I need help with what seems like a pretty simple integral for a Fourier Transformation. I need to transform $\psi \left( {0,t} \right) = {\exp^{ - {{\left( {\frac{t}{2}} \right)}^2}}}$ into $\psi(0,\omega)$ by solving:
$ \frac{1}{{2\pi }}\int_{ - \infty }^{ + \infty } {\psi \left( {0,t} \right){e^{ - i\omega t}}dt} $
So far I've written (using Euler's formula):
$\psi \left( {0,\omega } \right) = \frac{1}{{2\pi }}\int_{ - \infty }^{ + \infty } {\psi \left( {0,t} \right){e^{ - i\omega t}}dt} = \frac{1}{{2\pi }}\left( {\int_{ - \infty }^{ + \infty } {{e^{ - {{\left( {\frac{t}{2}} \right)}^2}}}\cos \omega tdt - i\int_{ - \infty }^{ + \infty } {{e^{ - {{\left( {\frac{t}{2}} \right)}^2}}}\sin \omega tdt} } } \right)$
$ \begin{array}{l} = \frac{1}{{2\pi }}\left( {{I_1} - i{I_2}} \right)\\ \end{array}$
I just don't recall a way to solve this integrals by hand. Wolfram Alpha tells me that the result of the first integral is ${I_1} = 2\sqrt \pi {e^{ - {\omega ^2}}}$ and for the second $I_{2}=0$. But on my notes I have ${I_1} = 2\sqrt \pi {e^{ - {{\left( {{\omega ^2}/2} \right)}^2}}}$.
Can anybody tell me how one can solve this type of integrals and if the result from Wolfram Alpha is accurate? Any help will be appreciated.