Fourier Transform of complicated product: $(1+x)^2 e^{-x^2/2}$ UPDATE: The problem reduces to this for me:
I ran into an issue with a change of variable integral:
I have $\int\limits_{-\infty}^{\infty}e^{-\frac{1}{2}(x+\frac{iy}{2})^{2}}dx$.
I do the change of variable $u = \frac{x + \frac{iy}{2}}{\sqrt{2}}du$.
This changes the integral to $\sqrt{2}\int\limits_{*}^{*}e^{-u^{2}}du$.  If my new bounds are the same as my old bounds, then I'm done since the integral is known to be $\sqrt{\pi}$.  But since I have done a complex translation I am not so sure.  Can I say that $\infty + i = \infty$ ?
Original Problem:
I need to find the Fourier transform of $g(x) = (x+1)^2e^{-x^2/2}$.
This is defined by:
$$\widehat{g}(y) = \int\limits_{-\infty}^{\infty}g(x)\,e^{-ixy}\,dx.$$
This works out as:
$$
\begin{eqnarray*}
\widehat{g}(y) &=& \int\limits_{-\infty}^{\infty}(x+1)^2\,e^{-x^2/2}\,e^{-iyx}\,dx\\
&=& \int\limits_{-\infty}^{\infty}(x+1)^2e^{-x^2/2 - ixy}\,dx.
\end{eqnarray*}$$
But I don't know how to evaluate this. Is using software appropriate for something like this?
 A: If I understand the comments correctly, the remaining problem seems to be to find the Fourier transform of $$\color{blue}{f(x) = e^{-x^2/2}}.$$
You can do this by completing the squares and substitution, as outlined in the comments, but I prefer this argument:
Let's compute formally first, deferring the justification of the interchange of differentiation and integration in $(!!)$ to the end of the answer:
$$
\begin{align*}
\frac{d}{dk} \widehat{f}(k) &= \frac{d}{dk} \int_{-\infty}^{\infty} e^{-x^2/2}e^{-ikx}\,dx \\
&\!\color{red}{\stackrel{(!!)}{=}}
\int_{-\infty}^\infty e^{-x^2/2}\, (-ix)\, e^{-ikx}\,dx \tag{!!} \\
&=
\int_{-\infty}^\infty i
\left( 
  \frac{d}{dx}e^{-x^2/2}
\right) \, 
e^{-ikx}\, dx \\
&= 
-\int_{-\infty}^\infty i e^{-x^2/2}(-ik)e^{-ikx}\,dx  
&& \text{(integration by parts)}\\
&= -k \,\hat{\!f}(k).
\end{align*}
$$
This gives us the ordinary differential equation $\hat{\!f}'(k) = -k \,\hat{f}(k)$ with the initial condition
$$
\hat{\!f}(0) = \int_{-\infty}^{\infty} e^{-x^2/2}\,dx = \sqrt{2\pi},
$$
which shows that 
$$\color{blue}{\hat{f}(k) = \sqrt{2\pi}\,e^{-k^2/2}}.$$

In order for you to be able to check your own work, here's my solution of the exercise (N. S. suggested a slightly different approach, but neither is simpler or more efficient, they come down to the same):
We have $f'(x) = -x\,f(x)$ and $f''(x) = (x^2-1)f(x)$ so that
$$g(x) = (x+1)^2 f(x) = f''(x) -2f'(x) + 2f(x),$$
hence
$$
\begin{align*}
\hat{g}(k) 
&= \widehat{f''}(k) - 2\widehat{f'}(k) + 2\widehat{f}(k) \\
&= (-k^2+2ik+2) \hat{f}(k)
\end{align*}
$$
which gives us
$$\color{red}{\hat{g}(k) = \sqrt{2\pi}\,(-k^2+2ik+2)\,e^{-k^2/2}},$$
where you can write $-k^2+2ik+2 = 1+(1+ik)^2$ if you prefer.

It remains to justify $(!!)$: Compute
$$
\begin{align*}
\frac{d}{dk}\hat{\!f}(k) & =
\lim_{h\to 0} \frac{1}{h} \left(\hat{\!f}(k+h)-\hat{\!f}(k)\right) \\
&= \lim_{h\to0} \int_{-\infty}^{\infty} e^{-x^2/2} \frac{e^{-ikx}}{h}\underbrace{\left(e^{-ihx}-1\right)}_{\large 2i e^{ihx/2}\sin{\frac{hx}{2}}}\,dx.
\end{align*}
$$
Using that $|\sin{\frac{hx}{2}}| \leq \dfrac{|hx|}{2}$ we see that the absolute value of the integrand is bounded above by $|x|e^{-x^2/2}$ which is clearly integrable, so we conclude by the dominated convergence theorem.  
