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Thread: Integral of e

  1. #1
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    Integral of e

    Hi,

    I've to show that

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

    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

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

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

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


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


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

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

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

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


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


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

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

    $\displaystyle = 0$.
    Be careful if $\displaystyle k = -n $.
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