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Math Help - Cauchy's integral theorem

  1. #1
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    Cauchy's integral theorem

    Use the parametrisation of the circle by  z = e^it  (0\leq t\leq 2\pi) to show that

    \int e^{\cos t} \cos(\sin t + t )dt = 0
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  2. #2
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    Quote Originally Posted by Richmond View Post
    Use the parametrisation of the circle by  z = e^{it}  (0\leq t\leq 2\pi) to show that

    \int\kern-20pt{\color{red}{\phantom\int}_0^{\ \ 2\pi}}\!\!\! e^{\cos t} \cos(\sin t + t )dt = 0
    Presumably this is a definite integral, from 0 to 2π?

    Use the fact that \cos x is the real part of e^{ix} to write the integral as \text{Re}\!\!\int_0^{2\pi}\!\!\! e^{\cos t}e^{i(\sin t+t)}dt = \text{Re}\!\!\int_0^{2\pi}\!\!\! e^{\cos t+i\sin t}e^{it}dt. Then let  z = e^{it}  and it becomes \text{Re}\!\oint_C ze^z\tfrac{dz}{iz} (C is the unit circle), so you can use Cauchy's theorem.
    Last edited by Opalg; April 28th 2009 at 09:34 AM. Reason: typo
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