13
$\begingroup$

I have to calculate integral $$\Large \int_0^1 e^{e^{e^x}} dx$$

Is this even possible?

$\endgroup$
17
  • 3
    $\begingroup$ You have to? Why? What's the context? $\endgroup$
    – Venus
    Jan 11, 2015 at 12:32
  • 18
    $\begingroup$ And as we see from the edits, the function grows rapidly ;) $\endgroup$ Jan 11, 2015 at 12:35
  • 3
    $\begingroup$ It depends on whether you interpret $\Large a^{b^{c^d}}$ as $((a^b)^c)^d$ or $a^{(b^{(c^d)})}.$ In one interpretation, you're just integrating an exponential function with a weird looking base. In the other interpretation, I don't think it's humanly possible. I tried it in two different computer programs, and each used a different interpretation. $\endgroup$
    – Randy E
    Jan 11, 2015 at 12:57
  • 1
    $\begingroup$ We know that the simpler case $e^{e^x}$ doesn't have an elementary antiderivative (see this). $\endgroup$
    – Winther
    Jan 11, 2015 at 13:00
  • 4
    $\begingroup$ If you are interested in trying to prove that no antiderivative exist you can try to apply Risch algorithm. See also this MSE question. $\endgroup$
    – Winther
    Jan 11, 2015 at 13:10

1 Answer 1

11
$\begingroup$

Take $u = e^{e^x}$. Thus, $ du = ue^x dx \implies dx = \frac{1}{u \text{log}(u)}du.$ Hence,

$$\int_{0}^{1} e^{e^{e^x}}\,dx = \int_{e}^{e^e} \frac{e^{u}}{u \text{log}(u) }\,du. $$By expanding in Taylor

$$\frac{e^{u}}{u \text{log}(u)}= \frac{1}{u\text{log}(u)}+ \frac{1}{\text{log}(u)}+\frac{u}{2!\cdot\text{log}(u)} + \frac{u^2}{3! \cdot\text{log}(u)}+ \mathcal{O}(u^3).$$ Hence,

$$\int \frac{e^{u}}{u \text{log}(u)}\,du = C + \text{log}\left(\text{log}(u)\right) +\sum \limits_{n = 0}^{\infty}\frac{1}{(n+1)!}\cdot\text{Ei}\left((n+1)\text{log}(u)\right),$$ where $\text{Ei}$ is the exponetial integral given by $$ \text{Ei}(x) = -\int_{-x}^{\infty}\frac{e^{-t}}{t}\,dt,$$ which is not an elementary function according to Risch algorithm.

$\endgroup$
5
  • $\begingroup$ This post was vandalised; I rolled it back $\endgroup$ Oct 26, 2020 at 11:24
  • $\begingroup$ @preferred_anon, thanks! $\endgroup$
    – Alex Silva
    Oct 26, 2020 at 11:31
  • $\begingroup$ Going back to $x$ makes it a little more digestible $$\int e^{e^{e^x}} \, dx=\sum _{n=0}^{\infty } \frac{\text{Ei}\left((n+1) e^x\right)}{(n+1)!}+x+C$$ $\endgroup$
    – Raffaele
    Oct 26, 2020 at 11:45
  • $\begingroup$ @AlexSilva Why did you approve the edit? $\endgroup$ Mar 28, 2023 at 18:04
  • $\begingroup$ Don't remember exactly. Long time ago. I think the answer had been vandalized. $\endgroup$
    – Alex Silva
    Mar 31, 2023 at 12:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .