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.
 A: The Fourier transform turns differentiation into multiplication and multiplication into differentiation, which is definitely related to the ODE
$$
                \frac{d}{dx}\left(e^{-ax^{2}}\right) = -ax\left(e^{-ax^{2}}\right),\\
                \left(e^{-ax^{2}}\right)|_{x=0} = 1.
$$
So the Fourier transform of $e^{-ax^{2}}$ must also be a Gaussian because the above ODE is transformed to a related ODE. Explicitly, if
$F_{a}(s)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ax^{2}}\,dx$, then
$$
               \frac{d}{ds}F_{a} = -\frac{1}{2a}F_{a}, \\ F_{a}(0)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ax^{2}}\,dx,
$$
which has unique solution
$$
         F_{a}(s) = F_{a}(0) e^{-x^{2}/4a}.
$$
A: Hint:
$$e^{-x^2/2} e^{-itx} e^{t^2/2} = e^{ \frac{(ix-t)^2}{2} }$$
A: 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.
A: 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.
