Define the Inverse Fourier Transform (IFT) as
$f(t) = \int_{-\infty}^{{\infty}}f(\omega) \exp(i \omega t) d\omega$
I require the IFT of a decaying Gaussian
$f(\omega) = \frac{\exp(-s^2 \omega^2/2)}{i \omega + 1/\tau}$
I call it decaying because of a very similar known IFT
$g(\omega) = \frac{1}{i \omega + 1/\tau}$
$g(t) = F[g(\omega)] = H(t) \exp(-t/ \tau)$
where $H(t)$ is the (Heaviside) step function.
Please help me find the IFT of $f(\omega)$
You have $$\int_{\mathbb{R}}\frac{\exp[-s^{2}\omega^{2}/2+i\omega{t}]}{i\omega+1/\tau}d\omega$$ The integrand has a pole at $\omega=i/\tau$. Here I assume $\tau>0$, so the pole is in the upper half plane. You take a contour which is an infinite semi-circle in the upper half plane. The residue at the pole is $$\lim_{\omega\rightarrow{i/\tau}}(\omega-i/\tau)\frac{\exp[-s^{2}\omega^{2}/2+i\omega{t}]}{i\omega+1/\tau}=\frac{\exp[s^{2}/2\tau-{t}/\tau]}{i}$$ So the integral is $$\int_{\mathbb{R}}\frac{\exp[-s^{2}\omega^{2}/2+i\omega{t}]}{i\omega+1/\tau}d\omega=2\pi{i}\frac{\exp[s^{2}/2\tau-{t}/\tau]}{i}=2\pi\exp[s^{2}/2\tau-{t}/\tau]$$