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Thread: fourier transform

  1. #1
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    fourier transform

    Hello all,

    I have a question regarding fourier transforms. I am considering the function f_{1} defined as:

    f_{1}(x)=e^{-x}\chi_{[0,1]}

    Now I would like to find the transform of f_{1}. I have done the following:

    (\digamma f_{1})(\gamma)=\digamma(f_{1}(\gamma))=\int_{0}^{1  }e^{-x}\chi_{[0,1]}e^{-2\pi ix\gamma}dx=\int_{0}^{1}e^{-x}\cdot 1\cdot e^{-2\pi ix\gamma}dx=\int_{0}^{1}e^{x}e^{-1-2\pi i\gamma}

    Treating e^{-1-2\pi i\gamma} as a konstant we get:

    \int_{0}^{1}e^{x} e^{-1-2\pi ix\gamma}dx=e^{-1-2\pi ix\gamma} \int_{0}^{1}e^{x}dx=\underline{e^{-1-2\pi ix\gamma}(e-1)}

    Could someone please run through the above and see if it is correct?

    Thank you very much.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by surjective View Post
    Hello all,

    I have a question regarding fourier transforms. I am considering the function f_{1} defined as:

    f_{1}(x)=e^{-x}\chi_{[0,1]}

    Now I would like to find the transform of f_{1}. I have done the following:

    (\digamma f_{1})(\gamma)=\digamma(f_{1}(\gamma))=\int_{0}^{1  }e^{-x}\chi_{[0,1]}e^{-2\pi ix\gamma}dx=\int_{0}^{1}e^{-x}\cdot 1\cdot e^{-2\pi ix\gamma}dx=\int_{0}^{1}e^{x}e^{-1-2\pi i\gamma}
    Certainly not: e^{-x}e^{-2\pi i x\gamma}=e^{-x(1+2i\pi\gamma)}. Go on from there.
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  3. #3
    Member
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    Feb 2010
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    Hello,

    Yes of course. Stupid mistake. Thanks.
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