Asymptotic expansion of the solution to $y^x=x^y$

Question

I was reading this article: https://en.wikipedia.org/wiki/Equation_x%CA%B8_%3D_y%CB%A3 on the solutions to $y^x=x^y$. It is stated that the non-trivial solution tends to $1$ as $$1+\frac{\log{x}}{x}+o\!\left (\frac{\log{x}}{x} \right)$$ when $x\rightarrow \infty$. I was wondering how to prove this, but I haven't found anything about it on the web. Could someone please help? Thanks

Answer 1

The non-trivial solution for $x>e$ is $$ y = \exp \left( { - W_0 \left( {-\frac{{ \log x}}{x}} \right)} \right) $$ where $W_0$ is the principal branch of the Lambert $W$-function. From the Maclaurin expansion of this function, we have $$ \exp \left( { - W_0 (z)} \right) = \exp \left( { - z + z^2 - \cdots } \right) = 1 - z + \frac{3}{2}z^2 + \cdots, $$ for $|z|<\frac{1}{e}$. Take $z=-\frac{{ \log x}}{x}$ and the result follows.