TITLE: What is the square root of a Fourier transform?
QUESTION [22 upvotes]: Given the Fourier transform defined like
$$
\mathcal F[f](\omega) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(t) e^{-iwt} \mathrm{d}t
$$
how could one define a square root $\mathcal G$ of the operator $\mathcal F$ so that 
$$
\mathcal G^2[f] = F[f]
$$
My idea is to literally take the square root of $\mathcal F$ like
$$
\mathcal G[f](\omega) = \sqrt{\mathcal F[f](\omega)} = \sqrt{\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(t) e^{-iwt} \mathrm{d}t} = \frac{1}{\sqrt{\sqrt{2 \pi}}} \int_{-\infty}^\infty \sqrt{f(t)} e^{-\frac{iwt}{2}} \mathrm{d}t
$$
Then the square of $\mathcal G[f]$ would be
$$
\mathcal G^2[f](\omega) = \frac{1}{\sqrt{\sqrt{2 \pi}}} \frac{1}{\sqrt{\sqrt{2 \pi}}} \int_{-\infty}^\infty \int_{-\infty}^\infty \sqrt{f(t)} \sqrt{f(t')} e^{-\frac{iwt}{2}} e^{-\frac{iw't'}{2}} dt' dt
$$
But this doesn't quite equate to $\mathcal F[f]$ again. So I guess my approach of literally taking the square root is a bit too naive. How then could one define the square root of a Fourier transform?

REPLY [13 votes]: I think what you are looking for is the Fractional Fourier Transform, which is a generalization of the Fourier transform to the $n^{\textrm{th}}$ power. They are a subset of the Linear Canonical Transforms.
For any value $\alpha=\frac{\pi}{2} \cdot n$, the definition of the Fractional Fourier Transform of a function $f$ is
$$
\mathcal{F}_\alpha[f](u) = 
\sqrt{1-i\cot(\alpha)} \cdot e^{i \pi \cot(\alpha) u^2} 
\int_{-\infty}^\infty 
e^{-i2\pi \left(\csc(\alpha) u x - \frac{\cot(\alpha)}{2} x^2\right)}
f(x)\, \mathrm{d}x
$$
For the square root, filling in $\alpha=\frac{\pi}{4}$ gives:
$$
\mathcal{F}_\alpha[f](u) = 
\sqrt{1-i} \cdot e^{i \pi  u^2} 
\int_{-\infty}^\infty 
e^{-i2\pi \left(\sqrt{2} u x - \frac{1}{2} x^2\right)}
f(x)\, \mathrm{d}x
$$
For integer multiples of $\pi$, the cotangent and cosecant functions diverge, so you need to take the limit instead.