# Questions concerning the Integration of Integer Tetration

I've been interested in finding the antiderivative of integer tetration, a function defined as iterative exponentiation. Integer tetration is written as $^n$$x where ^1$$x =x$, $^2$$x =x^x, ^3$$x =$ $x^{\scriptscriptstyle x^{x}}$ and so forth where $n=1,2,3\ldots$ (Further info can be found on Wikipedia. One solution to the problem is found on the MathWorld page (see equation #10), but the given solution is difficult to evaluate; accordingly, I am searching for an simpler representation than the result given. (Note that my questions are located at the bottom of this post, and number theorists are encouraged to skip to that point... everything leading up to the questions is background and context.)

Using Wolfram Alpha I found the Puiseux series of $x^x$ to be the following:$$1+ x\log(x)+\frac{1}{2}x^2\log^2(x) + \frac{1}{6}x^3\log^3(x)\ldots$$ $$= \sum_{n=0}^{\infty} \frac{x^n \log^n(x)}{n!} \qquad 1.$$ As such, the antiderivative can be found simply by integrating the series: $$\int x^xdx = \int \bigg(\sum_{n=0}^{\infty} \frac{x^n \log^n(x)}{n!}\bigg)dx = \sum_{n=0}^{\infty}\bigg(\frac{1}{n!}\int x^n \log^n(x)dx\bigg)$$ We then note the following: (again, pulled from Wolfram Alpha) $$\int x^n \log^n(x)dx = \frac{\Gamma(n+1,-(n+1)\log(x))(-n-1)^{-n} }{(n+1)} \qquad 2.$$ All that is left is to substitute in Equation 2 and convert the denominator the gamma function. $$\sum_{n=0}^{\infty}\bigg(\frac{1}{n!}\int x^n \log^n(x)dx\bigg) = \sum_{n=0}^{\infty}\bigg(\frac{\Gamma(n+1,-(n+1)\log(x))(-n-1)^{-n} }{\Gamma(n+2)}\bigg)$$ This solution can be checked by substituting in x=1, yielding the Sophomore's Dream: $$\sum_{n=0}^{\infty}\bigg(\frac{\Gamma(n+1,-(n+1)\log(1))(-n-1)^{-n} }{\Gamma(n+2)}\bigg) = \sum_{n=0}^{\infty}\bigg(\frac{\Gamma(n+1,0)(-n-1)^{-n} }{\Gamma(n+2)}\bigg)$$ $$= \sum_{n=0}^{\infty}\bigg(\frac{\Gamma(n+1)(-n-1)^{-n} }{\Gamma(n+2)}\bigg) = \sum_{n=0}^{\infty}\bigg((-1)^n(n+1)^{n+1}\bigg)$$ $$= 1-\frac{1}{4}+\frac{1}{27}-\frac{1}{64}\,\ldots \, = -\sum_{n=1}^{\infty}(-n)^{-n} = \int_{0}^{1}x^xdx$$ Further, any other definite integral of $x^x$ can be calculated the same way. (Note that $F'(x)$ is $x^x$) $$\int_{0}^{r}x^xdx = F(r) - \lim_{a\to 0}F(a)$$ $$= F(r) - \sum_{n=0}^{\infty}\lim_{a\to 0}\bigg(\frac{\Gamma(n+1,-(n+1)\log(a))(-n-1)^{-n} }{\Gamma(n+2)}\bigg) = F(r)$$

