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

Is this even possible?

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    $\begingroup$ You have to? Why? What's the context? $\endgroup$ – Venus Jan 11 '15 at 12:32
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    $\begingroup$ And as we see from the edits, the function grows rapidly ;) $\endgroup$ – Hagen von Eitzen Jan 11 '15 at 12:35
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    $\begingroup$ 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 '15 at 12:57
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    $\begingroup$ We know that the simpler case $e^{e^x}$ doesn't have an elementary antiderivative (see this). $\endgroup$ – Winther Jan 11 '15 at 13:00
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    $\begingroup$ 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 '15 at 13:10

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.


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