ans12 - Math 185. Sample Answers to Problem Set #12 Page...

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Math 185. Sample Answers to Problem Set #12 Page 265 7. Since the integrand is even, it will suffice to find the Cauchy principal value of the integral. Let f ( z ) = ze iz / (( z 2 + 1)( z 2 + 4)) . For R > 2 , integration around the contour consisting of the real interval [ - R,R ] and the upper half C R of the circle | z | = R yields the equation Z R - R xe ix dx ( x 2 + 1)( x 2 + 4) = 2 πi ± Res z = i f ( z ) + Res z =2 i f ( z ) ² - Z C R f ( z ) dz (since the poles of f ( z ) inside of the given contour occur at i and 2 i ). Since ³ ³ ³ ³ z ( z 2 + 1)( z 2 + 4) ³ ³ ³ ³ R ( R 2 - 1)( R 2 - 4) for all z C R with R > 4 , and since the above upper bound tends to 0 when R → ∞ , Jordan’s lemma applies and we have lim R →∞ Z C R f ( z ) dz = 0 . We also note that Res z = i ze iz ( z 2 + 1)( z 2 + 4) = ie i 2 2 i ( i 2 + 4) = i/e 6 i = 1 6 e and Res z =2 i ze iz ( z 2 + 1)( z 2 + 4) = 2 ie 2 i 2 (4 i 2 + 1)(4 i ) = 2 i/e 2 - 12 i = - 1 6 e 2 . Thus, we have lim R →∞ Z R - R xe ix dx ( x 2 + 1)( x 2 + 4) = 2 πi ´ 1 6 e - 1 6 e 2 µ = πi 3 ´ 1 e - 1 e 2 µ . Taking imaginary parts then gives Z -∞ x sin xdx ( x 2 + 1)( x 2 + 4) = π 3 ´ 1 e - 1 e 2 µ . 10. Let f ( z ) = ( z + 1) e iz / ( z 2 + 4 z + 5) . This function is analytic everywhere except for poles at - 2 ± i . Integrating around the same contour as in the previous problem then gives Z R - R ( x + 1) cos x x 2 + 4 x + 5 dx = 2 πi Res z = i - 2 f ( z ) - Z C R f ( z ) dz 1
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2 for all R > | 2 - i | . If R > 4 then ± ± ± ± z + 1 z 2 + 4 z + 5 ± ± ± ± R + 1 R 2 - 4 R - 5 , and this bound tends to 0 as R → ∞ . Therefore, Jordan’s lemma gives lim R →∞ Z C R f ( z ) dz = 0 . Also,
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This note was uploaded on 09/01/2010 for the course MATH 185 taught by Professor Lim during the Fall '07 term at Berkeley.

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ans12 - Math 185. Sample Answers to Problem Set #12 Page...

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