# Is there a formula for $\int_0^1 x\uparrow\uparrow n dx$

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

• What is the accepted definition for fractional tetration? Also possible duplicate of How to integrate $\int\underbrace{x^{x^{\cdot^{\cdot^x}}}}_ndx$ Sep 12 at 15:49
• I would suggest visiting this link from MathOverflow: mathoverflow.net/questions/259278/… Sep 14 at 2:21
• @ArjunVyavaharkar How could you integrate fractional tetration as this would be fascinating for users like us? Thanks. Sep 15 at 1:50
• @hwood Thanks for the bounty. Wouldn’t others also deserve it and the accept? Sep 19 at 21:04

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,

$$\textstyle\displaystyle{^nx=\sum_{k=0}^{n}\frac{(k+1)^k}{(k+1)!}\ln^k(x)+\sum_{k=n+1}^{\infty}a_{n,k}\ln^k(x)}$$

Where,

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

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

$$=\textstyle\displaystyle{\int_{0}^{1}\left(\sum_{k=0}^{n}\frac{(k+1)^k}{(k+1)!}\ln^k(x)+\sum_{k=n+1}^{\infty}a_{n,k}\ln^k(x)\right)dx}$$

$$\textstyle\displaystyle{=\sum_{k=0}^{n}\frac{(k+1)^k}{(k+1)!}\int_{0}^{1}\ln^k(x)dx+\sum_{k=n+1}^{\infty}a_{n,k}\int_{0}^{1}\ln^k(x)dx}$$

$$\textstyle\displaystyle{=\sum_{k=1}^{n+1}\frac{k^{k-2}}{k!}+\sum_{k=n+1}^{\infty}\frac{a_{n,k}}{k+1}}$$

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 ~$$0.9977$$:

https://www.desmos.com/calculator/e56wdxumxv

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.

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

$$n=5$$

$$n=6$$

$$n=7$$

$$n=8$$

$$n=9$$

$$n=10$$

$$n\ge 11?$$

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!