In quantum mechanics, I'm given a wave packet that is described by the Gaussian \begin{align*} \psi(x,t=0) = \sqrt{a} \exp\Big(\frac{-x^2} { 2a}\Big). \end{align*} I need to compute the Fourier transform of this. By definition, the Fourier transform of a function $f(x)$ is given by \begin{align*} \mathcal{F}\left\{f \right\} (k) = \int_{- \infty}^{\infty} dx f(x) \exp(-2 \pi i k x). \end{align*} Applying this to the given Gaussian gives me \begin{align*} \mathcal{F}\left\{\psi \right\} (k) = \int_{- \infty}^{\infty} dx \sqrt{a}\exp\Big(\frac{-x^2} { 2a}\Big)\exp\Big(\frac{-x^2} { 2a}\Big). \end{align*} I completed the square in the exponential, and got \begin{align*} \mathcal{F}\left\{\psi \right\} (k) = \sqrt{a} \exp(-2 \pi^2 k^2 a) \int_{-\infty}^{\infty} dx \exp\Big(- \frac{1}{2a} (2a i k \pi + x)^2\Big). \end{align*} However, now I still don't know what to do with the integral \begin{align*} \int_{-\infty}^{\infty} dx \exp\Big(- \frac{1}{2a} (2a i k \pi + x)^2\Big). \end{align*} Do I use a substitution or is there some other trick to evaluate this?
EDIT: I wish to show that $$ I = \int_{-\infty}^{\infty} dx \exp\Big(- \frac{1}{2a} (2a i k \pi + x)^2\Big) = \int_{-\infty}^{\infty} dx \exp\Big(- \frac{1}{2a} x^2\Big). $$ If I can show this somehow, then I can use polar coordinates to evaluate it, by using $I^2$, then integrating over $dx$ and $dy$ and using polar coordinates.
Partial work. Let define the following function: $$g(\xi)=\int_{\mathbb{R}}\exp\left(-\frac{x^2}{2a}\right)\exp(-i\xi x)\,\mathrm{d}x.$$ Notice that: $$g'(\xi)=-i\int_{\mathbb{R}}x\exp\left(-\frac{x^2}{2a}\right)\exp(-i\xi x)\,\mathrm{d}x=-\frac{\xi}{a}g(\xi).$$ I used derivation under the integral and then integration by parts. Moreover, using substitution, one has: $$g(0)=\sqrt{2\pi a}.$$ $g$ satisfies an ODE, you can easily solve.