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(I'm reposting this from math.stackexchange as I didn't get an answer there and thought it might be "advanced" enough for this site.)

As one may know, the Fourier Transform

$$F[f](\nu) = \int_{-\infty}^{\infty} f(t) e^{-2\pi i \nu t} dt$$

can be iterated, and this iteration generalized to fractional iteration count via

$$F^h[f](\nu) = \sqrt{1 - i \cot\left(\frac{\pi h}{2}\right)} e^{i\pi \cot\left(\frac{\pi h}{2}\right) \nu^2} \int_{-\infty}^{\infty} e^{-i 2\pi \left( \csc\left(\frac{\pi h}{2}\right) \nu t - \frac{\cot\left(\frac{\pi h}{2}\right)}{2} t^2\right)} f(t) dt$$.

This can be interpreted, in a sense, as "rotating" $f$ between the time and frequency domain. In particular, $h = 1/2$ corresponds to a half-way rotation -- half-way between time and frequency. $h = -1$ is the inverse Fourier transform. $h = 2$ (i.e. $F \circ F$) is the "mirror-image" transform, that is, the time reversal of $f$, or $f(-t)$ (equivalent to a 180-degree rotation, "flipping the function over"). The above formula may be written without the $\frac{\pi}{2}$ factors, which parameterizes it more directly in terms of the angle as an actual radian angle, with $\frac{\pi}{2}$ radians, i.e. a right angle, being the angle at which the frequency domain lies to the time one.

But what is the interpretation and meaning of the above when $h = i$, that is, a single imaginary iteration of the transform? What does the "spectrum" given by $F^i[f]$ mean? More generally, what happens with $h = 2i$, $h = 3i$, etc.?

I note that for such positive-imaginary $h$, the kernel in the integrand grows quickly because the exponent becomes real (note that $\csc(ix) = -i\ \mathrm{csch}(x)$ and $\cot(ix) = -i \coth(x)$ and these get multiplied by $-i 2\pi$, turning them real), suggesting a strong decay of $f$ is required to get convergence, unless perhaps we can abuse analytic continuation from the real iteration counts in other cases somehow, though I think that might not yield a unique result. However it does seem to work better for negative-imaginary $h$, i.e. $h = -i$, $h = -2i$, etc. .

An example

Let

$$f(t) = \begin{cases} 1, \mbox{if } |t| \le 1 \\ 0, \mbox{otherwise} \end{cases}$$,

i.e. the "boxcar" function, or a fixed-width pulse in the time domain. Since the support is compact, the integral will converge. If we take the usual Fourier transform, $F^1[f] = F[f]$, we get a sinc function as the result. We can interpret this as a frequency spectrum of the original $f$, and it contains mostly low-frequency content, with little ripples of higher frequencies.

But, suppose we take the imaginary Fourier transform -- one imaginary iteration of $F$, or $F^i$. What we get cannot be expressed in elementary terms, however we can use the "imaginary error function" to get a "closed form" of sorts:

$$F^i[f](\nu) = \frac{\sqrt{1 – \coth\left(\frac{\pi}{2}\right)}}{2 \sqrt{\coth\left(\frac{\pi}{2}\right)}} e^{\pi \left(\coth\left(\frac{\pi}{2}\right) – \mathrm{csch}\left(\frac{\pi}{2}\right) \mathrm{sech}\left(\frac{\pi}{2}\right)\right) \nu^2} \\ \left(\mathrm{erfi}\left(\frac{2\pi \coth\left(\frac{\pi}{2}\right) – 2\pi \mathrm{csch}\left(\frac{\pi}{2}\right) \nu}{2 \sqrt{\pi \coth\left(\frac{\pi}{2}\right)}}\right) – \mathrm{erfi}\left(\frac{-2\pi \coth\left(\frac{\pi}{2}\right) – 2\pi \mathrm{csch}\left(\frac{\pi}{2}\right) \nu}{2 \sqrt{\pi \coth\left(\frac{\pi}{2}\right)}}\right)\right)$$

Now the $1 - \coth\left(\frac{\pi}{2}\right)$ under the one square root sign is negative, so $F^i[f]$ is purely imaginary at each $\nu$. At $0$, it is about $3.338i$, and grows very quickly as we move toward the positive and negative $\nu$ directions. What kind of "spectrum" is this? What is it telling us about the boxcar function?

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