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Complex Analysis Integral by Substitution

  1. Dec 8, 2011 #1
    Im trying to take the integral, using substitution, of [tex]\int_0^1\frac{2\pi i[cos(2\pi t) + isin(2\pi t)]dt}{cos(2\pi t)+isin(2\pi t)}[/tex]

    So I set [tex]u=cos(2\pi t)+isin(2\pi t)[/tex][tex]du=2\pi i[cos(2\pi t) + isin(2\pi t)]dt[/tex]

    Yet when I change the endpoints of the integral I get from 1 to 1, which doesn't make sense. Since the fraction is equal to 2πi I could just take the integral directly and get 2πi, yet for some reason doing it by substitution gives me 0. This integral was a pull back from the integral of 1/z around a closed curve, so it seems that for some reason by doing the u-du substitution the integral "forgets" that there is a singularity on the interior of the curve, what is going on here?
     
    Last edited: Dec 8, 2011
  2. jcsd
  3. Dec 8, 2011 #2
    Setting [itex]u=\cos(2\pi t)+i\sin(2\pi t)[/itex] for [itex]0\leq t\leq 1[/itex] changes the path of integration from a straight line over the real axis to a closed contour about the origin with radius one. So your integral is equivalent to:

    [tex]\mathop\oint\limits_{|u|=1} \frac{du}{u}=2\pi i[/tex]
     
  4. Dec 8, 2011 #3

    Dick

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    And the endpoints of the integration of du/u are 1 and 1. But that doesn't mean the integral is zero. The antiderivative of 1/u is log(u) and that's multivalued. You can't just insert the limits without considering the path.
     
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