how to integrate

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

Thanks in advance.

$\color{green }{\text{my attempt}}$ : 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 $$\int _0^1x^xdx=\sum_{k=0}^\infty \frac{(-1)^k}{k!}\frac{1}{(k+1)^k}\int_0^{\infty} e^tt^k\,dt =\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!}\frac{1}{(k+1)^k}\Gamma(k+1)=\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)^{k+1}} = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^n}$$

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

  • 28
    $\begingroup$ Can you integrate $x^x$? $\endgroup$ – Mariano Suárez-Álvarez Jun 23 '13 at 20:19
  • 2
    $\begingroup$ 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.) $\endgroup$ – Alexander Gruber Jun 23 '13 at 20:35
  • 15
    $\begingroup$ 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. $\endgroup$ – apnorton Jun 23 '13 at 20:39
  • 1
    $\begingroup$ @ Mariano Suárez-Alvarez:yes i can integrate $x^x$ $\endgroup$ – M.H Jun 23 '13 at 21:06
  • 3
    $\begingroup$ @MaisamHedyelloo, ok. Can you do the next one? :-) $\endgroup$ – 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 $x$)

%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)

  • 1
    $\begingroup$ 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 $\endgroup$ – 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.

  • 1
    $\begingroup$ Nice find! Wolfram never fails to surprise me. $\endgroup$ – Simply Beautiful Art Dec 23 '16 at 20:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.