how can u get this result?

• Oct 14th 2008, 10:22 AM
szpengchao
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:

$\int_{-\infty}^{\infty}{|\phi|^{2}dx}=1$
• Oct 14th 2008, 10:56 AM
Peritus
$
\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:

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

we know that:

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

thus:

$\int\limits_{ - \infty }^\infty {\frac{1}
{{\sqrt {\pi \left( {1 + t^2 } \right)} }}e^{ - \frac{{x^2 }}
{{4\left( {1 + t^2 } \right)}}} } = 1$
• Oct 14th 2008, 11:23 AM
szpengchao
i got something different.
[quote=Peritus;203126] $
\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)}}}
$