TITLE: Is there a formula for $\int_0^1 x\uparrow\uparrow n dx$
QUESTION [9 upvotes]: $$\int_0^1  x\uparrow\uparrow n dx$$
I was working on an answer to this question on integrating $x^{x^x}$ and I was thinking is there a particular formula/rule/pattern in integration if $x\uparrow\uparrow n$ ($x$ to the power of $x \ n$ times) from $0$ to $1$. Integrating $x^x$ we use the Taylor series to arrive a
$$\int_0^1 x^xdx=\int_0^1 e^{x\ln(x)}dx =\int_0^1 \sum^\infty_{n=0}\frac{x^n\ln^n(x)}{n!}dx=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^n}=0.78343\ldots
$$
similarly integrating $x^{x^x}$ the taylor series would be $$x^{x^x}=\sum^\infty_{n=0}\frac{\ln^n(x)}{n!}x^{x^n}=\sum^\infty_{n=0}\frac{\ln^n(x)}{n!}\Bigg(\sum_{k=0}^\infty\frac{x^k\ln^k(x)}{k!}\Bigg)^n$$
and integrating $x\uparrow\uparrow 3$ the series would be
$$x^{x^{x^x}}=\sum^\infty_{n=0}\frac{\ln^n(x)}{n!}x^{x^{x^n}}=\sum^\infty_{n=0}\frac{\ln^n(x)}{n!}\Bigg(\sum_{k=0}^\infty\frac{\ln^k(x)}{k!}x^{x^k}\Bigg)^n=
\sum^\infty_{n=0}\frac{\ln^n(x)}{n!}\Bigg(\sum_{k=0}^\infty\frac{\ln^k(x)}{k!}\Big(\sum^\infty_{m=0}\frac{x^m\ln^m(x)}{m!}\Big)^k\Bigg)^n
$$
So is there a solution to the integral?
$$\int_0^1  x\uparrow\uparrow n dx$$
Can it be simplified? Can one find series solutions for all $n$? What if $n\not\in \mathbb{N}$?
I have been interested in this function for quite a long time so thank you for your time

REPLY [3 votes]: I cannot give an answer as detailed as @Tyma Gaidash's, but what I can give you is an approximation.
The answer to the integral, $$y(x)=\int_0^1 {}^x t\; \textrm{d}t$$ is approximately $$\dfrac{y_\textrm{sup}(x)+y_\textrm{inf}(x)}{2}+\dfrac{y_\textrm{sup}(x)-y_\textrm{inf}(x)}{2} \cos(\pi x)$$ where $y_\textrm{sup}$ and $y_\textrm{inf}$ are both exponential functions. The exact details are in the following graph, which has an $R^2$ of ~$0.9977$:
https://www.desmos.com/calculator/e56wdxumxv