# extra_13 - Extra Problems Sheet 13 Stephen Taylor June 3,...

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Unformatted text preview: Extra Problems Sheet 13 Stephen Taylor June 3, 2005 1 (a). Evaluate Z x 2 x 4 + x 2 + 1 dx We note this function has simple poles at e i/ 3 and e 2 i/ 3 . Since the integrand is even, we find Z x 2 x 4 + x 2 + 1 dx = 1 2 Z R x 2 x 4 + x 2 + 1 dx (1) We now define = Re it where (0 t ) and = [- R, R ] and consider the integral Z z 2 z 4 + z 2 + 1 dz = Z z 2 z 4 + z 2 + 1 dz + Z R x 2 x 4 + x 2 + 1 dx We first show that R z 2 z 4 + z 2 +1 dz 0 as R . Through standard analysis and inequality 3.4 of Chapter 1 we find, Z z 2 z 4 + z 2 + 1 dz Z R 2 e 2 it R 4 e 4 it + R 2 e 2 it + 1 iRe it dt Z R 3 | R 4 e 4 it + R 2 e 2 it + 1 | dt Z R 3 R 4- R 2- 1 dt = R 3 R 4- R 2- 1 dt We compute the two residues interior to to obtain a value for the second integral Res( e i/ 3 ) = lim z e i/ 3 ( z- e i/ 3 ) z 2 z 4 + z 2 + 1 dz = 1 4- i 3 12 Res( e 2 i/ 3 ) = lim z e 2 i/ 3 ( z- e 2 i/...
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## This note was uploaded on 10/19/2009 for the course MATH 814 taught by Professor Cong during the Three '09 term at University of Adelaide.

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extra_13 - Extra Problems Sheet 13 Stephen Taylor June 3,...

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