On the Fourier transform of $f(x)=e^{-x^2+2x}$ So, I have the $f(x)=e^{-x^2+2x}$ and to take the FT of it, I complete the square:
\begin{equation}
f(x)=e^{-x^2+2x \pm1}=e^{-(x-1)^2}e
\end{equation}
Then, by knowing that the FT of $g(x)=e^{-x^2}$ is:
\begin{equation}
\mathcal{F}[g(x)](k)=\hat{g}(k)=\frac{e^{-k^2/4}}{\sqrt{2}}
\end{equation}
I make use of the property:
\begin{equation}
f(x)=g(x-a) \Rightarrow \mathcal{F}[f(x)](k)=e^{-iax}\hat{g}(k)
\end{equation}
I can see that for a=1:
\begin{equation}
\hat{f}(k)=ee^{-ik}\hat{g}(k)=\frac{e^{-(k+2i)^2/4}}{\sqrt{2}}
\end{equation}
But I see on the book that the correct answer is:
\begin{equation}
\hat{f}(k)=\frac{e^{-(k-2i)^2/4}}{\sqrt{2}}
\end{equation}
Why is that minus over there? I cannot see my mistake :/
Thank you!
 A: Let the Fourier transform be defined as
\begin{align}
f(\omega) = \frac{1}{2\pi} \, \int_{-\infty}^{\infty} f(x) \, e^{-i \omega x} \, dx
\end{align}
Now, for $f(x) = e^{- x^{2} + 2x}$ the following is developed.
\begin{align}
2 \pi \, f(\omega) &= \int_{-\infty}^{\infty} e^{- (x^2 - (2 - i\omega) x)} \, dx \\
&= e^{-\left(1 - \frac{i \omega}{2}\right)^{2}} \, \int_{-\infty}^{\infty} e^{- \left(x - 1 + \frac{i \omega}{2}\right)^{2}} \, dx \\
&= e^{-\left(1 - \frac{i \omega}{2}\right)^{2}} \, \int_{-\infty}^{\infty} e^{- u^{2}} \, du \hspace{10mm} u=x-1+\frac{i\omega}{2} \\
&= 2 \, e^{-\left(1 - \frac{i \omega}{2}\right)^{2}} \, \int_{0}^{\infty} e^{- u^{2}} \, du \\
&= e^{-\left(1 - \frac{i \omega}{2}\right)^{2}} \, \int_{0}^{\infty} e^{- t} \, t^{-1/2} \, dt \hspace{10mm} t = u^{2} \\
&= \sqrt{\pi} \, e^{-\left(1 - \frac{i \omega}{2}\right)^{2}}.
\end{align}
This leads to
\begin{align}
f(\omega) = \frac{1}{2\sqrt{\pi}} \, e^{-\left(1 - \frac{i \omega}{2}\right)^{2}}.
\end{align}

Method 2:
Using the properties
\begin{align}
f(t-t_{0}) &= f(\omega) \, e^{- i \omega t_{0}} \\
f(e^{- t^{2}}) &= \sqrt{\pi} \, e^{- \frac{\omega^{2}}{4}}
\end{align}
then
\begin{align}
F\{ e^{-x^{2}+ 2x}; \omega\} &= e \, F\{ e^{-(x-1)^{2}}; \omega\} = e^{1 - i \omega} \, F\{e^{-x^{2}}; \omega\} = \sqrt{\pi} \, e^{1 - i \omega - \frac{\omega^{2}}{4}} = \sqrt{\pi} \, e^{\left(1 - \frac{i \omega}{2}\right)^{2}}
\end{align}
- as it appears the definitions of the properties used here remove the general minus sign of the exponential. 
A: I propose another method that does not use integration. Problem is in the change of variable when completing the square : because of the imaginary unit $i$, the new variable is complex and thus the integral must be done carefully in $\mathbb{C}$.
As $f$ is smooth (i.e. $f\in\mathcal{C}^\infty\left(\mathbb{R};\mathbb{R}\right)$), $\hat{f}$ is also smooth. Since we have
$$f'=-2(x-1)f$$
we obtain thanks to the properties of the Fourier transformation
$$ik\hat{f}=-2i\hat{f}'+2\hat{f}$$
i.e.
$$\hat{f}'=-\frac{(ik+2)}{2i}\hat{f}.$$
Then, on can easily check that a solution of this EDO is given by
$$\hat{f}(k)=C\mathrm{e}^{-k^2/4+ik}\quad\quad C\in\mathbb{R}.$$
You can seek the constant C by letting $k\to0$ (use the continuity under integral criterion to link $\hat{f}(0)$ to $f$).
