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Math Help - fourier transform of f(x)=e^(-ax^2) sin(bx)

  1. #1
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    fourier transform of f(x)=e^(-ax^2) sin(bx)

    i need to find the fourier transform of f(x)=e^{-ax^2} sin(bx)

    i get F(\lambda) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty} e^{-ax^2} sin(bx) e^{i \lambda x} dx (by definition)

    now sin(bx) = \frac{e^{ibx} - e^{-ibx}}{2i}
    so F(\lambda) =  \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty} e^{-ax^2} \[\frac{e^{ibx} - e^{-ibx}}{2i}\] e^{i \lambda x} dx
    = \frac{1}{\sqrt{2\pi}} \frac{1}{2i} \int^{\infty}_{-\infty} e^{-ax^2 + ibx + i\lambda x} - e^{-ax^2 - ibx + i\lambda x} dx

    im stuck as to where to go from here! judging by our notes the next step should be
    = \frac{1}{\sqrt{2\pi}} \frac{1}{2i} \sqrt{\frac{\pi}{a}} \int^{\infty}_{-\infty} e^{-\frac{(b+\lambda)^2}{4a}} - e^{-\frac{(b-\lambda)^2}{4a}}
    but i have no idea why this should be (and not even sure its the right step!)
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  2. #2
    Super Member girdav's Avatar
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    Re: fourier transform of f(x)=e^(-ax^2) sin(bx)

    You will be able to conclude if you know the value of I(a,b):=\int_{-\infty}^{+\infty}e^{-ax^2}e^{ibx}dx. If you fix a, differentiate this with respect to b, and integrate by parts, you will find a differential equation.
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  3. #3
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    Re: fourier transform of f(x)=e^(-ax^2) sin(bx)

    im not sure what you mean here.why would i differentiate it with respect to b?
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  4. #4
    Super Member girdav's Avatar
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    Re: fourier transform of f(x)=e^(-ax^2) sin(bx)

    Fix a and put G(t)=\int_{-\infty}^{+\infty}e^{-ax^2}e^{itx}dx. We have G'(t)=\int_{-\infty}^{+\infty}e^{-ax^2}e^{itx}ixdx=i\left[-e^{-ax^2}\frac 2ae^{itx}\right]_{-\infty}^{+\infty}+\frac{2i}a\int_{-\infty}^{+\infty}e^{-ax^2}e^{itx}itdx hence G'(t)=-\frac{2t}aG(t).
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    Re: fourier transform of f(x)=e^(-ax^2) sin(bx)

    im a little rusty on integration by parts, i know its \int u dv = uv - \int v du but i cant figure out what you have specified as u and dv above. also im still unsure as to how this helps me??
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  6. #6
    Super Member girdav's Avatar
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    Re: fourier transform of f(x)=e^(-ax^2) sin(bx)

    I took u(x)=xe^{-ax^2} and v'(x)=e^{itx}. Now you can solve the previous differential equation, since you can compute G(0).
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