Thread: how can u get this result?

1. 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:

$\displaystyle \int_{-\infty}^{\infty}{|\phi|^{2}dx}=1$

2. $\displaystyle \left| {\frac{1} {{\pi ^{\frac{1} {4}} \sqrt {1 + it} }}e^{ - \frac{{x^2 }} {{2(1 + it)}}} } \right|^2 = \frac{1} {{\sqrt {\pi \left( {1 + t^2 } \right)} }}e^{ - \frac{{x^2 }} {{4\left( {1 + t^2 } \right)}}}$

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:

$\displaystyle \int\limits_{ - \infty }^\infty {\frac{1} {{\sqrt {\pi \left( {1 + t^2 } \right)} }}e^{ - \frac{{x^2 }} {{4\left( {1 + t^2 } \right)}}} } = 1$

3. 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..