Help solving $\frac{1}{{2\pi}}\int_{-\infty}^{+\infty}{{e^{-{{\left({\frac{t}{2}} \right)}^2}}}{e^{-i\omega t}}dt}$ 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.

If you complete the square in the argument of the exponentials, $$ -\frac{1}{4}(t^2 + 4i \omega t) \to -\frac{1}{4}(t^2+4i\omega t -4\omega^2) -\omega^2 = -\frac{1}{4}(t+i2\omega)^2-\omega^2. $$ After a change of variables $u=\frac{t}{2}+i\omega$, the integral becomes $$2e^{-\omega^2}\int_{-\infty -i\omega}^{\infty -i\omega} e^{-u^2}\ du,$$ which, apart from the pesky factors of $-i\omega$ in the bounds, is a standard Gaussian integral equal to $\sqrt{\pi}$. As a physicist, I'm inclined to just sweep these factors under the rug, but we can do better: if we form a contour in the complex plane along the paths $(-\infty,i0)\to\ (\infty,i0)\to(\infty,-i\omega)\to(-\infty,-i\omega)\to(-\infty,i0),$ the integrand kills the contributions moving in the imaginary direction, and the overall integral is zero since $e^x$ has no poles. So, our integral is equal to the one along the real axis, and we can discard the $-i\omega$ terms.
