I'm a PhD student who is starting to work right now in the well-established field of ultra-fast optics. The thing is that, in most of the papers I have been reading during the past few days, there is a Fourier transform pair that is normally taken for granted and serves as the basis for more complex calculations. However, and despite my efforts, I haven't been able to find any proof of such pair.
In general, it is stated that:
\begin{equation}
\mathcal{F}\{e^{(jt^2)/(2\alpha)}\} \propto e^{(-j\alpha\omega^2)/2}
\end{equation}
where $j$ is the imaginary unit, as is the usual convention in engineering.
To put it briefly, the Fourier transform of a complex exponential that depends quadratically with time is proportional to a complex eponential that depends quadratically with angular frequency.
I know that this is in fact the case when a exponential with a REAL quadratic argument (i.e. a scaled version of the Normal distribution) is considered:
\begin{equation}
\mathcal{F}\{e^{-t^{2}/(2\alpha)} \} \propto e^{-\sigma\omega^{2}/2}
\end{equation}
However, I don't see an easy way to extend that result to the case when the imaginary unit is included in the argument of the exponential.
Any comments would be very much appreciated, thanks!
Best Answer
$$ \begin{align} \int_{-\infty}^\infty e^{i\frac{t^2}{2a}}e^{-ixt}\,\mathrm{d}t &=\sqrt{a}\int_{-\infty}^\infty e^{i\frac{t^2}2}e^{-ixt\sqrt{a}}\,\mathrm{d}t\tag{1}\\ &=\sqrt{a}\,e^{-iax^2/2}\int_{-\infty}^\infty e^{i\frac{(t-x\sqrt{a})^2}2}\,\mathrm{d}t\tag{2}\\ &=\sqrt{a}\,e^{-iax^2/2}\int_{-\infty}^\infty e^{i\frac{t^2}2}\,\mathrm{d}t\tag{3}\\ &=\sqrt{a}\,e^{-iax^2/2}e^{i\pi/4}\int_{-\infty}^\infty e^{-\frac{t^2}2}\,\mathrm{d}t\tag{4}\\ &=\sqrt{2\pi a}\,e^{i(\pi-2ax^2)/4}\tag{5} \end{align} $$ Explanation:
$(1)$: substitute $t\mapsto t\sqrt{a}$
$(2)$: complete the square
$(3)$: substitute $t\mapsto t+x\sqrt{a}$
$(4)$: contour integration
$(5)$: $\int_{-\infty}^\infty e^{-t^2/2}\,\mathrm{d}t=\sqrt{2\pi}$
To justify $(4)$, consider the contour $[-R,R]\cup Re^{[0,1]i\pi/4}\cup[R,-R]e^{i\pi/4}\cup Re^{-[3,4]i\pi/4}$ :
The integral of $e^{iz^2/2}$ along this contour is $0$. The integral along the large arcs vanishes as $R\to\infty$. The integral along the horizontal line tends to $\int_{-\infty}^\infty e^{it^2/2}\,\mathrm{d}t$, while the integral along the diagonal line is the negative of $e^{i\pi/4}\int_{-\infty}^\infty e^{-t^2/2}\,\mathrm{d}t$ .