$$\int_0^1 x\uparrow\uparrow n dx$$ 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


There is no generally accepted definition for $^nx$ when $n\not\in\mathbb{N}$. So I am not going to talk about it.

From Wolfram Alpha we have that,



$$\textstyle\displaystyle{ a_{n,k} = \begin{cases} 1, & k = 0 \\ \frac{1}{n!}, & n = 1 \\ \frac{1}{k}\sum_{j=1}^{n}ja_{n,k-j}a_{n-1,k-1}, & \text{otherwise} \end{cases}} $$

So, $\textstyle\displaystyle{\int_{0}^{1}{^nx}dx}$




If you want want an explicit formula for $a_{n,k}$ Then forgive me I don't know. I don't even know how is the series representation of $^nx$ derived. So yeah


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 ~$0.9977$:



I do not have a solution, but I do have an algorithm for finding such solutions using simple Taylor Series of an $n\in\Bbb N$ height power tower at $x=1$. These power towers usually have rational coefficients. If you do the Taylor series about another point, then you get a polynomial of natural logarithms. Here are a few examples.

$$\int x^x dx$$ using the Lehmer-Comtet constants.

$$\int x^{x^x}dx$$ using A179230 and the Exponential Generating function of $$(x+1)^{(x+1)^{x+1}}$$

$$\int x^{x^{x^x}}dx$$







$n\ge 11?$

Inverse of $n=\infty$ case

Note that $x^\frac{1}{x}$ is the inverse of $x\uparrow\uparrow\infty=\lim_{n\to\infty} \,n\text{ times}\big\{x^{x^{x^…}}$ for the Infinite Power Tower convergence interval and not the analytic continuation.

Here is the general solution and one assuming the interval of convergence works: $$\int x\uparrow\uparrow n dx =\int\sum_{k=0}^\infty\frac{\left[\frac{d^k}{dx^k} (x\uparrow\uparrow n)\right]_{x=1}\ (x-1)^k}{k!} dx= \sum_{k=1}^\infty\frac{\left[\frac{d^{k-1}}{dx^{k-1}} (x\uparrow\uparrow n)\right]_{x=1}\ (x-1)^k}{k!}\mathop=^\text{definite integral}_\text{from 0 to 1} -\sum_{k=1}^\infty\frac{\left[\frac{d^{k-1}}{dx^{k-1}} (x\uparrow\uparrow n)\right]_{x=1}\ (-1)^k}{k!} $$

If you know the convergence intervals, I would love to really know so that we could apply the same to similar problems.

I suggest using @Arjun’s suggestion for the non-natural number tetration. Please correct me and give me feedback!


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