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?

  • $\begingroup$ First you have to make sense of the function $x^{x^{x...}}$. What exactly would convergence mean in this case? $\endgroup$ – Gregory Grant Feb 8 '16 at 18:29
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    $\begingroup$ There's a rigorous theory of infinite sums and of infinite products. You'd need something similar for infinite exponentiation before you can even talk about integrating it. $\endgroup$ – Gregory Grant Feb 8 '16 at 18:29
  • $\begingroup$ I have a feeling your function is identically $1$ wherever it is defined. $\endgroup$ – Gregory Grant Feb 8 '16 at 18:30
  • $\begingroup$ That is a well-defined function whose domain (where it converges) has to do with $e$. I'll see if I can find a link. @GregoryGrant $\endgroup$ – Akiva Weinberger Feb 8 '16 at 18:33
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    $\begingroup$ @GregoryGrant It converges for $e^{-e}\le x\le e^{1/e}$. Wikipedia, Relevant MathWorld $\endgroup$ – Akiva Weinberger Feb 8 '16 at 18:38

$$\large \int_{a}^{b} x^{x^{x^{x\,\cdots}}} \, dx = - \int_{a}^{b} \frac{W\left(-\ln(x) \right)}{\ln(x)}\, dx $$ 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


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.


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