Showing that Gaussians are eigenfunctions of the Fourier transform I'm having a bit of trouble on this problem:

I've tried to evaluate the integral directly (using the trick from multivariable calculus where you "square" the integral and convert to polar coordinates), but that hasn't gotten me anywhere. Does anyone have a suggestion on where to start?
Just for context, this is for complex analysis.

Expand $\exp(itx)$ using Euler's and note the symmetry of the bounds.  Explicitly we have
$
\begin{align}
  I &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\exp\left(\frac{-x^{2} - 2itx}{2}\right)\,\mathrm{d}x \\
  &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\exp\left(-\frac{x^{2}}{2}\right)\big(\cos(tx) - i\sin(tx)\big)\,\mathrm{d}x \\
  &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\sum_{n\geqslant 0}(-1)^{n}\frac{(tx)^{2n}}{(2n)!}\exp\left(-\frac{x^{2}}{2}\right)\,\mathrm{d}x \\
  &( \text{ Note that $\sin(tx)$ is an odd function so $\displaystyle\int_{-\infty}^{\infty}\sin(tx)\exp\left(-\frac{x^{2}}{2}\right)\,\mathrm{d}x = 0$ )} \\
\end{align}
$
From here switch the order of the summation and integral sign ( justified by the Dominated Convergence Theorem ) and one can evaluate the resulting integral via various ways; gamma function, differentiating under the integral sign, etc.
