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$$ \int_0^1 x^{2^{x^{2^{x^{\ldots}}}}} ~~ dx = ~~?$$

                                                         An Image of the function (until a certain amount of iterations)              
                                                                 enter image description here


What I've tried so far:

$$ x^{2^{x^{2 \ldots}}} \overbrace{=}^{\text{def}} y \\ x^{2^{y}} = y \\ 2^y \ln (x) = \ln(y)$$

Now applying the chain rule:

$$\begin{align} \\&2^y \cdot \frac{dy}{dx} \cdot \ln(2) \cdot \ln(x) + \frac{1}{x} \cdot 2^y \ln(y) = \frac{dy}{dx} \cdot \frac1y \\ &\implies \frac{dy}{dx} \left (2^y \cdot \ln(x) \cdot \ln(2) - \frac1y \right) = - \frac1x 2^y \cdot \ln(y) \\ &\implies \frac{dy}{dx} = \left ( \frac{- \frac1x 2^y \cdot \ln(y)}{2^y \cdot \ln(x) \cdot \ln(2) - \frac1y} \right) \end{align}$$

The problem that this looks like a pretty harsh equation to solve for $dy$ and then substitute it back to the integral, with the fact that we need to find the value of $x$ in terms of $y$.

Any help would be appreciated!

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  • $\begingroup$ Are you looking for an exact solution or a numerical approximation? $\endgroup$ – open problem Jan 23 at 23:21
  • $\begingroup$ @openproblem Well, there is no problem approximate it using a script or plugging it some kind of a site like Wolfram-Alpha so I guess the problem is to find an exact solution for this. Of course, if it is not possible (I don't know if it is, but maybe -and probably- I don't have all the tools or knowledge to solve this) then I would assume an approximation would do the job! $\endgroup$ – user12686317 Jan 23 at 23:23
  • $\begingroup$ Where does the problem come from? Could you give some context? $\endgroup$ – NN2 Jan 23 at 23:37
  • $\begingroup$ The inner function in the photo seems not correct (why there is $x^x$). Besides, I plot the inner function and didn't obtain the same plot you gave. $\endgroup$ – NN2 Jan 23 at 23:46
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    $\begingroup$ @NN2 This is not the recurrence relation, for example for $f_2$ you will get $f_2(x) = (f_1 (x))^{x^2} = (x^2)^{x^2}$ this is not equal to $x^{2^{x^2}}$. I don't have a source of context unfortunately! $\endgroup$ – user12686317 Jan 23 at 23:58
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We can transform this to an integral of an elementary function: \begin{align} \int_0^1ydx&=[xy]_0^1-\int_0^1xdy\\ &=1-\int_0^1y^{1/(2^y)}dy.\\ \end{align} However, I don't think there is a closed-form to this integral since Wolfram Alpha doesn't show anything. I tried to use some series expansion, but that didn't seem to work.

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    $\begingroup$ Awesome this is an improvement because we got rid of this ugly infinite power tower! $\endgroup$ – user12686317 Jan 24 at 0:01
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    $\begingroup$ Series work quite well. In any manner, (+1) for your answer. $\endgroup$ – Claude Leibovici Jan 24 at 8:12
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Consider $$I=\int_0^1 y^{2^{-y}}\,dy=\int_0^\frac 12 y^{2^{-y}}\,dy-\int_1^\frac 12 y^{2^{-y}}\,dy=I_1-I_2$$ Each integrand will be developed as a series expansion around the lower bound to order $(n+1)$ and termwise integrated. This does not lead to very complicated expressions for the integrands and the result of integration is just a polynomial in $\log(2)$.

For example, for $n=3$, the integrands are $$y-y^2 \log (2) \log (y)+\frac{1}{2} y^3 \left(\log ^2(2) \log ^2(y)+\log ^2(2) \log (y)\right)+O\left(y^4\right)$$ $$1+\frac{y-1}{2}+(y-1)^2 \left(-\frac{1}{8}-\frac{\log (2)}{2}\right)+(y-1)^3 \left(\frac{1}{16}+\frac{\log ^2(2)}{4}\right)+O\left((y-1)^4\right)$$

and the result of integration is $$\frac{5127+72 \log ^4(2)-36 \log ^3(2)+339 \log ^2(2)-64 \log (2)}{9216}$$ As a function of order $n$, some results $$\left( \begin{array}{cc} n & I_1-I_2 \\ 1 & 0.56250000 \\ 2 & 0.57249702 \\ 3 & 0.56967702 \\ 4 & 0.56945295 \\ 5 & 0.56962947 \\ 6 & 0.56968187 \\ 7 & 0.56969293 \\ 8 & 0.56969494 \\ 9 & 0.56969522 \\ 10 & 0.56969526 \\ 11 & 0.56969527 \\ 12 & 0.56969528 \end{array} \right)$$

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