My result for this integral is as follows: $$\int_{1}^{e^{\frac{1}{e}}} x^{x^{x^{....}}} = (e^{\frac{1}{e}})e - e - \frac{1}{2} - \sum_{k=1}^{\infty} \left( \frac{\gamma((k+2),(k))}{{k}^{(k+2)}\Gamma(k+2)} \right)$$
If anyone is curious about how I got this result, I did a writeup in LaTeX: https://www.overleaf.com/read/krbpcdytgdqj
I've been trying to solve this integral for a long time, and I finally managed to do it. However, the answer is still really bulky. First, is this answer correct, and secondly, is there some way to rewrite the solution in a more elegant/compact way?
To make it a bit shorter, why not to use $$I=\int x^{x^{x^{x^{…}}}}\, dx=-\int \frac{W(-\log (x))}{\log (x)}\,dx$$ Now let $x=e^{-t}$ $$I=-\int \frac{W(t)}{t\, e^t}\,dt$$ Now, using the infinite expansion $$W(t)=\sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!}t^n $$ $$I=-\sum_{n=1}^\infty \int \frac{(-n)^{n-1} }{n!}\,e^{-t}\, t^{n-1}\,dt=\sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!} \Gamma (n,t)=\sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!} \Gamma (n,-\log(x))$$ $$\int_{1}^{e^{\frac{1}{e}}} x^{x^{x^{x^{…}}}}\, dx=\sum_{n=1}^\infty (-n)^{n-2} \left(1-\frac{\Gamma \left(n,-\frac{1}{e}\right)}{\Gamma (n)}\right)$$
The numerical result is $$\int_{1}^{e^{\frac{1}{e}}} x^{x^{x^{x^{…}}}}\, dx=0.6186854900626491368967505548225992845913259921723$$