Integrating the Fabius function $q(x) = \int_0^x F(u) du$

Question

There are any formulas for the integral of the Fabius function?

Context_____________

After reading the following question: Smooth functions that resemble random walks, I started thinking about some examples and the Fabius function comes into my mind: it is a smooth function that have lobes that switches from possitive to negative values following the Thue–Morse sequence, which is deterministic, but looks pretty random for the untrained viewer so I think about the changing patterns as seen in binary on the Thue–Morse sequence could easily represent an specific sequence of random variables that when added together will resemble a realization of an one dimensional Random walk.

So following this idea, I think the integrating the Fabius function $F(x)$ will lead to a function $q(x) = \int_0^x F(u) du$ that when seen from "far away" (zooming out enough, or conversely, scaling the variable $x$ by some positive quantity $a>0$ such we see $q\left(\frac{x}{a}\right)$) will look like a realization of a random walk, but keeping the function as a smooth one.

Attempts_________

I have tried seen it numerically in Desmos after doing some simplifications (approximating the Thue-Morse sequence and using $\sin(x)^4$ as the lobes), but is not powerfull enough to take the plots of the integrated function. Analitically, the Fabius function has some specific properties that have leads to some interpolation formulas for the lobes, which could be approximated as shown in the links here, but I don't have the skills to find the values of the integrated function. Maybe someone could do it or know some reference to a paper were it is done.

Given $F(x)$ the Fabius function, find $q(x)$ in: $$q(x) = \int\limits_0^x F(u)\ du$$

Answer 1

The Fabius function satisfies the “functional differential equation” $$ F'(x) = 2 F(2x) \, , $$ therefore is $$ \int_0^x F(u) \, du = \int_0^{x/2} 2 F(2t) \, dt = \int_0^{x/2} F'(t)\, dt = F(x/2) \, . $$