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Complex Analysis - Maths KU

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2<br />

6.7 Applications 379<br />

s = γ + Re iθ , π/2 ≤ θ ≤ 3π/2, then ds = Rie iθ dθ =(s − γ)idθ, and so,<br />

�<br />

1<br />

2πi<br />

��<br />

1 �<br />

�<br />

2π �<br />

CR<br />

CR<br />

e st F (s) ds = 1<br />

� 3π/2<br />

e<br />

2πi π/2<br />

γt+Rteiθ<br />

F (γ + Re iθ )Rie iθ dθ<br />

e st � �<br />

�<br />

3π/2<br />

F (s) ds�<br />

1<br />

�<br />

�<br />

� ≤ �e<br />

2π<br />

γt+Rteiθ�� �<br />

� �F (γ + Re iθ ) � � � �Rie iθ��dθ. (11)<br />

π/2<br />

To find an upper bound for the expression in (11) we examine the three moduli<br />

of the integrand of the right-hand side.First,<br />

↓ since � �eiRt sin θ � � =1<br />

�<br />

�<br />

�e γt+Rteiθ� �<br />

� �<br />

� = �e γt �<br />

Rt(cos θ+i sin θ) �<br />

e � = e γt e Rt cos θ .<br />

Next, for |s| sufficiently large, we can write<br />

�<br />

� Rie iθ � � = |s − γ| � �i � � ≤|s| + |γ| < |s| + |s| =2|s| and |sF (s)| 0we<br />

conclude that lim CR R→∞<br />

estF (s) ds = 0.Finally, as R →∞we see from (10)<br />

that<br />

� γ+i∞<br />

e st n�<br />

F (s) ds = Res � e st �<br />

F (s), sk .<br />

γ−i∞<br />

∗ See Problem 52 in Exercises 6.6.<br />

k=1<br />

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