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

Question

how to integrate

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

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

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

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
if see this question please explain me why some user vote down or close my question ? is my question wrong ? or are there other reasons? — M.H, Jun 23 '13 at 21:19
Maisam, I don't think your question is wrong . That there is no closed form for the integral should not be considered a reason for the question to be closed ... — Georges Elencwajg, Jun 23 '13 at 21:25
@MaisamHedyelloo Could you tell how do you compute the integral of $x^x$ and what is the result? I can only get this series. — Start wearing purple, Jun 23 '13 at 22:16
@O.L. :if put bound for this integral then we can compute it ? — M.H, Jun 23 '13 at 22:28
@MaisamHedyelloo Don't understand. Again: can you find $\int x^x dx$ in explicit form? (series doesn't count since it is even more difficult to compute than the original integral). If not, then "it is difficult to understand what is asked here". I vote to close. — Start wearing purple, Jun 23 '13 at 22:31
@PeterTamaroff I don't think that mathematicians can really ask what the point of a computation is... Sometimes it's not apparent until years down the road. — apnorton, Jun 23 '13 at 22:43
@O.L. Just because you can't compute something explicitly doesn't mean a question about it is bad. — apnorton, Jun 23 '13 at 22:43
@MaisamHedyelloo At your link I only find the very same series. — Start wearing purple, Jun 23 '13 at 22:49
@MaisamHedyelloo Sorry, I've read "can" instead of "cant". But then I iterate my question. If you know that we cannot integrate $x^x$, what is the point of asking more complicated question containing $x^x$ as a special case? — Start wearing purple, Jun 23 '13 at 22:56
@O.L.:no problem thanks .but you can assume this integral is bounded then solve it — M.H, Jun 23 '13 at 23:00
I don't even understand the expression. What are $n$, $t$, and $h$? — dfeuer, Jun 24 '13 at 00:45
@AlexanderGruber, shouldn't it be $ x^{x^{x^{.\cdot^.}}} $? Using the grouping your second expression is equivalent to (x^x)^{x^{x + 1}} = x^{x^{x^{x+1} + 1}. — Jon Claus, Jun 24 '13 at 02:22
@dfeuer, I think the OP is trying to specify that the total amount of $x$s in the power tower is $n$. It's not a very concise explanation, which is probably why there are so many downvotes. — Jon Claus, Jun 24 '13 at 02:23
If you repeatedly use $e^x$’s Maclaurin series or $\ln^n(x)$ Stirling number expansion, you can get an expansion of $n-1$ sums. It it straightforward enough that someone should have done it elsewhere already. — Тyma Gaidash, Aug 18 '23 at 02:10

Gottfried Helms · Answer 1 · 2013-06-23T23:33:00.843

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 — , Jun 23 '13 at 22:47

score 6 · Answer 2 · answered Oct 22 '13 at 21:06

6

MathWorld has already done for this.

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

answered Oct 22 '13 at 21:06

doraemonpaul

16,178
3
31
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1

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

Тyma Gaidash · Answer 3 · 2021-06-15T14:57:42.720

1

I have found this solution: $$\mathrm{\int{x^{x^{x^…}}dx=\sum_{n=1}^\infty (-1)^nn^{n-2} Q\left(n,-ln(x)\right)+C}}$$

This works as a definite integral for $$\mathrm{e^{-\frac1e}\le x\le e^\frac 1e}$$

Here is my

work

edited Jun 15 '21 at 14:57

answered Jun 15 '21 at 14:24

Тyma Gaidash

12,081

Note: You can get an expansion of $n−1$ sums using the super root series expansion near complex $z=1$ and integral of inverse functions. To be added. Additionally, if we could find the general derivative pattern here, we could extend the convergence radius to include values closer to $z=0$. – Тyma Gaidash Aug 17 '23 at 12:27

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

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