Computing an inverse Fourier Transform / Solving the free particle Schrödinger equation with a gaussian wave packet as initial condition I was asked to solve the following parabolic PDE
$$\partial_t \psi(x,t) = i \frac{\Delta}{2m} \psi(x,t)$$ with initial condition  $\psi(t,0 ) = \frac{1}{\sqrt{2 \pi a^2}}e^{-\frac 1 4 ({\frac{x}{a})^2}}$, where $x\in\mathbb{R}^3$.
Fourier transforming I  obtained that ${\partial_t \hat{\psi}(\xi,t) = -i \frac{|\xi|^2}{2m} \hat{\psi}(x,t)}$ so 
$$\hat{\psi}(\xi,t) = e^{-\frac{i}{2m} |\xi|^2 t} \hat{\psi}(\xi,0) = e^{-\frac{i}{2m}|\xi|^2 t} \frac{2 a^2}{\sqrt{\pi}}e^{-\frac{1}{2}(\sqrt{2}a\xi)^2}$$

Now I have to compute the Inverse Fourier transform of this abominable expression $e^{-\frac{i}{2m}|\xi|^2 t} \frac{2 a^2}{\sqrt{\pi}}e^{-\frac{1}{2}(\sqrt{2}a\xi)^2}$, is there some trick or cunning way to compute it?

 A: In this case the integral that defines the reverse Fourier transform can be done as a Gaussian integral in the naive fashion. 
Technically, you need to look in to complex analysis to justify the shift in the imaginary direction of the integral path in the complex plain inherent in the integration of a Gaussian function with complex mean value. Short version, the fact that the Gaussian function is analytic for all finite $z$ allows you to shift the integration contour up or down by constructing a rectangular path that the desired path is one side of and the new path is the opposite side. As long as you can show that the integral along the ends of the rectangle vanish when the rectangle's length is taken to infinity, then the two paths will have to have an equal integral because it encloses no poles in the integrand (see: residue theorem).
If the integrand can be written as $\exp(- a x^2 + 2bx - c)$, with $a$, $b$, and $c$ any complex numbers, then as long as the real part of $a$ is positive, it should converge (hint: complete the square in the exponential's argument).
