πR R 2 | R 2 - 1 | 2 = 0 (8) Since z 2 (1+ z 2 ) 2 = z 2 ( z + i ) 2 ( z - i ) 2 , C positively encloses a second order pole at z = i . Therefore, Z 0 x 2 (1 + x 2 ) 2 dx = lim R →∞ 1 2 Z Γ z 2 (1 + z 2 ) 2 dz = lim R →∞ 1 2 Z Γ+ C R z 2 (1 + z 2 ) 2 dz = lim R →∞ 1 2 I C z 2 (1 + z 2 ) 2 dz = 1 2 2 πi Res ˆ z 2 ( z + i ) 2 ( z - i ) 2 ; i ! = πi lim z i 1 1! d dz z 2 ( z + i ) 2 = πi lim z i ( z + i ) 2 2 z - z 2 2( z + i ) ( z + i ) 4 = πi ( - 4)(2 i ) - ( - 1)2(2 i ) 16 = π 4 (9) Note: We could have closed the contour below the real axis. That contour would negatively enclose the pole at z = - i , but the final answer would not be different. 2
Figure 1: The contour and poles for problem 2. Problem 3 (20 points) Evaluate the following improper integral using an appropriate contour Z 0 x x 4 + 4 dx. Solution 3 Let I Z 0 x x 4 + 4 dx (10) Note that the integrand is odd about x = 0, so could not use the contour C from problem 2 since P Z -∞ x x 4 + 4 dx = 0 (11) On the other hand, since i 4 = 1, it is useful to create a closed contour integral by adding to I a contour integral along the imaginary axis (where z = iy ) plus another contour integral for closing purposes. Consider I C z z 4 + 4 dz (12) where C Γ 1 + C R + Γ 2 (see Figure 2) with Γ 1 ≡ { z = x + iy : x = 0 to x = R, y = 0 } (13) C R n z = re : r = R, θ = 0 to θ = π 2 o (14) Γ 2 ≡ { z = x + iy : x = 0 , y = R to y = 0 } (15) With this contour, 3

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R iR C R Γ 1 Γ 2 Figure 2: The contour and two of the poles in Problem 3.
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