No standard Puiseux series can be found for $^n$$x when n>2, although Wolfram Alpha tells me that "Generalized Puiseux series" exists for these functions. For ^3$$x$ the series is as such: $$x + x^2\log^2(x) + \frac{1}{2}x^3\log^3(x)[1+\log(x)] + \frac{1}{6}x^4\log^4(x)[1+3\log(x)+\log^2(x)]\ldots$$ Notice that this series closely resembles Equation 1, except for the additional powers of $\log(x)$. If we temporarily exclude the coefficients on these additional terms we find that the $n^{th}$ term of the series is multiplied by $\sum_{n=0}^{n-2}\log^n(x)$ (except for the first term). As a result, we can represent the series as follow, where the additional coefficients are represented as $C_{kn}$ $$\sum_{n=2}^{\infty}\bigg[\frac{x^n\log^n(x)}{\Gamma(n)}\sum_{k=0}^{n-2}\bigg(C_{kn}\log^k(x)\bigg)\bigg] = \sum_{n=2}^{\infty}\sum_{k=0}^{n-2}\bigg(\frac{C_{kn}\log^{k+n}(x)x^n}{\Gamma(n)}\bigg) \qquad 3.$$ Listing the coefficients generated, we get $[1],[1,1],[1,3,1],[1,7,6,1]\ldots$
A quick search on the OEIS revealed that these are in fact the Stirling numbers of the second kind, written as $S_n^k$ or $S(n,k)$. Substituting this into the series representation above, we get the following series for $^3$$x:$$x + \sum_{n=2}^{\infty}\sum_{k=0}^{n-2}\bigg(\frac{S_{n-1}^{k+1}\log^{k+n}(x)x^n}{\Gamma(n)}\bigg) = x + \sum_{n=0}^{\infty}\sum_{k=0}^{n}\bigg(\frac{S_{n+1}^{k+1}\log^{k+n+2}(x)x^{n+2}}{\Gamma(n+2)}\bigg)$$Going through a similar process to the one outlined for x^x, \int ^3$$xdx$ can be shown to be the following equation: $$\frac{x^2}{2} + \sum_{n=0}^{\infty}\sum_{k=0}^{n}\bigg(\frac{\Gamma[n+3+k,-(n+3)\log(x)]S_{n+1}^{k+1}}{\Gamma(n+2)(-(n+3))^{n+3+k}}\bigg)$$
Likewise, we find that $$\int_{0}^{r} {^3x}dx = \frac{r^2}{2} + \sum_{n=0}^{\infty}\sum_{k=0}^{n}\bigg(\frac{\Gamma[n+3+k,-(n+3)\log(r)]S_{n+1}^{k+1}}{\Gamma(n+2)(-(n+3))^{n+3+k}}\bigg)$$ The hard part is when we get to the generalized Puiseux series for $^4x$, which is: $$1+x\log(x)+\frac{1}{2}x^2\log^2(x)[1+2\log(x)]+\frac{1}{6}x^3 \log^3(x) [1+9 \log(x)+3\log^2(x)]+\ldots$$ $$***$$ Here is where the real questions start. This series has the first few powers of $x\log(x)$ as $x^x$, namely $1+x\log(x)$. I wonder if this is related to the fact that the function $y = (x^x)^y$ is the same as $y = (^{2n}x)^y$? Equivalently, $y=x^y$ is the same as $y = (^{2n-1}x)^y$. We can see this same symmetry in that for all $^{2n-1}x$ I checked (up to $^7x$) the first two terms are $x+x^2\log^2(x)$, while for all $^{2n}x$ I checked the first two terms are $1+x\log(x)$. Thus, my first question is somewhat broad... Is there any simple explanation for this symmetry?
