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Complexifying the integral of the secant function

  1. Jul 4, 2014 #1
    Since learning about being able to complexify differential equations (I am doing the MIT OCW course by Arthur Mattuck), I have tried to apply to this to particular problems in integration as well. Specifically, I have tried to integrate the secant of some function to see if it would lead to the same accepted answer as when you used trignometric identities and u substitutions. What I have done is:


    ∫secΘdΘ = ∫(1/cosΘ)dΘ = Re{∫(1/e)dΘ}


    ∫(1/e)dΘ = ∫e-iΘdΘ = -e/i = -1/(i(cosΘ+isinΘ))

    = -1/(-sin+icosΘ) = -1(-sin-icosΘ)/(sin2-(-cos2Θ)) = sinΘ+icosΘ

    Re(sinΘ+icosΘ) = sinΘ

    I know sinΘ is not the correct answer, but I do not understand why I cannot do the math this way. In the video I watched for complexifying integrals, I watched the professor do:

    ∫cosΘdΘ = Re{∫edΘ}

    and the professor got the correct answer doing this, so I do not understand why I cannot do this with the secant.

    I have limited access to teachers right now, but I did visit one teacher at a local college. He said I cannot complexify the equation as I have done, rather that I should have done this:

    cosΘ = (e+e-iΘ)/2

    However, he did not explain why, and before I could ask he went off and explained how to do the integral in the "traditional way." I am not interested in the traditional way; I am trying to find another way to do the integral, and am trying to find the hole in my knowledge regarding complexifying problems.

    Any help would be appreciated. Thank you.
     
  2. jcsd
  3. Jul 4, 2014 #2

    mfb

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    2016 Award

    Staff: Mentor

    Why do you expect this to be true?

    I see this step:
    ∫(1/cosΘ)dΘ = ∫(1/Re(e))dΘ

    What did you do afterwards?
     
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