Find $\int e^{e^x}dx$

Question

My question is to find $$\int e^{e^x}dx$$ I have tried integration by parts, substitution and simply by inspection, but nothing seems to work! I have been told that it is impossible to integrate, but someone else told me that it is possible.

If it's possible would you mind giving me a hint how to approach this, and if it's impossible, would you mind informing me of it with perhaps an explanation why?

Thank you for your help.

Answer 1

Let $x = \ln u, dx = \frac 1u du$

$\int \frac {e^u}{u} \ du$

This cannot be integrated into elementary functions.

But, a special function exists to describe just these sorts of integrals.

https://en.wikipedia.org/wiki/Exponential_integral

$\int \frac {e^u}{u} \ du = \operatorname{Ei}(u)+ C = \operatorname{Ei}(e^x) + C$

However, if you want to stay away from this exponential integral, it is pretty easy to integrate as a power series.

$e^u = 1 + u + \frac {u^2}{2!} + \cdots\\ \frac {e^u}{u} = \frac 1u + 1 + \frac {u}{2!} + \cdots$

$\int \frac {e^u}{u}\ du = \ln u + C + \sum_\limits{n=1}^{\infty} \frac {u^n}{n(n!)}$

Reversing the substitution:

$x + C + \sum_\limits{n=1}^{\infty} \frac {e^nx}{n(n!)}$

Answer 2

In terms of the exponential integral $\operatorname{Ei}(y):=\int_{-\infty}^y\frac{e^t}{t}dt$, your integral is $\operatorname{Ei}(e^x)+C$ (substitute $t=e^x$).

Answer 3

$$I=\int e^{e^x}dx=\int\sum_{n=0}^\infty\frac{e^{nx}}{n!}dx=\sum_{n=0}^\infty\frac{e^{nx}}{n.n!}+C$$ That is- assuming we can justify interchanging the integral and summation