The [Fabius function][1] $F\colon\mathbb R\to[-1,1]$ may be defined as the unique solution of the functional integral equation 
$F(x)=\int_0^{2x}F(t)\,dt$ for all real $x$ such that $F(1)=1$. 

The [recent MO post][2] provided a link to the [MathSE question][3], asking to confirm a conjectured "non-recursive, self-contained formula for the Fabius function". The MO post has been overall negatively received and may get closed. I think the mentioned MathSE question may be of interest. 

On this page, whereas the mentioned conjectured formula will not be confirmed so far, a simpler non-recursive, explicit formula for the Fabius function will be offered, which is expressed in terms similar to, but simpler than, those in the conjectured formula. 

So, a question yet remains: Can one use the simpler formula [below](https://mathoverflow.net/a/351005) to confirm the conjecture on MathSE? Or maybe one could do that otherwise?


  [1]: https://en.wikipedia.org/wiki/Fabius_function
  [2]: https://mathoverflow.net/questions/350989/conjectured-formula-for-the-fabius-function
  [3]: https://math.stackexchange.com/questions/3283519/conjectured-formula-for-the-fabius-function