Is the integral of x^x from 0 to a transcendental?

[Saracen Rue]

Just a quick thought I had:

If we have the following integral: $\int_0^ax^xdx$ where $a>0$, is there any way to tell if all real number results will be transcendental? And if not $a∈R$, would it be possible if we restrict it to only integers?

 

[fresh_42]

I don't think so. Even for much more common numbers as e.g. $\pi^\pi , \zeta(5)$ or the Euler-Mascheroni constant etc. it is not know, usually not even if they were irrational at all, not to mention transcendental. There are some series of transcendental numbers known, if the theorem of Gelfond-Schneider is applicable, but usually each known transcendental number has its own proof, because it is given in a very individual way. The integral results in a family of numbers so if we assumed to have it proven in some case, it would probably apply to many cases. However, the numbers themselves are of no mathematical interest that I know of, so I doubt that someone has tried. But I'm not sure. Maybe a closer look on the proof of Gelfond-Schneider reveals a chance to extend it to integrals.

 

[haruspex]

Just a quick thought I had:

If we have the following integral: $\int_0^ax^xdx$ where $a>0$, is there any way to tell if all real number results will be transcendental? And if not $a∈R$, would it be possible if we restrict it to only integers?

Clearly not true ∀a∈ℝ. There must be some a between 1 and 2 for which the integral equals 1.

 

[FactChecker]

The integral is a continuous function of a, so it will take on all real values between the lower and upper limits of its range. Many of those will not be transcendental.

 

[Saracen Rue]

Yeah it's clearly not for $a∈R$, sorry for making that mistake it was rather late when I made this post. However I'm still curious for $a∈Z$ because the x^x is in itself a transcendental function.

 

[fresh_42]

Well, it's an integer for $a=0$ and complex for $a <0$ (acc. to WA). So let's say $a \in \mathbb{N}$. Unfortunately Gelfond-Schneider requires an algebraic, irrational exponent, so its proof is probably of no help. I would look at it anyway, it might be of help here.