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# Quiz6 - sin cos Re 2 2 2 2 2 2 1 1 i R iR i i i e e d d R e...

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Quiz 6 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 R R x dx x - = + because the integrand is odd. ( 29 ( 29 ( 29 ( 29
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Unformatted text preview: 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-+ = ā + + ā« ā« as R ā ā because ( 29 sin in the interval, that is the integrand becomes exponentially small. Therefore ( 29 ( 29 2 2 cos 1 x dx e x ā-ā = + ā« ....
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