Let $x\in \mathbb R.$ Define $g:\mathbb R\to \mathbb R$ as $g(y)=e^{-\pi y^2 +2\pi yx}$ for all $y\in \mathbb R.$
My Question is: What is the Fourier transform of $g$? In other word, how to compute the integral $\hat{g}(w)=\int_{\mathbb R} g(\xi) e^{-2\pi i \xi w} d\xi, (w\in \mathbb R)$?
Let me use $y$ instead of $\xi$, since it is easier to type. If you complete a square in the exponent, your integral is
$$\int \limits _{\Bbb R} \Bbb e ^{-\pi y^2 + 2 \pi y x - 2 \pi \Bbb i y w} \Bbb d y = \int \limits _{\Bbb R} \Bbb e ^{-\pi y^2 + 2 \pi y (x - \Bbb i w)} \Bbb d y = \int \limits _{\Bbb R} \Bbb e ^{-\pi y^2 + 2 \pi y (x - \Bbb i w) - \pi (x - \Bbb i w)^2 + \pi (x - \Bbb i w)^2} \Bbb d y = \\ \Bbb e ^{\pi (x - \Bbb i w)^2} \int \limits _{\Bbb R} \Bbb e ^{-\pi \big(y - (x - \Bbb i w) \big) ^2} \Bbb d y = \Bbb e ^{\pi (x - \Bbb i w)^2} \lim \limits _{R \to \infty} \int \limits _{-R} ^R \Bbb e ^{-\pi \big(y - (x - \Bbb i w) \big) ^2} \Bbb d y = \\ \Bbb e ^{\pi (x - \Bbb i w)^2} \lim \limits _{R \to \infty} \int \limits _{-R - (x - \Bbb i w)} ^{R - (x - \Bbb i w)} \Bbb e ^{-\pi u^2} \Bbb d u .$$
The last integral is to be interpreted as an integral in the complex plane along the horizontal line segment joining $-x + \Bbb i w - R$ and $-x + \Bbb i w + R$ (in this very direction, i.e. from left to right). To this line segment (call it $L$) add three more ones, namely: a segment $L_1$ joining $-x + \Bbb i w + R$ to $-x + R$, a segment $L_2$ joining $-x + R$ to $-x - R$, and a segment $L_3$ joining $-x -R$ to $-x + \Bbb i w - R$ (sketching these on paper will help you). These four segments traversed in precisely the direction given above will form a closed contour $C = L \cup L_1 \cup L_2 \cup L_3$ (a rectangle with the edge $L_2$ on the real axis - a fact which will be important later). Since the function $\Bbb e ^{- \pi u^2}$ is holomorphic on $\Bbb C$, Cauchy's theorem implies that its integral along $C$ will be $0$, so $\int \limits _L = - \int \limits _{L_1} - \int \limits _{L_2} - \int \limits _{L_3}$ (keep in mind that we want $\lim \limits _{R \to \infty} \int \limits _L$).
To see what happens on $L_1$ parametrize it by $t \mapsto -x - \Bbb i t w + R, \ t \in [-1, 0]$ and note that
$$\int \limits _{L_1} \Bbb e ^{- \pi u^2} \Bbb d u = \int \limits _{-1} ^0 \Bbb e ^{- \pi (-x - \Bbb i t w + R)^2} \Bbb d (-x - \Bbb i t w + R) = -\Bbb i w \int \limits _{-1} ^0 \Bbb e ^{- \pi (-x - \Bbb i t w + R)^2} \Bbb d t = \\ -\Bbb i w \int \limits _{-1} ^0 \Bbb e ^{- \pi [(R - x)^2 - t^2 w^2 -2 \Bbb i t w (R - x)]} \Bbb d t .$$
This integral does not have a nice closed form, but remember that we only want its limit when $R \to \infty$; in order to compute this one, we shall "slip" $\lim \limits _{R \to \infty}$ inside the integral with the aid of Lebesgue's dominated convergence theorem. To use it, all we have to do is to show that the integrand is bounded with respect to $t$ and $R$; remembering that $|\Bbb e ^z| = \Bbb e ^{\text{Re} \ z}$ we can write:
$$| \Bbb e ^{- \pi [(R - x)^2 - t^2 w^2 -2 \Bbb i t w (R - x)]} | = \Bbb e ^{- \pi [(R - x)^2 - t^2 w^2]} \le \Bbb e ^{\pi t^2 w^2} \le \Bbb e ^{\pi w^2}$$
because $\max \limits _{t \in [0,1]} t^2 = 1$.
We may now apply Lebesgue's theorem to obtain that
$$\lim \limits _{R \to \infty} -\Bbb i w \int \limits _{-1} ^0 \Bbb e ^{- \pi [(R - x)^2 - t^2 w^2 -2 \Bbb i t w (R - x)]} \Bbb d t = -\Bbb i w \int \limits _{-1} ^0 \lim \limits _{R \to \infty} \Bbb e ^{- \pi [(R - x)^2 - t^2 w^2 -2 \Bbb i t w (R - x)]} \Bbb d t$$
and noting that $\lim \limits _{R \to \infty} \Bbb e ^{- \pi [(R - x)^2 - t^2 w^2]} = 0$ and the imaginary part of the integrand is an imaginary exponential, therefore bounded, the limit of the whole integrand is $0$, so $\lim \limits _{R \to 0} \int \limits _{L_1} = 0$.
A similar reasoning applies to showing that $\lim \limits _{R \to 0} \int \limits _{L_2} = 0$.
We discover, then, that $\lim \limits _{R \to 0} \int \limits _L = - \lim \limits _{R \to 0} \int \limits _{L_2}$ or, putting all the missing terms,
$$\Bbb e ^{\pi (x - \Bbb i w)^2} \int \limits _{\Bbb R} \Bbb e ^{-\pi \big(y - (x - \Bbb i w) \big) ^2} \Bbb d y \Bbb e ^{\pi (x - \Bbb i w)^2} = \Bbb e ^{\pi (x - \Bbb i w)^2} \lim \limits _{R \to \infty} \int \limits _{-R - (x - \Bbb i w)} ^{R - (x - \Bbb i w)} \Bbb e ^{-\pi u^2} \Bbb d u = \\ \Bbb e ^{\pi (x - \Bbb i w)^2} \lim \limits _{R \to \infty} \int \limits _{-R - x} ^{R - x} \Bbb e ^{-\pi u^2} \Bbb d u = \Bbb e ^{\pi (x - \Bbb i w)^2} \int \limits _{-\infty} ^\infty \Bbb e ^{-\pi u^2} \Bbb d u = \frac {\Bbb e ^{\pi (x - \Bbb i w)^2}} {\sqrt \pi} \int \limits _{-\infty} ^\infty \Bbb e ^{-t^2} \Bbb d t = \frac {\Bbb e ^{\pi (x - \Bbb i w)^2}} {\sqrt \pi} \sqrt \pi = \color{blue} {\Bbb e ^{\pi (x - \Bbb i w)^2}} .$$
(Assuming that you know why $\int \limits _{-\infty} ^\infty \Bbb e ^{-t^2} \Bbb d t = \sqrt \pi$. Also, note that having chosen a half-circle instead of a rectangle for your contour would have horribly complicated the solution, therefore choosing the most convenient contour is a skill the training of which should not go neglected.)