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)
