part of example 6 is in here
http://books.google.com/books?id=tVXXD8sJ7uwC&printsec=frontcover&dq=Basic+Partial+Differential+Equations&hl=en&sa=X&ei=b-V6VNLJGYKSoQSP2oHQBA&ved=0CCUQ6AEwAQ#v=onepage&q=Basic%20Partial%20Differential%20Equations&f=false
page 425

I've never even heard of that phrase before....
Maybe try English:SE?

so we should have something like this
$\hat{(e^{-1 / 2 x^2})}(\xi) = e^{-1 / 2 \xi^2}$ ?

Since by corollary 2
$\hat{\left( x^n f(x) \right)} = i^n \frac{\mathrm{d}^n}{\mathrm{d \xi^n}} \hat{f}(\xi)$ if $n=2$ because we have $x^2$ in $f(x)$, we have
$\hat{\left( x^2 f(x) \right)} = i^2 \frac{\mathrm{d}^2}{\mathrm{d \xi^2}} \hat{f}(\xi)$

From the example we have
$\hat{(e^{-1 / 2 x^2})}(\xi) = e^{-1 / 2 \xi^2}$

$-\xi e^{\frac{-1}{2}\xi^2} $

product rule leaving the $\xi$ alone and then deal with $e^{\frac{-1}{2}\xi^2} $. Then leave $e^{\frac{-1}{2}\xi^2} $ alone deal with $-\xi$

ah here

Absolutely integrable
A real or complex-valued function defined on $(-\infty, \infty)$ is said to be absolutely integrable on $(-\infty, \infty)$ if \int_{-R}^{R} \mid f(x) \mid dx$ exists for all $R>0$ and $\int_{-\infty}^{\infty} \mid f(x) \mid dx \equiv$ as limit R approaches infinity $\int_{-R}^{R} \mid f(x) \mid dx < \infty$

ah here

Absolutely integrable
A real or complex-valued function defined on $(-\infty, \infty)$ is said to be absolutely integrable on $(-\infty, \infty)$ if $\int_{-R}^{R} \mid f(x) \mid dx$ exists for all $R>0$ and $\int_{-\infty}^{\infty} \mid f(x) \mid dx \equiv$ as limit R approaches infinity $\int_{-R}^{R} \mid f(x) \mid dx < \infty$