Recursive Integration over Piecewise Polynomials: Closed form? Is there a closed form to the following recursive integration?
$$
f_0(x) =
\begin{cases}
1/2 & |x|<1 \\
0 & |x|\geq1
\end{cases}
\\
f_n(x) = 2\int_{-1}^x(f_{n-1}(2t+1)-f_{n-1}(2t-1))\mathrm{d}t
$$
It's very clear that this converges against some function and that quite rapidly, as seen in this image, showing the first 8 terms:

Furthermore, the derivatives of it have some very special properties.
Note how the (renormalized) derivatives consist of repeated and rescaled functions of the previous degree which is obviously a result of the definition of the recursive integral:

EDIT
I found the following likely Fourier transform of the expression above. I do not have a formal proof but it holds for all terms I tried it with (first 11).
$$ \mathcal{F}_x\left[f_n(x)\right](t)=\frac{1}{\sqrt{2\pi}}\frac{2^n \sin \left(2^{-n} t\right)}{t} \prod _{k=1}^n \frac{2^{k} \sin \left(2^{-k} t\right)}{t} $$
Here an image of how that looks like (first 10 terms in Interval $[-8\pi,8\pi]$):

With this, my question alternatively becomes:
What, if there is one, is the closed form inverse fourier transform of
$\mathcal{F}_x\left[f_n(x)\right](t)=\frac{1}{\sqrt{2\pi}}\frac{2^n \sin \left(2^{-n} t\right)}{t} \prod _{k=1}^n \frac{2^{k} \sin \left(2^{-k} t\right)}{t}$,
especially for the case $n\rightarrow\infty$?

As a side note, it turns out, that this particular product is a particular Browein integral (Wikipedia) using as a sequence $a_k = 2^{-k}$ which exactly sums to 1. The extra term in the front makes this true for the finite sequence as well. In the limit $k \to \infty$, that term just becomes $1$, not changing the product at all. It is therefore just a finite depth correction.

Suppose $f$ is a fixed point of the iterations. Then
$$f(x) = 2\int_{-1}^x\big(f(2t+1)-f(2t-1)\big)\,\mathrm{d}t,$$
which, upon differentiating both sides by $x$, implies that
$$f'(x) = 2\big(f(2x+1)-f(2x-1)\big).$$
I'll assume that $f$ vanishes outside $[-1,1]$, which you can presumably prove from the initial conditions. Then we get
$$f'(x) = \begin{cases}
2f(2x+1) & \text{if }x\le0, \\
-2f(2x-1) & \text{if }x>0.
\end{cases}$$
This is pretty close to the definition of the Fabius function. In fact, your function would be $\frac{\text{Fb}'(\frac{x}{2}+1)}{2}$
The Fabius function is smooth but nowhere analytic, so there isn't going to be a nice closed form for your function.
