Sol9 - MAE294B/SIO203B Methods in Applied Mechanics Winter Quarter 2010 http/maecourses.ucsd.edu/mae294b Solution IX 1 The first integrand has no

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Unformatted text preview: MAE294B/SIO203B: Methods in Applied Mechanics Winter Quarter 2010 http://maecourses.ucsd.edu/mae294b Solution IX 1 The first integrand has no singularities. Deform the contour slightly below the real axis and then write sin z = 1 2 ( e i z- e- i z ) to give Z c- e 2i z + 2- e- i z 4 z 2 d z . The last term decays on a semicircle in the lower half-plane and since the integrand has no singu- larity below the contour there is no contribution. The first two terms decay on a semicircle in the upper half-plane. The origin is a pole of order 2 and the residue can be found from- 1- 2i z + ··· + 2 4 z 2 = 1 4 z 2- i 2 z + ··· . For the second integral, make the substitution x = log x and find I = 4 Z ∞ u ( log u ) 2 ( u 2- 1 ) 2 d u . The most convenient contour to take is the keyhole contour with the branch cut for the logarithm along the negative real axis with the branch real and positive on the positive real axis. The numer- ator needs an extra log term because the leading-order logarithm cancels. That’s not enough as itator needs an extra log term because the leading-order logarithm cancels....
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This note was uploaded on 09/22/2010 for the course MAE MAE294B taught by Professor Mae294b during the Spring '09 term at UCSD.

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Sol9 - MAE294B/SIO203B Methods in Applied Mechanics Winter Quarter 2010 http/maecourses.ucsd.edu/mae294b Solution IX 1 The first integrand has no

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