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Thread: how can u get this result?

  1. #1
    Senior Member
    Joined
    Feb 2008
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    321

    how can u get this result?

    $\displaystyle \phi(x,t)=\frac{1}{\pi^{\frac{1}{4}}(1+it)^{\frac{ 1}{2}}}e^{-\frac{x^{2}}{2(1+it)}}$

    how to show this:

    $\displaystyle \int_{-\infty}^{\infty}{|\phi|^{2}dx}=1$
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  2. #2
    Senior Member Peritus's Avatar
    Joined
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    $\displaystyle
    \left| {\frac{1}
    {{\pi ^{\frac{1}
    {4}} \sqrt {1 + it} }}e^{ - \frac{{x^2 }}
    {{2(1 + it)}}} } \right|^2 = \frac{1}
    {{\sqrt {\pi \left( {1 + t^2 } \right)} }}e^{ - \frac{{x^2 }}
    {{4\left( {1 + t^2 } \right)}}}
    $

    now if suppose that this function represents some normal distribution with:

    $\displaystyle \begin{gathered}
    \mu = 0 \hfill \\
    \sigma = \sqrt {2\left( {1 + t^2 } \right)} \hfill \\
    \end{gathered} $

    we know that:

    $\displaystyle
    \int\limits_{ - \infty }^\infty {e^{ - \frac{{\left( {x - \mu } \right)^2 }}
    {{2\sigma ^2 }}} } dx = \sigma \sqrt {2\pi }
    $

    thus:

    $\displaystyle \int\limits_{ - \infty }^\infty {\frac{1}
    {{\sqrt {\pi \left( {1 + t^2 } \right)} }}e^{ - \frac{{x^2 }}
    {{4\left( {1 + t^2 } \right)}}} } = 1$
    Last edited by Peritus; Oct 14th 2008 at 11:16 AM.
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  3. #3
    Senior Member
    Joined
    Feb 2008
    Posts
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    i got something different.

    [quote=Peritus;203126]$\displaystyle
    \left| {\frac{1}
    {{\pi ^{\frac{1}
    {4}} \sqrt {1 + it} }}e^{ - \frac{{x^2 }}
    {{2(1 + it)}}} } \right|^2 = \frac{1}
    {{\sqrt {\pi \left( {1 + t^2 } \right)} }}e^{ - \frac{{x^2 }}
    {{\left( {1 + t^2 } \right)}}}
    $
    Please check who is wrong..
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