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.

You have $e^{-x^2/2} e^{-itx}$.  The exponent is
$$
\begin{align}
-\frac{x^2}{2} - itx & = -\frac 1 2 (x^2 + 2itx) \\[10pt]
& = -\frac{1}2 \left((x^2+2itx+ (it)^2) - (it)^2\right) \tag{completing the square} \\[10pt]
& = -\frac 1 2 \left( (x+it)^2 - t^2\right)
\end{align}
$$
So you're integrating
$$
e^{-(1/2)(x+it)^2} \cdot \underbrace{{}\ \ e^{t^2/2}\ \ {}}_{\text{no $x$ appears here}}
$$
The factor in which no $x$ appears can be pulled out of the integral when the variable with respect to which you're integrating is $x$.
Then the integral is $\text{constant}\cdot\int_{-\infty}^\infty e^{-(1/2)(x - \text{something})^2} \,dx$.  The last thing to do is show that the value of the integral doesn't depend on the "something".  You can write $u = (x-\text{something})$ and $du=dx$, etc.
