Prove that $\int_{-\infty}^{\infty} e^{{-x^2}/2} e^{-2itx}dx = \sqrt {2\pi} e^{{-(2t)^2}/2} $. Why $\int_{-\infty}^{\infty} e^{{-x^2}/2} e^{-2itx}dx = \sqrt {2\pi} e^{{-(2t)^2}/2} $. 
This value is needed in a problem of Hermite orthogonal functions. I have solved the original using this value but wandering how to prove it?
 A: Another method.  First do it for
$$
\int_{-\infty}^\infty e^{-x^2/2}dx = \sqrt{2\pi}
$$
then complete the square to do it for
$$
\int_{-\infty}^\infty e^{-x^2/2}e^{-2zx}dx = \sqrt{2\pi}\;e^{2z^2}
$$
with real $z$.  Then observe that both sides are analytic functions of the complex variable $z$ which agree for real $z$.  Therefore they agree for all complex $z$.
added
"complete the square"
$$
-x^2/2-2zx = -x^2/2-2zx-2z^2+2z^2
=-(x+2z)^2/2+2z^2
$$
Thus, substitute $y=x+2z$ (with $z$ constant), $dy = dz$, to get
$$
\int_{-\infty}^\infty e^{-x^2/2}e^{-2zx}dx
-\int_{-\infty}^\infty e^{-(x+2z)^2/2}e^{2z^2}dx
=e^{2z^2}\int_{-\infty}^\infty e^{-y^2/2}dy = e^{2z^2}\sqrt{2\pi}
$$
A: $$
\int_{-\infty}^{\infty} e^{{-x^2}/2} e^{-2itx}dx=\int_{-\infty}^{\infty} e^{{-x^2}/2} \cos{2tx}\:dx-i\int_{-\infty}^{\infty} e^{{-x^2}/2} \sin{2tx}\:dx
$$
Since $e^{{-x^2}/2} \sin{2tx}$ is odd
$$
I(t)=\int_{-\infty}^{\infty} e^{{-x^2}/2} e^{-2itx}dx=\int_{-\infty}^{\infty} e^{{-x^2}/2} \cos{2tx}\:dx
$$
Then
\begin{align}
I'(t)&=-2\int_{-\infty}^{\infty} xe^{{-x^2}/2} \sin{2tx}\:dx
\\
&=2e^{{-x^2}/2} \sin{2tx}\Bigg|_{-\infty}^{\infty}-4t\int_{-\infty}^{\infty} e^{{-x^2}/2} \cos{2tx}\:dx
\\
&=-4tI(t)
\end{align}
So $I(t)=Ce^{-2t^2}$. Since 
$$
I(0)=\int_{-\infty}^{\infty} e^{{-x^2}/2}\:dx=\sqrt{2}\int_{-\infty}^{\infty} e^{{-u^2}}\:du=\sqrt{2\pi}
$$
We have $C=\sqrt{2\pi}$, and
$$
\int_{-\infty}^{\infty} e^{{-x^2}/2} e^{-2itx}dx=\sqrt{2\pi}e^{-2t^2}
$$
