The Fabius function $F\colon\mathbb R\to[-1,1]$ 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 provided a link to the MathSE question, 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 to confirm the conjecture on MathSE? Or maybe one could do that otherwise?

The Fabius function $F\colon\mathbb R\to[-1,1]$ 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 provided a link to the MathSE question, 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 to confirm the conjecture on MathSE? Or maybe one could do that otherwise?

The Fabius function $F\colon\mathbb R\to[-1,1]$ 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 provided a link to the MathSE question, 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.

The Fabius function $F\colon\mathbb R\to[-1,1]$ 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 provided a link to the MathSE question, 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 to confirm the conjecture on MathSE? Or maybe one could do that otherwise?