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    Hi,

    I'm trying to compute the fourier transform for:

      e^{\frac{-x^2}{(2\sigma)^2}}

    This makes the integrand when doing the transform:

     e^{\frac{-x^2}{(2\sigma)^2}} e^{-i \omega x} = e^{\frac{-x^2}{(2\sigma)^2} - i\omega x}

    There's a hint saying to complete the square in the exponent, so let

     S = \frac{-x^2}{(2\sigma)^2} - i\omega x

     = \frac{-1}{4 \sigma^2}(x+h)^2 + k   \mathrm{\ for\ some\ } h, k \in \mathbb {C}

     S = \frac{-1}{4 \sigma^2}(x^2 + 2xh + h^2) + k

     = \frac{-x^2}{4 \sigma^2} - \frac{2xh}{4 \sigma^2} - \frac{h^2}{4 \sigma^2} + k

    c.f. coefficients gives:

     -i \omega = \frac{-2h}{4 \sigma^2} = \frac{-h}{2 \sigma^2}

     \Rightarrow h = 2i \omega \sigma^2

     \frac{-h^2}{4 \sigma^2} + k  = 0 \Rightarrow k = \frac{h^2}{4 \sigma^2} = \frac{-4 \omega^2 \sigma^4}{4 \sigma^2} = - \sigma^2 \omega^2

     \Rightarrow S = \frac{-1}{4 \sigma^2}(x + 2i\omega \sigma^2)^2 - \sigma^2 \omega^2

     = \frac{-1}{4 \sigma^2}[(x + 2i\omega \sigma^2)^2  - 4 \sigma^4 \omega^2]

    However in the solution, on completing the square they obtain:

     \frac{-1}{2 \sigma^2}[(x+ \sigma^2 i \omega)^2 + \sigma^4 \omega^2]

    But I can't see where I've gone wrong.

    Any advice?

    Thanks
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    (Original post by Chemhistorian)
     \Rightarrow S = \frac{-1}{4 \sigma^2}(x + 2i\omega \sigma^2)^2 - \sigma^2 \omega^2

     = \frac{-1}{4 \sigma^2}[(x + 2i\omega \sigma^2)^2  - 4 \sigma^4 \omega^2]
    You've made a sign error while factorising here. I'll look for the other errors now.

    Edit:

    Your solution yields: \dfrac{-x^2}{(2\sigma)^2} - i\omega x

    Their solution yields: \dfrac{-x^2}{2\sigma^2} - i\omega x

    Error on their part?
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    (Original post by Chemhistorian)
    ..
    (Original post by Phichi)
    ..
    If you look at the coefficient of x^2 in the given ("correct") answer, you'll see it's -1/2\sigma^2, but the original coefficient (in the OP) is -1/(2\sigma)^2.

    So I suspect the error probably comes right at the start - probably before what's actually been posted.
 
 
 
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