Fourier transform of Hermite polynomial via generating function I've spent a lot of times trying to show that
$$
\mathcal{F}[e^{-x^2/2} G(x,t)] = e^{-k^2/2} G(k, -it)
$$
with $G(x,t)$ being the generating function of Hermite polynomial,
$$
G(x,t) = e^{2tx - t^2}
$$
My attempt
We want to show that
\begin{align*}
  \mathcal{F}[e^{-x^2/2} G(x,t)] 
    &= e^{-k^2/2} G(k, -it) \\
    &= e^{-k^2/2} e^{-2itk+t^2} \\
    &= e^{-\frac{k^2}{2} - 2itk + t^2}
\end{align*}
Computing the Fourier transform:
\begin{align*}
    \mathcal{F}[e^{-x^2/2} G(x,t)] 
      &= \mathcal{F}[e^{-x^2/2} e^{2tx - t^2}] \\
      &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-x^2/2} e^{2tx - t^2} e^{-ikx} dx
\end{align*}
We know that
\begin{align*}
    \sqrt{2\pi} 
      &= \sqrt{\frac{\pi}{1/2}} \\
      &= \int_{-\infty}^{+\infty} e^{-\frac{1}{2}x^2} dx
  \end{align*}
I've been trying to factor the exponential isolating something like $-\frac{1}{2}x^2$.
In fact I should get something like $e^{-\frac{1}{2}x^2} e^{-\frac{k^2}{2} + 2itk + t^2}$ in the integrant so I can get rid of the $\frac{1}{\sqrt{2\pi}}$ and get the desired result but I can't do it.
 A: We can use the following relation,
\begin{equation*}
  \int_{-\infty}^{\infty} e^{-ax^2 + 2bx} dx = \sqrt{\frac{\pi}{a}} e^{\frac{b^2}{a}}
\end{equation*}
Therefore,
\begin{align*}
  \mathcal{F}[e^{-x^2/2} G(x,t)] 
    &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty}e^{-x^2/2} e^{2tx - t^2} e^{-ikx} dx \\
    &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty}e^{-x^2/2} e^{2tx} e^{-ikx} e^{-t^2} dx \\
    &= e^{-t^2} \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty}e^{-x^2/2} e^{2tx} e^{-ikx} dx
\end{align*}
We can rewrite the exponential argument as a 2nd degree polynomial tor the exponential to fit the above relation,
\begin{align*}
  -\frac{x^2}{2} + 2tx - ikx 
    &= -\frac{1}{2}x^2 + (2t - ik)x \\
    &= -\frac{1}{2}x^2 + 2(t - \frac{ik}{2})x
\end{align*}
We have $a = \frac{1}{2}$ et $b = t-\frac{ik}{2}$.
Therefore,
\begin{align*}
  \mathcal{F}[e^{-x^2/2} G(x,t)] 
    &= e^{-t^2} \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty}e^{-x^2/2} e^{2tx} e^{-ikx} dx \\
    &= e^{-t^2} \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-\frac{1}{2}x^2 + 2(t - \frac{ik}{2})x} dx \\
    &= e^{-t^2} \frac{1}{\sqrt{2\pi}} \sqrt{2\pi} e^{2(t-\frac{ik}{2})^2} \\
    &= e^{-t^2} e^{2(t^2 - 2\frac{ikt}{2} - \frac{k^2}{4})} \\
    &= e^{-t^2} e^{2t^2} e^{-2ikt} e^{-\frac{k^2}{2}} \\
    &= e^{-\frac{k^2}{2}} e^{-2ikt} e^{t^2} \\
    &= e^{-\frac{k^2}{2}} G(k, -it)
\end{align*}
