how to find fourier transform of $\exp(-x^2/2)$ How can we find the fourier transform of $e^\frac{-x^2}{2}$ where -$\infty $ < x < $\infty $.
I tried applying the standard formulae but ended up in un defined form..
 A: [Expanding on the comments by T.A.E. and Stefan Smith]. One of  interesting features of $f(x)=e^{-x^2/2}$ is that differentiation has the same effect as multiplication by the independent variable (up to minus sign): $$f'(x) = -xf(x) \tag{1}$$
Recall that under the Fourier transform, differentiation becomes multiplication by independent variable, and vice versa (with multiplicative constants dependent on your convention for Fourier transform). E.g., with the convention $\hat f(\xi) = \int_{-\infty}^{\infty} e^{-i\xi x}f(x)\,dx$, identity (1) transforms into 
$$ i \xi \hat f(\xi) =  -i \frac{d}{d\xi}\hat f(\xi) \tag{2}$$
Thus, $\hat f$ solves the same first-order ODE as $f$ itself. This implies $\hat f$ is a constant multiple of $e^{-\xi^2/2}$, and you can find that multiple from $\hat f(0)=\int f(x)\,dx=\sqrt{2\pi}$.
A: The way I do it is completing squares for the integral.
$$
\int_{-\infty}^{\infty} e^{ix\omega}e^{\frac{-x^2}{2}} \,dx =\int_{-\infty}^{\infty} e^{-\frac{1}{2}(x+i\omega)^2}e^{-\frac{\omega^2}{2}} \,dx  \\=\sqrt{2\pi}e^{-\frac{\omega^2}{2}}
$$
Where I used the fact that the gaussian integral $$ \int_{-\infty}^{\infty} e^{\frac{-(x+b)^2}{a}} \,dx = \sqrt{a\pi}$$ even when b is complex. When b is real, is easy to prove this with a change of variables because the limits doesn't change. For the imaginary case, this is proven using complex integration on the complex plane as you can see in this answer.
