Is it possible to express $\int_{0}^{1-\epsilon}\left(\sqrt{1-x^2}^{\sqrt{1-x^2}^{\cdots}}\right) dx$ in elementary functions?

Question

let $\epsilon >0$, I tried to evaluate $\int_{0}^{1}\left(\sqrt{1-x^2}^{\sqrt{1-x^2}^{\cdots}}\right) dx$ , using the fact $x= \cos t$ yield to have integrand using $\sin $ function seems is not easy to get such closed form by this variable change , For one iteration by means $\int_{0}^{1}\left(\sqrt{1-x^2}^{\sqrt{1-x^2}}\right) dx$ we have the integrand converge approximately to $\frac{\sqrt{3}}{{2}}$, Now for some odd iterations we have $l=0.89..$ and for even iterations we have $l=0.9..$ , Now if we fixe $\epsilon$ at at some small value such that $x$ lie at a least between $(0,0.99782)$ to get convergence according to below comment by @Oscar Lanzi and @Sangchul Lee , Now my question here is : Is it possible to express the titled integral in elementary functions ?

Edit: I edited the question according to the two below montioned comments to assure convergence also my Goal was to express the titled integrand tower in elementary functions

No. The integrand itself does not converge over the entire range of integration. — Oscar Lanzi, Jul 14 '20 at 22:11
Check this posting and/or this article, for instance, for the convergence of infinite power tower. — Sangchul Lee, Jul 14 '20 at 22:48
It is $\int_0^1\frac{x(x+1)e^x}{\sqrt{e^{2xe^x}-1}},dx$, highly unlikely to have a closed form. — metamorphy, Jul 15 '20 at 05:00
If the question is about the integral $$I = \int_0^{\sqrt {1 - e^{-2 e}}} \sqrt {1 - x^2} ! \uparrow \uparrow ! \infty , dx,$$ then, since $\int f^{-1}(x) , dx = \int y , df(y)$, $$I = \int_{1/e}^1 \frac {y^{2/y - 1} (1 - \ln y)} {\sqrt {1 - y^{2/y}}} dy \approx .836333 \hspace{1.5px} .$$ — Maxim, Jul 31 '20 at 12:06

score 1 · Answer 1 · answered Jul 30 '20 at 17:40

If we let

$$y=\lim_{n\to\infty}(\sqrt{1-x^2}\uparrow\uparrow n)$$

then we have

$$y=(\sqrt{1-x^2})^y\implies x=\sqrt{1-y^{2/y}}$$

and the integral can be represented more neatly as

$$I=\exp(-e)\sqrt{1-\exp(-2\exp(1-e))}+\int_{\exp(-e)}^1\sqrt{1-y^{2/y}}~\mathrm dy$$

where $I$ is the original integral over the largest subinterval of $[0,1]$ where it converges.

Although it is highly unlikely a closed form exists, this form is easy to numerically compute and gives

$$I\simeq0.807316213493$$

The same approach, however, is not so nice if one attempts to find the limits of the even and odd heights over the entire interval $[0,1]$, since one needs to solve for $x$ from

$$y=\sqrt{1-x^2}~\widehat~~(\sqrt{1-x^2}~\widehat~~y)$$

and this will require multiple branches of the Lambert W function.

I don't know why this question is not well received here – zeraoulia rafik Jul 30 '20 at 20:11 — zeraoulia rafik, Jul 30 '20 at 20:11
It doesn't seem to be a "research question" I suppose. – Simply Beautiful Art Jul 31 '20 at 04:35 — Simply Beautiful Art, Jul 31 '20 at 04:35

Is it possible to express $\int_{0}^{1-\epsilon}\left(\sqrt{1-x^2}^{\sqrt{1-x^2}^{\cdots}}\right) dx$ in elementary functions?

1 Answers1