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Math Help - integral sin(x)exp(nx)

  1. #1
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    integral sin(x)exp(nx)

    \int sin(x)exp(-inx), where i is the imaginary unit and n is an entire.
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    Super Member girdav's Avatar
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    Compute \int e^{ix}e^{-inx}dx and take the imaginary part.
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  3. #3
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    Quote Originally Posted by girdav View Post
    Compute \int e^{ix}e^{-inx}dx and take the imaginary part.
    sin(x)e^{-inx} = sin(x)(cos(nx)-isin(nx)) has a real part and a imaginary part, while if i take the imaginary part of e^{ix}e^{-inx} i'll have no real part. Is that correct?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by girdav View Post
    Compute \int e^{ix}e^{-inx}dx and take the imaginary part.
    Actually this doesn't work. Consider briefly that it will return a real answer, which is obviously not correct.

    I'd do it by parts. Here's the first step:
    \int sin(x)e^{-inx}dx = \frac{1}{-in}sin(x)e^{-inx} - \int \frac{1}{-in}cos(x)e^{-inx}dx

    Do the second integral by parts and you will have the form:
    \int sin(x)e^{-inx}dx = ~...~ + \frac{1}{n^2} \int sin(x)e^{-inx}dx

    from which you can solve for \int sin(x)e^{-inx}dx.

    -Dan
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  5. #5
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    Unless I am misreading something I dont think this is purely real.

    \int e^{ix}e^{-inx}dx

    \int e^{-(n-1)ix}dx=\frac{i}{(n-1)}e^{-(n-1)ix}

    You could also use

    \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}
    Last edited by TheEmptySet; June 6th 2011 at 05:37 PM.
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  6. #6
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    Quote Originally Posted by girdav View Post
    Compute \int e^{ix}e^{-inx}dx and take the imaginary part.
    The title of the original question is misleading. The above method works for \int \sin(x) e^{-nx} \, dx, but not for the posted question, viz. \int \sin(x) e^{-inx} \, dx.
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  7. #7
    Super Member girdav's Avatar
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    If we don't want to do it by parts, we can write e^{-inx} as \cos (nx)-i\sin (nx). Now we have to compute \int \sin x \cos(nx) dx and \int \sin x\sin(nx)dx.
    For the first we use  \sin x \cos(nx) =\frac{\sin((n+1))x-\sin((n-1)x)}2 and for the second \sin x\sin (nx)=\frac{\cos((n-1)x)-\cos ((n+1)x)}2.
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