<s>[INST] Integrating an Infinite exponent tower $$ \int_0^1 x^{2^{x^{2^{x^{\ldots}}}}} ~~ dx =  ~~?$$
                                                         
             
                                                                 

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!
 [/INST] 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
$$\left(
\begin{array}{cc}
 n & I_1-I_2 \\
 1 & 0.56250000 \\
 2 & 0.57249702 \\
 3 & 0.56967702 \\
 4 & 0.56945295 \\
 5 & 0.56962947 \\
 6 & 0.56968187 \\
 7 & 0.56969293 \\
 8 & 0.56969494 \\
 9 & 0.56969522 \\
 10 & 0.56969526 \\
 11 & 0.56969527 \\
 12 & 0.56969528     
\end{array}
\right)$$
</s>