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Thread: [SOLVED] Fourier transform

  1. #1
    Forum Admin topsquark's Avatar
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    [SOLVED] Fourier transform

    So I'm just sitting in the chair being bored and I decided to review Fourier transforms. I pick what I thought would be an easy function to transform: e^{-x^2}, so I have
    g(k) = \int_{-\infty}^{\infty}e^{-x^2} e^{-ikx}~dx

    = e^{-k^2/4}\int_{-\infty}^{\infty}e^{-(x - ik/2)^2}~dx

    I have (somewhere) in my references how to do these, but I can't find it. (I really need to organize my references!) My initial impluse is to do a complex integration. Let z = x + ik/2. Then
    g(k) = \int_{-\infty + ik/2}^{\infty + ik/2}e^{-z^2}~dz
    which is a line integral on the whole line y = ik/2. I can't quite figure out a good contour for this.

    Thanks!
    -Dan
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  2. #2
    Super Member wingless's Avatar
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    I think you made a typo there, e^{-k^2/4}\int_{-\infty}^{\infty}e^{-(x - ik/2)^2}~dx makes \int_{-\infty}^{\infty}e^{-x^2} e^{\color{red}+\color{black}ikx}~dx

    I used,

    e^{-k^2/4}\int_{-\infty}^{\infty}e^{-(x - ik/2)^2}~dx

    e^{-k^2/4}\int_{-\infty}^{\infty}e^{-(x + ik/2)^2}~dx

    Using your substitution, we get

    e^{-k^2/4}\int_{-\infty+ ik/2}^{\infty+ ik/2}e^{-u^2}~du

    I don't think those ik/2's will mean anything even when summed by infinity. So I get

    e^{-k^2/4}\int_{-\infty}^{\infty}e^{-u^2}~du

    e^{-k^2/4}\sqrt{\pi}

    I'm not sure about this result. I'll check it using contour integration as soon as I have time, (if someone doesn't find my error, lol)
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  3. #3
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    Quote Originally Posted by topsquark View Post
    So I'm just sitting in the chair being bored and I decided to review Fourier transforms. I pick what I thought would be an easy function to transform: e^{-x^2}, so I have
    g(k) = \int_{-\infty}^{\infty}e^{-x^2} e^{-ikx}~dx
    There's a good trick for doing this. Let g(w) = \int_{-\infty}^{\infty}e^{-x^2} e^{-iwx}\,dx. Differentiate under the integral sign to get g'(w) = \int_{-\infty}^{\infty}-ixe^{-x^2} e^{-iwx}\,dx = \int_{-\infty}^{\infty}{\textstyle \frac{d}{dx}(\frac12e^{-x^2})} e^{-iwx}\,dx. Now integrate by parts and you find that g'(w) = -\textstyle\frac12wg(w), from which g(w) = g(0)e^{-w^2/4}. But g(0) = \int_{-\infty}^{\infty}e^{-x^2}\,dx = \sqrt\pi.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Opalg View Post
    There's a good trick for doing this. Let g(w) = \int_{-\infty}^{\infty}e^{-x^2} e^{-iwx}\,dx. Differentiate under the integral sign to get g'(w) = \int_{-\infty}^{\infty}-ixe^{-x^2} e^{-iwx}\,dx = \int_{-\infty}^{\infty}{\textstyle \frac{d}{dx}(\frac12e^{-x^2})} e^{-iwx}\,dx. Now integrate by parts and you find that g'(w) = -\textstyle\frac12wg(w), from which g(w) = g(0)e^{-w^2/4}. But g(0) = \int_{-\infty}^{\infty}e^{-x^2}\,dx = \sqrt\pi.
    Oooh! I like that one.

    -Dan
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  5. #5
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    Quote Originally Posted by Opalg View Post
    There's a good trick for doing this. Let g(w) = \int_{-\infty}^{\infty}e^{-x^2} e^{-iwx}\,dx. Differentiate under the integral sign to get g'(w) = \int_{-\infty}^{\infty}-ixe^{-x^2} e^{-iwx}\,dx = {\color{red}i} \, \int_{-\infty}^{\infty}{\textstyle \frac{d}{dx}(\frac12e^{-x^2})} e^{-iwx}\,dx. Now integrate by parts and you find that g'(w) = -\textstyle\frac12wg(w), from which g(w) = g(0)e^{-w^2/4}. But g(0) = \int_{-\infty}^{\infty}e^{-x^2}\,dx = \sqrt\pi.
    Nice solution. i (ha ha) think you'll find a small typo (in red). The final result is unaffected.
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