# Quiz6 - R R x dx x-= âˆ because the integrand is odd 29 29...

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EGM 4313 NAME:_____________________________ Spring 2009 Quiz 6 Open book/Closed notes/No calculators or computers 25 minute time limit Evaluate ( 29 ( 29 2 2 cos 1 x dx x -∞ + . Solution: Consider the integral ( 29 2 2 1 iz C e dz z + Ñ on the contour C shown. There is a second order pole at z=I with residue given by ( 29 ( 29 ( 29 ( 29 ( 29 2 2 2 2 3 2 2 lim lim lim 1 2 iz iz iz iz z i z i z i d e d e ie e z i dz dz z i z i z i z i e - = = - ÷ ÷ + + + + = - Therefore ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 Re 2 2 2 2 2 2 2 2 2 0 cos sin 2 2 1 1 1 1 i R R iz i i C R R x x e i e dz i dx i dx d e e z x x R e θ π - - = × - = = + + ÷ + + + + Ñ ( 29 ( 29 2 2 sin 0 1
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Unformatted text preview: R R x dx x-= + âˆ« because the integrand is odd. ( 29 ( 29 ( 29 ( 29 sin cos Re 2 2 2 2 2 2 1 1 i R iR i i i e e d d R e R e-+ = â†’ + + âˆ« âˆ« as R â†’ âˆž because ( 29 sin in the interval, that is the integrand becomes exponentially small. Therefore ( 29 ( 29 2 2 cos 1 x dx e x âˆž-âˆž = + âˆ« ....
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## This note was uploaded on 12/15/2011 for the course EGM 4313 taught by Professor Mei during the Fall '08 term at University of Florida.

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