Solution to $xe^{e^x}$

Question

The problem $xe^{e^x}=e$ came up another day and I wondered if it were solvable.

My attempt was the following substitution,$$x=W(u)$$$$W(u)e^{e^{W(u)}}=e$$Where I used a Lambert W identity to get $$W(u)e^{\frac u{W(u)}}=e$$And attempted to solve.

I got this far:$$-\frac1{W(u)}e^{-\frac u{W(u)}}=-\frac1e$$However, I couldn't continue. Its so darn close... but alas, its not quite there.

So I've come to you guys for help, well knowing that most of you will say "no solution" of closed form, but that's ok. I will even accept answers that attempt to continue or start from the beginning to try out a different path for the solution.

I also considered the following: If you could get into a form where adding/subtracting/multiplying/dividing/whatever will cancel a part of the equation with a previous form, then that'll be great e.g.$$f(u)W(u)=g(u)$$$$xf(u)=W(u)$$Divide the two and you get $$\frac{W(u)}x=\frac{g(u)}{W(u)}$$$$W^2(u)=xg(u)$$Noting that you can switch around with substitutions as long as you are consistent with your substitutions.

I also realize that in the beginning, I could have used the substitution $x=-W(x)$ to flip the resulting fraction to put the $W$ on top and repeat the Lambert identity process, only to create$$-W(u)[\frac{u}{W(u)}]^{1/u}=e$$ Which is still not solvable!

Answer 1

Is it solvable means two different things :

A solution exists and it can be calculated numerically, so in this sense it is solvable.

In terms of special functions I would say that if any solution exists it is always possible to express it as some expression involving special functions. It is a question of how much effort and what special function you want to employ. The question is whether that effort is useful to you or not.

Answer 2

$$xe^{e^x}=e$$

Your equation is an equation of elementary functions. It's an algebraic equation in dependence of $x$ and $e^{e^x}$. Because the terms $x,e^{e^x}$ are algebraically independent, we don't know how to rearrange the equation for $x$ by only elementary operations (means elementary functions).
I don't know if the equation has solutions in the elementary numbers.

Your equation cannot be solved in terms of Lambert W but in terms of Hyper Lambert W:

$$G(1;x)=e$$ $$x=HW(1;e)$$

So we have a closed form for $x$, and the representations of Hyper Lambert W give some hints for calculating $x$.

Galidakis, I. N.: On solving the p-th complex auxiliary equation $f^{(p)}(z)=z$. Complex Variables 50 (2005) (13) 977-997

Galidakis, I. N.: On some applications of the generalized hyper-Lambert functions. Complex Variables and Elliptic Equations 52 (2007) (12) 1101-1119

Answer 3

Here is how to find an exact form. Substitute $w=ex$

$$xe^{e^x}=e\iff w=e^{-e^{ew}}$$

Now use:

Convert $\displaystyle\frac1b\sum_{n=1}^\infty\frac{(b e^a)^n}{n!}\operatorname B_{n-1}(an)$ to integral using $\displaystyle\operatorname B_n(x)=\frac{n!}{2\pi i}\oint\frac{e^{x(e^t-1)}}{t^{n+1}}dt$:

which solves $e^{ae^{bx}}=x$. Unfortunately, the series expansions diverge, but an integral representation does work by letting $a=-1,b=e$. Therefore:

$$\bbox[2px,border:3px groove blue]{ \begin{align} xe^{e^x}=e \implies x &=-\frac1{2\pi}\int_{-\pi}^\pi e^{it}\ln\left(1-e^{1-it-e^{e^{it}}}\right)dt\\ &=\frac1{2\pi}\int_{-\pi }^\pi\frac{e^{e^{it}+it}+1}{e^{e^{e^{it}}-1}-e^{-i t}}dt\\ &=\textstyle{\frac e\pi\int_0^\pi \frac{e^{e^{\cos(t)}\cos(\sin(t))}(\cos(e^{\cos(t)}\sin(\sin(t)))+e^{\cos(t)}\cos(t+\sin(t)-e^{\cos(t)}\sin(\sin(t)))-e(\cos(t)+e^{\cos(t)}\cos(2t+\sin(t))))}{e^2+e^{2e^{\cos(t)}\cos(\sin(t))}-2e^{e^{\cos(t)}\cos(\sin(t))+1}\cos(t+e^{\cos(t)}\sin(\sin(t)))}dt} \end{align}}$$

shown here.FullSimplify[ComplexExpand[Re[(E^(I t) (1 + E^(I t + E^(I t))))/(2 (-1 + E^(-1 + I t + E^E^(I t))) Pi)]]] was used to expand the integrand in Mathematica.