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Thread: indefinite integral

  1. #1
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    indefinite integral

    How to evaluate
    $\displaystyle \int_{-\infty}^{\infty} \exp(-x^2) \exp(-i\alpha x) dx = \int_{-\infty}^{\infty} \exp(-x^2-i\alpha x) dx$
    , the Fourier transform of $\displaystyle \exp(-x^2)$
    Thanks
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  2. #2
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    Re: indefinite integral

    Try completing the square inside the exponent.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: indefinite integral

    Quote Originally Posted by hurz View Post
    How to evaluate
    $\displaystyle \int_{-\infty}^{\infty} \exp(-x^2) \exp(-i\alpha x) dx = \int_{-\infty}^{\infty} \exp(-x^2-i\alpha x) dx$
    , the Fourier transform of $\displaystyle \exp(-x^2)$
    Thanks
    First we define...

    $\displaystyle f(x)= e^{-x^{2}}$ (1)

    $\displaystyle F(\alpha)= \mathcal{F} \{f(x)\}= \int_{-\infty}^{+\infty} e^{-x^{2}}\ e^{-i \alpha x}\ dx $ (2)

    ... and then, remembering that is...

    $\displaystyle \mathcal{F} \{f^{'} (x) \}= i\ \alpha\ \mathcal{F} \{f(x)\}$ (3)

    ... from (1), (2) and (3) we derive...

    $\displaystyle F^{'} (\alpha) = - \frac{\alpha}{2}\ F(\alpha)$ (4)

    The (4) is an ordinary first order DE and its general solution is found separating the variables...

    $\displaystyle \frac{d F}{F} = - \frac{\alpha}{2}\ d \alpha \implies \ln F = - \frac{\alpha^{2}}{4} + \ln c \implies F = c\ e^{- \frac{\alpha^{2}}{4}} $ (5)

    Taking into account the known result $\displaystyle F(0)= \sqrt {\pi}$ we obtain the final result...

    $\displaystyle \int_{-\infty}^{+\infty} e^{-x^{2}}\ e^{-i \alpha x}\ dx = \sqrt {\pi}\ e^{- \frac{\alpha^{2}}{4}} $ (6)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Last edited by chisigma; Jun 23rd 2011 at 03:50 PM. Reason: some errors corrected... see my previous post...
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  4. #4
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    Re: indefinite integral

    Quote Originally Posted by chisigma View Post
    First we define...

    $\displaystyle f(x)= e^{-x^{2}}$ (1)

    $\displaystyle F(\alpha)= \mathcal{F} \{f(x)\}= \int_{-\infty}^{+\infty} e^{-x^{2}}\ e^{-i \alpha x}\ dx $ (2)

    ... and then, remembering that is...

    $\displaystyle \mathcal{F} \{f^{'} (x) \}= i\ \alpha\ \mathcal{F} \{f(x)\}$ (3)

    ... from (1), (2) and (3) we derive...

    $\displaystyle F^{'} (\alpha) = -2\ \alpha\ F(\alpha)$ (4)

    The (4) is an ordinary first order DE and its general solution is found separating the variables...

    $\displaystyle \frac{d F}{F} = -2\ \alpha\ d \alpha \implies \ln F = - \alpha^{2} + \ln c \implies F = c\ e^{- \alpha^{2}} $ (5)

    Taking into account the known result $\displaystyle F(0)= \sqrt {\pi}$ we obtain the final result...

    $\displaystyle \int_{-\infty}^{+\infty} e^{-x^{2}}\ e^{-i \alpha x}\ dx = \sqrt {\pi}\ e^{- \alpha^{2}} $ (6)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    I'm sorry, can you ive details about how to derive (4) ?
    What does $\displaystyle F^{'} (\alpha)$ mean? The derivate of the fourier transform?
    Thanks
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  5. #5
    MHF Contributor chisigma's Avatar
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    Re: indefinite integral

    Quote Originally Posted by hurz View Post
    I'm sorry, can you ive details about how to derive (4) ?

    What does $\displaystyle F^{'} (\alpha)$ mean? The derivate of the fourier transform?

    Thanks


    All right!... that's useful because there is an error in my previous post!... it is...



    $\displaystyle \mathcal{F} \{f^{'}(x)\} = - 2\ \int_{- \infty}^{+ \infty} x\ e^{-x^{2}}\ e^{-i \alpha x}\ dx = $



    $\displaystyle = - 2\ i \frac{d}{d \alpha}\ \int_{- \infty}^{+ \infty} e^{-x^{2}}\ e^{-i \alpha x}\ dx = - 2\ i\ F^{'} (\alpha) $ (1)



    ... so that from (1)...



    $\displaystyle F^{'} (\alpha) = \frac{1}{2 i}\ \mathcal{F} \{f^{'}(x)\} = - \frac{\alpha}{2}\ F(\alpha)$ (2)



    ... and the (2) with the 'initial condition' $\displaystyle F(0)= \sqrt{\pi}$ gives the result...



    $\displaystyle \int_{- \infty}^{+ \infty} e^{-x^{2}}\ e^{-i \alpha x}\ dx = \sqrt{\pi}\ e^{-\frac{\alpha^{2}}{4}}$ (3)



    Sorry for my previous error!...



    Kind regards



    $\displaystyle \chi$ $\displaystyle \sigma$
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