Integrate $x$ to the power $x$... to the power $x$... infinitely

Question

This came across my mind, integrating $x$ to the power $x$ infinitely, I couldn't find anything on it.

$$\Large \int x^{x^{x^{x\,\cdots}}} \, dx$$

How would you go about this?

Answer 1

$$\large \int_{a}^{b} x^{x^{x^{x\,\cdots}}} \, dx = - \int_{a}^{b} \frac{W\left(-\ln(x) \right)}{\ln(x)}\, dx $$ Where W is the Lambert W function.

The integral is convergent on $\quad e^{-e}\leq a\leq e^{1/e} \quad \text{and} \quad e^{-e}\leq b\leq e^{1/e}$

There is no closed form with a finite number of standard functions.

Example of serie expansion aroud $\quad x\sim 1 \quad$ : $$\int x^{x^{x^{x\,\cdots}}} \, dx \sim \ln(x)+\ln^2(x)+\ln^3(x)+\frac{29}{24}\ln^4(x)+\frac{53}{30}\ln^5(x)+O\left( \ln^6(x) \right)+\text{constant}$$

This integral can be seen on page 12, Eq.(12:6) in the paper : https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function

Answer 2

You cannot have a closed form but you can have this nice expansion

$$x^{x^{x^{x\,\cdots}}}=-\frac{W(-\ln(x))}{\ln(x)}=\sum\limits_{n=0}^{+\infty} \frac{(n+1)^n}{(n+1)!}\ln^{n}(x)=1+\ln(x)+\frac{3^2}{3!}\ln^2(x)+\frac{4^3}{4!}\ln^3(x)+...$$

Knoebel, 1981, Exponentials reiterated

Eisenstein, 1844, Entwicklung von $a^{a^{a^{.^{.^{.}}}}}$

This gives:

$$\int \sum\limits_{n=0}^{+\infty} \frac{(n+1)^n}{(n+1)!}\ln^{n}(x) \mathrm{d}x = \sum\limits_{n=0}^{+\infty} \frac{(-1)^n(n+1)^n}{(n+1)!} \Gamma (n+1, -\log(x))$$

where $\Gamma (m, x)$ is the incomplete gamma function.

Observe that you can use this expansion in this form only

$$\int_{a}^{b} x^{x^{x^{x\,\cdots}}}=\sum\limits_{n=0}^{+\infty} \frac{(-1)^n(n+1)^n}{(n+1)!} (\Gamma (n+1, -\log(a))-\Gamma (n+1, -\log(b)))$$

where $e^{-\frac{1}{e}}<a,b<e^{\frac{1}{e}}$ since this is the region of convergence for the series and the incomplete gamma must be paired. So it does not cover the entire region where $-\frac{W(-\ln(x))}{\ln(x)}$ is defined.