My next question concerns the series for $^4x$. We can represent this in a very similar form to Equation 1, except for the additional powers of $\log(x)$, reminiscent of the additions to Equation 1. Thus, we get the form $$\sum_{n=0}^{\infty} \bigg[\frac{x^n \log^n(x)}{n!}\sum_{k=0}^{n-1}\bigg(C_{kn}\log^k(x)\bigg)\bigg] = 1 + \sum_{n=1}^{\infty} \bigg[\frac{x^n \log^n(x)}{n!}\sum_{k=0}^{n-1}\bigg(C_{kn}\log^k(x)\bigg)\bigg]$$ $$= 1 + \sum_{n=1}^{\infty}\sum_{k=0}^{n-1}\bigg(\frac{x^n \log^{n+k}(x)C_{kn}}{n!}\bigg) = 1 + \sum_{n=0}^{\infty}\sum_{k=0}^{n}\bigg(\frac{x^{n+1} \log^{n+1+k}(x)C_{kn}}{\Gamma(n+2)}\bigg)$$ Having already worked with $^2x$ and $^3x$ integrating this series would not be difficult, but I can't figure out what the constants are. Thus, my main question is:
What is the pattern for the numbers $[1,2],[1,9,3],[1,28,36,4],[1,75,245,110,5],[1,186,1290,1410,300,6]...$ where each number is some $C_{kn}$? I have noticed that the first member of each set is 1, and the second term in each set is $nS_n^2$ (where the first set is n=2, the second is n=3, and so forth and $S_n^k$ are the Stirling numbers of the second kind). The nth set also contains n members (using the definition of n above), and the final member of each set is n. Other than these observations, I can't find a way to represent these numbers.
My final question is "How do we calculate a generalized Puiseux series, specifically for $^nx$ where $n>2$? I feel that if I had more knowledge on how these series are calculated I could gain insight into my other questions, but I cannot find much literature on finding generalized Puiseux series, let alone literature elementary enough for me to understand and apply to tetration.
(Note: If I should split this post into three seperate posts for each question I can, but I included them all here to avoid posting excessively. If I should edit my tags to reach a wider audience I will do so, although I am not sure what specific fields of mathematics my questions fall under)

• I've few concerns. First, the series $1 - 1/4 + 1/9 -1/16 ...$ seems to me wrong; according to your given definition I'd expect that this is $1-1/4+1/27-1/64+...$. Second, "(...) no puisieux series ... for $\; ^2 x$ when $n>2$ (...)" - how does $n$ enter here the scene? Third for me it was difficult to sort out the points, where the question goes about the series of $\; ^n x$ itself and that of the integrals - perhaps it would be helpful/it would make it less difficult if that could be sorted more explicitely (but perhaps just my own weakness, no offense intended) – Gottfried Helms Sep 21 '15 at 5:33
• One more suggestion: after the $\log(x)$ has a prominent role and occurs everywhere in the formulae, and also its powers (but always as a cofactor of $x$ with the same power, I'd propose to use a simple symbol for it, say $u$ - when I tried to reproduce your formulae in Pari/GP I found this much more readable and in general easier to uncover patterns. – Gottfried Helms Sep 21 '15 at 5:38
• @GottfriedHelms Ah, good catches! Glad to see someone actually read through my work :D I rushed through $x^x$ as it can be found solved elsewhere on the internet in many places, and solving for $^nx$ where $n>2$ has been my field of study... as such it appears I made a few mistakes! I also like the recommendation to simplify the formulas by making a substitution for the $\log(x)$ function... good call. I'll probably re-write this post to be much clearer here in the next few days (too busy in life to re-write it all now... so much detail in there!), and hopefully then I can get some answers... – Brevan Ellefsen Sep 21 '15 at 6:16
• @GottfriedHelms I should note that I suspect Mathematica is calculating the "generalized Puiseux series" by simply expressing $^nx$ as $e$ to some power, and is then applying the Taylor series of $e^x$, although the logarithms don't go away for $^3x$ and higher tetration due to the nature of the derivatives of higher tetrations... Just a thought though, I haven't had a lot of spare time recently to continue work into my studies, hence me posting here to see if others could help me along for a bit. Hopefully this answers one of my three questions though, simplifying my post! – Brevan Ellefsen Sep 21 '15 at 6:20
• An essay seemingly perfectly adressing your question for the coefficients $c$ in the power series of $\;^n x$ might be this one: go.helms-net.de/math/tetdocs/TTetrationExactEntries_short.htm This goes up to $n=3$, unfortunately I've not done up to $n=4$, and I do not yet have a "1-to-1" representation for your question. However, I arrive at all your series by some very systematic matrix-multiplication-scheme hich should explain the pattern by a recursive procedure. – Gottfried Helms Sep 21 '15 at 10:09