Area enclosed by a function containing two power towers: $f(x)=(−\ln(x↑↑(2k)))↑↑(2k+1)$ I've been considering functions involving power towers lately and come across the following function: $$f(x)=(−\ln(x↑↑(2k)))↑↑(2k+1)$$ $$\text{Where }k∈\mathbb{Z} ^+$$
In the image below we can see some examples of what this function looks like for the first $5$ values of $k$: 

It can be visually seen from the graphs that while the area under each graph does become larger for increasing values of $k$, the rate at which the area is increases decrease. However, I went ahead and took the liberty of doing the calculations anyway. I found the difference in area between $ f_{k=2}(x)$ and $ f_{k=1}(x)$, and then between $f_{k=3}(x)$ and $f_{k=2}(x)$, and so on. I then took these values and plotted them as such:

(Note: The graphs of $f_{k=10}(x)$ and $f_{k=11}(x)$ weren't included previously, they are only being used here now to help with the accuracy of the curve fitting) 
Each point was plotted in the following manor: $(1, \int_0^1(f_{k=2}(x)-f_{k=1}(x))dx), (2, \int_0^1(f_{k=3}(x)-f_{k=2}(x))dx), \text{etc}.$
The points were found to fit the general relationship of $y_A \approx m_1 x_A^{a_1x_A+b_1}+c_1$ extremely well, have an $R^2$ value very close to $1$. This leads me to believe that this relationship appropriately represents the difference in area between different values of $k$ in $f(x)$. Therefore, because this regression curve approaches $0$ as $x$ approaches $\infty$, the area enclosed by the graph of $(x)=(−\ln⁡(x↑↑(2k)))↑↑(2k+1)$ must approach a constant value as $k \to \infty$.
So, we know it does approach a constant value, we just need to find out what said value is. In other words, we need to evaluate $$\int_0^1 ((−\ln(x↑↑(2k)))↑↑(2k+1))dx \text{ as } k \to \infty$$
 A: By numerical integration, assuming $k$ is a integer, $$\lim_{k\rightarrow\infty}\int_0^1~^{2k+1}\left(-\log\left(~^{2k}x\right)\right)\approx0.57951$$ It is very likely that a closed form isn't possible
More info on the function:
Assuming $k$ is an integer, we see that the graph of $~^{2k+1}\left(-\log\left(~^{2k}x\right)\right)$ have $4$ distinct regions due to the bifurcation:

The location of the regions is outlined below:
We first note that the power tower $~^\infty x$ have a bifurcation for $x<e^{-e}\approx0.066$

with the blue line being $~^{2n}x$ and the yellow being $~^{2n+1}x$
We can obtain a closed form solution for $~^\infty x$ when $x>e^{-e}$, given by Lambert W function:
$$~^\infty x=\frac{W\left(-\log x\right)}{-\log x}$$
from solving $x^z=z$, however afaik there isn't any closed form after the bifurcation, which poses an issue in finding a closed form for the integral.
The intersection of the second and third region is given by $x=e^{-e}$. The intersection between the first and second, third and forth regions, occurs when $\lim_{k\rightarrow\infty}-\log\left(~^{2k}x\right)=e^{-e}$
The solutions to this are the solutions to $$~^{2k}x=e^{-e^{-e}}$$, which are $$x^{e^{-e^{-e}}}=e^{-e^{-e}}$$ and $$x^{x^{e^{-e^{-e}}}}=e^{-e^{-e}}$$
The first solution is given by $e^{-e^{-e+e^{-e}}}\approx0.932$ while the second, given by Mathematica, is $e^{e^{e^{-e}} \left(-e-e^{-e}\right)} \left(-\frac{1}{W_{-1}\left(-e^{-e-e^{-e}}\right)}\right)^{e^{e^{-e}}}\approx0.011$
Mathematica code:
PowerTower[a_, k_Integer] := Power @@ Table[a, {k}];
InfiniteTower[x_?NumericQ] := 
  Piecewise[{{ProductLog[-Log[x]]/-Log[x], 
     E^-E < x}, {a /. FindRoot[x^x^a == a, {a, x}], x < E^-E}}];
Plot[{x^
  InfiniteTower[x], InfiniteTower[x]}, {x, 0, 1}]
Plot[InfiniteTower[-Log[x^InfiniteTower[x]]], {x, 0, 1}]
NIntegrate[InfiniteTower[-Log[x^InfiniteTower[x]]], {x, 0, 1}]
ListPlot[Table[{n, 
   NIntegrate[
    PowerTower[-Log[PowerTower[x, 2 n + 2]], 2 n + 3], {x, 0, 
     1}]}, {n, 1, 50, 1}]]

Note: The $R^2$ value being close to $1$ in this case doesn't quite provide much insight as all the differences are already quite close to $0$ and any function that approximately passes through the first few points and asymptotically reaches $0$ would have a similarly small $R^2$ value as the integral converges rather quickly(after some trails functions of the form $ae^{bn}$ seems to fit the best).

