Integrate : $\int e^{x+e^{x+e^x}}\, dx$

Question

Question : $$\int e^{x+e^{x+e^x}} dx$$

source : Integration Techniques and Tricks University of Miami Mathematics Union Trevor Birenbaum

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My attempt:

can be rewritten as $$\int e^x (e^{e^x})^{(e^{e^x})} dx$$

let $z=e^{e^x} \implies dz = e^{e^x} e^x dx$

so the integral transforms to $$\int \frac{z^z}{z}dz$$

now is it even possible to solve this

Answer 1

NOTE: The answer has been improved using Тyma's suggestion in the comment section.

First, a couple of observations. The function $z^z/z$ seems to be continuous. I checked it using the Desmos graphing calculator. The picture is given below:

Second, the integral $\int z^z$ dz is called the Bernoulli integral and seems similar to your problem. I recommend watching this YT video that solves it because I used a similar approach. With that in mind, here is how I solved the problem.

Let us first reformulate your integral:

$$I=\int{z^z\over z}dz=\int{z^{-1}z^z}dz \tag 1$$

Keeping in mind that $z^{z}=e^{\ln(z^{z})}=e^{{z}\ln(z)}$, we write:

$$I = \int z^{-1}e^{{z}\ln(z)}dz \tag 2$$

The function $e^x$ has a nice Taylor expansion:

$$e^x = \sum_{n=0}^\infty { x^n\over n!} \tag 3$$

where $n \in \mathbb{N}_0$. This means we can reformulate the integral as:

$$I = \int z^{-1}\sum_{n=0}^\infty { \big({{z}\ln(z)}\big)^n\over n!} dz \tag 4$$

The sum and the factorial do not have to be inside the integral since the integrated function is continuous:

$$I = \sum_{n=0}^\infty {1\over n!} \int z^{-1}{ {{z}^n \ln(z)}^n} dz \tag 5$$

We can rewrite this as:

$$I = \sum_{n=0}^\infty {1\over n!} \int { {{z}^{n-1} \ln(z)}^n} dz \tag 6$$

Wolfram solves this. The result is:

$$I = \sum_{n=0}^\infty {1\over n!} {\ln(z)^n \big( -n \ln(z) \big)^{-n} \Gamma \big( n+1,-n \ln(z) \big) \over n} + const \tag 7$$

where $\Gamma (x,y)$ is the incomplete gamma function. Resubstituting $z=e^{e^x}$ we get:

$$I = \sum_{n=0}^\infty {1\over n!} {e^{nx} \big( -n e^x \big)^{-n} \Gamma \big( n+1,-n e^x \big) \over n} + const \tag 8$$

I think this is as far as we can get with solving this integral. Note that I didn't check if the series converges.