# how to integrate $\int\underbrace{x^{x^{\cdot^{\cdot^x}}}}_ndx$

how to integrate

$$\int\underbrace{x^{x^{\cdot^{\cdot^x}}}}_ndx$$ $\color{red}{\text{or how to calculate this integral when its bounded}}$
$$\color{red}{\int_0^1\underbrace{x^{x^{\cdot^{\cdot^x}}}}_ndx}$$

$\color{green }{\text{my attempt}}$ : its easy to integrate $\int x^xdx$ $$\int{x^xdx} = \int{e^{\log x^x}dx} = \int{\sum_{k=1}^{\infty}\frac{x^k\log^k x}{k!}}dx= \sum_{k=0}^\infty \frac{1}{k!}\int x^k(\log x)^k\,dx \Rightarrow$$ substitute ${u = -\log x}$ then $$\int x^xdx=\sum_{k=0}^\infty \frac{(-1)^k}{k!}\int e^{u(k+1)}u^k\,du=\sum_{k=0}^\infty \frac{(-1)^k}{k!}\frac{1}{(k+1)^k}\int e^{u(k+1)}[(k+1)u]^k\,du.$$ Ii substitute $t = (k+1)u$ and $$\sum_{k=0}^\infty \frac{(-1)^k}{k!}\frac{1}{(k+1)^k}\int e^tt^k\,dt$$ if i put bound for this integral we have $$\int _0^1x^xdx=\sum_{k=0}^\infty \frac{(-1)^k}{k!}\frac{1}{(k+1)^k}\int_0^{\infty} e^tt^k\,dt =\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!}\frac{1}{(k+1)^k}\Gamma(k+1)=\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)^{k+1}} = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^n}$$

$$\int_0^1\underbrace{x^{x^{\cdot^{\cdot^{x}}}}}_ndx=\int_0^1e^{\log\underbrace{x^{x^{\cdot^{\cdot^{x}}}}}_n}dx=\sum_{k=0}^\infty\frac{1}{k!}\int_0^1\biggl(\underbrace{x^{x^{\cdot^{\cdot^{x}}}}}_{n-1}\biggr)^k(\log x)^k~dx$$

• Can you integrate $x^x$? – Mariano Suárez-Álvarez Jun 23 '13 at 20:19
• Do you mean $((((x^x)^x)^x)^x)^x$ etc., as opposed to, say, $(x^x)^{(x^x)^{x^x}}$? (One must stipulate this sort of thing when it comes to exponentiation, it's not associative.) – Alexander Gruber Jun 23 '13 at 20:35
• Why so many downvotes? This is a legitimate question, especially if someone does not know that the expression in question doesn't have an elementary anti-derivative. – apnorton Jun 23 '13 at 20:39
• @ Mariano Suárez-Alvarez:yes i can integrate $x^x$ – M.H Jun 23 '13 at 21:06
• @MaisamHedyelloo, ok. Can you do the next one? :-) – Mariano Suárez-Álvarez Jun 23 '13 at 21:26

If you replace $x$ by $x+1$ then you have $\int (x+1)^{(x+1)^{...^{(x+1)}}} dx$ and the exponential-tower has an interesting power series, whose coefficients at the leading terms become constant in spite of increasing height $h$ (that $n$ in your formula) . Then you can integrate termwise to have a power series for the integral.
For instance, the tower of iteration height $h=6$ has the power series $$1 + x + x^2 + 3/2 \cdot x^3 + 7/3 \cdot x^4 + 4 \cdot x^5 + 283/40 \cdot x^6 + 4321/360 \cdot x^7 + O(x^8)$$ where the first terms up to $4 \cdot x^5$ stay constant for all higher iterations/exponential towers. (I do not know the range of convergence at the moment, maybe it is $\small \eta-1 \approx 0.4446...$ where $\small \eta = \exp(\exp(-1))$ because of the range of convergence for the exponential-tower of infinite height).

The termwise integration gives $$x + 1/2 \cdot x^2 + 1/3 \cdot x^3 + 3/8 \cdot x^4 + 7/15 \cdot x^5 + 2/3 \cdot x^6 + 283/280 \cdot x^7 + O(x^8)$$ for the indefinite integral of the height $h=6$ exponential tower.

I get, using Pari/GP by the "explicite" integration (and substitution $x+1$ for $x$)

intnum(x=-1,0,(1+x)^(1+x)^(1+x)^(1+x)^(1+x)^(1+x))
%379 = 0.710658941398


which should be the correct value wrt to truncation to the shown digits. The use of the power series gives the value $0.710452400137$ which is inaccurate from the fourth digit (although I also applied Euler-summation for the diverging terms which have alternating signs), so the power series should be in principle usable also for higher exponential towers for small integration bounds and a more improved summation-procedure.

Unfortunately, there's a replacement $x \to (x+1)$ inside the integral and I do not know whether this is legitimate (I'm nearly illiterate with integration)

• Sure, the substitution $x \to x+1$ is totally legitimate. $\int_a^b f(x)\,\mathrm dx = \int_{a-1}^{b-1}f(x+1)\,\mathrm dx$. More generally: integration by subsitution – Rahul Jun 23 '13 at 22:47

MathWorld has already done for this.

Please see the formula (10) in http://mathworld.wolfram.com/PowerTower.html.

• Nice find! Wolfram never fails to surprise me. – Simply Beautiful Art Dec 23 '16 at 20:40