Can we use contour integration to compute the Fourier transform of Gaussian?

Question

Let $f(x)=e^{-x^2}.$ It is known that the Fourier transform of Gaussian is a Gaussian.

My question: Can we use complex analysis (contour integration etc..) to compute

$$\hat{f}(\xi)= \int_{\mathbb R} e^{-x^2} e^{ix\xi} dx \ (\xi \in \mathbb R)$$

Motivation: My motivation comes from trying to justify:

$$\begin{eqnarray*} \int_\mathbb{R} e^{-\left(\frac{1}{t^2}+i\frac{1}{2t}\right)x^2}e^{-i2\pi x\xi} dx & = & \int_\mathbb{R} e^{-\frac{\sqrt{4+t^2}}{2t^2}x^2}e^{-i2\pi x \left(e^{-\frac{i}{2}\arctan\left(\frac{t}{2}\right)}\xi\right)} dx \\ \end{eqnarray*} $$

which came about from an Answer to a previous (now deleted) Question.

Answer 1

We have $\exp[-(x-\frac12 \xi i)^2]=e^{-x^2}e^{\frac14 \xi^2} e^{\xi xi}$.

Consider the countour $\gamma$ connecting points $-R, -R-\frac12 \xi i, R-\frac12 \xi i, R$, where $R>0$. We know that the function $$ f(z)=\exp(-z^2) $$ has no singularities inside this countour, so $\int_\gamma f(z)dz=0$. The integral \begin{align*} &\int_{R-\frac 12 \xi i}^R \exp(-z^2) dz \\ &= \int_{-\frac 12 \xi}^0 \exp(-(R+xi)^2) idx\\ &=\exp(-R^2)\int_{-\frac 12 \xi}^0 \exp x^2 \exp (-2Rxi) idx\\ &= 0 \ \text{ as } \ R\to \infty \end{align*} Where the integral is bounded. Similarly, $$ \int_{-R-\frac 12 \xi i}^{-R} \exp(-z^2) dz\to 0. $$ As a result, \begin{align*} 0 &= \int_\gamma f(z)dz\\ 0 &= \int_{-\infty}^\infty \exp (-x^2) dx - e^{\frac14 \xi^2} \int_{-\infty}^\infty e^{-x^2}e^{\xi xi}dx \end{align*}

so, $$ \mathcal F f(\xi)=\sqrt \pi \exp(-\xi^2/4). $$