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Math Help - Integral of e

  1. #1
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    Integral of e

    Hi,

    I've to show that

    \int_0^{2\pi}e^{-i(k+n)\theta}d\theta=0

    Thanks for any help.
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  2. #2
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    Quote Originally Posted by Arczi1984 View Post
    Hi,

    I've to show that

    \int_0^{2\pi}e^{-i(k+n)\theta}d\theta=0

    Thanks for any help.
    \int_0^{2\pi}{e^{-i(k + n)\theta}\,d\theta} = -\frac{1}{i(k + n)}\left[e^{-i(k + n)\theta}\right]_0^{2\pi}

     = -\frac{1}{i(k + n)}\left[e^{-2\pi i(k + n)} - e^{0i(k + n)}\right].


    Assuming that -(k + n) is an integer, these two terms are actually equal (since an integer multiple of 2\pi represents a number of revolutions about the unit circle back to an angle of 0).


     = -\frac{1}{i(k + n)}\left[1 - 1\right]

     = -\frac{1}{i(k + n)}\left[0\right]

     = 0.
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Prove It View Post
    \int_0^{2\pi}{e^{-i(k + n)\theta}\,d\theta} = -\frac{1}{i(k + n)}\left[e^{-i(k + n)\theta}\right]_0^{2\pi}

     = -\frac{1}{i(k + n)}\left[e^{-2\pi i(k + n)} - e^{0i(k + n)}\right].


    Assuming that -(k + n) is an integer, these two terms are actually equal (since an integer multiple of 2\pi represents a number of revolutions about the unit circle back to an angle of 0).


     = -\frac{1}{i(k + n)}\left[1 - 1\right]

     = -\frac{1}{i(k + n)}\left[0\right]

     = 0.
    Be careful if  k = -n .
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