Calculate integral $\int_0^1 e^{e^{e^x}} dx$

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

Is this even possible?

Solution 1:

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.