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Math Help - how can u get this result?

  1. #1
    Senior Member
    Joined
    Feb 2008
    Posts
    321

    how can u get this result?

    \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:

    \int_{-\infty}^{\infty}{|\phi|^{2}dx}=1
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  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
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    <br />
\left| {\frac{1}<br />
{{\pi ^{\frac{1}<br />
{4}} \sqrt {1 + it} }}e^{ - \frac{{x^2 }}<br />
{{2(1 + it)}}} } \right|^2  = \frac{1}<br />
{{\sqrt {\pi \left( {1 + t^2 } \right)} }}e^{ - \frac{{x^2 }}<br />
{{4\left( {1 + t^2 } \right)}}} <br />

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

    \begin{gathered}<br />
  \mu  = 0 \hfill \\<br />
  \sigma  = \sqrt {2\left( {1 + t^2 } \right)}  \hfill \\ <br />
\end{gathered}

    we know that:

    <br />
\int\limits_{ - \infty }^\infty  {e^{ - \frac{{\left( {x - \mu } \right)^2 }}<br />
{{2\sigma ^2 }}} } dx = \sigma \sqrt {2\pi } <br />

    thus:

    \int\limits_{ - \infty }^\infty  {\frac{1}<br />
{{\sqrt {\pi \left( {1 + t^2 } \right)} }}e^{ - \frac{{x^2 }}<br />
{{4\left( {1 + t^2 } \right)}}} }  = 1
    Last edited by Peritus; October 14th 2008 at 12:16 PM.
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  3. #3
    Senior Member
    Joined
    Feb 2008
    Posts
    321

    i got something different.

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