On the Fourier transform of $f(x)=e^{-x^2+2x}$

Question

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!

Answer 1

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}

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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.

Answer 2

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 $f$).