It is to be know that:
$$f(t) = \exp\left(\dfrac{-a\,t^2}{2}\right) \xrightarrow{\textbf{FT}} \hat{f}(\omega) = \sqrt{\dfrac{2\,\pi}{a}}\,\exp\left(\dfrac{-\omega^2}{2\,a}\right)$$
But what about $$f(t) = \exp\left(\dfrac{-a\,(t-b)^2}{2}\right)$$
Intuitively I would claim the Fourier transform looks somewhat similar, but that's what I end up with:
$$\begin{align}\hat{f}(\omega) =& \int \exp\left(\dfrac{-a\,(t-b)^2}{2}\right)\,\exp(-i\,\omega\,t)\,\mathrm{dt} \\[12pt] \hat{f}(\omega) =& \int \exp\left(-a \left( \dfrac{(t^2-2\,t\,b + b^2}{2} + \dfrac{i\,\omega\,t}{a}\right)\right)\,\mathrm{dt}\\[12pt] \hat{f}(\omega) =& \int\exp\left(-\dfrac{a}{2}\left(t^2- t\,\left(2\,b - \dfrac{2\,i\,\omega}{a}\right) + b^2\right)\right)\,\mathrm{dt}\\[12pt] \hat{f}(\omega) =& {\Large\int}\exp\left( -\dfrac{a}{2}\left(\left(t- \dfrac{2b-\dfrac{2\,i\,\omega}{a}}{2}\right)^{\textstyle 2} - \left(\dfrac{2b-\dfrac{2\,i\,\omega}{a}}{2}\right)^{\textstyle2} + b^2\right) \right)\,\mathrm{dt}\\[12pt] \hat{f}(\omega) =& \exp\left(-\dfrac{a}{2}\,\left(-\left(b - \dfrac{i\,\omega}{a}\right)^{\textstyle 2} + b^2\right)\right){\large\int}\exp\left(-\dfrac{a}{2}\,\left( t - \dfrac{b-i\,\omega}{a}\right)^{\textstyle 2}\right)\,\mathrm{dt}\\[12pt] \hat{f}(\omega) =& \exp\left(-\dfrac{a}{2}\,\left(-\left(b - \dfrac{i\,\omega}{a}\right)^{\textstyle 2} + b^2\right)\right)\,\sqrt{\dfrac{2\,\pi}{a}}\end{align} $$
Now this looks like something totally different not derivable from the formula above. Did I turned wrong?
You're correct. The last term is
\begin{equation} \sqrt{\frac{2\pi}{a}}exp(-\frac{(\omega+iab)^2}{2a})exp(-\frac{ab^2}{2})=exp(-\frac{ab^2}{2})\hat{f}(\omega+iab), \end{equation}
which is still a gaussian function with translation and scaling.