How to find the inverse of $\frac{1}{x^2}e^x$

Question

From what I know the function $$f(x)=xe^{x}$$ doesn't have an inverse function $f^{-1}$ that can be written in a closed form. However his inverse function $f^{-1}$ does exists and it's "famous": it's the Lambert W function. Wonderful, but then what about finding the inverse function of functions similar to $f$? For example: in the case of the function $$g(x)=\frac{1}{x}e^x$$ turns out that its inverse can be written using the $W$ function: $$g^{-1}(x)=-W\left(-\frac{1}{x}\right)$$ But what about other similar functions? In particular I am interested in the function: $$h(x)=\frac{1}{x^2}e^x$$ what is the inverse of this function? Can we express it via $W$ as we have done for $g(x)$?

Answer 1

First, let's rewrite the expression: $$ \begin{align} h &= \frac{1}{x^2}e^x \\ \frac{1}{h} &= x^2e^{-x} \\ \frac{1}{\sqrt h} &= \pm x e^{-x/2} \\ \pm\frac{1}{2\sqrt h} &= -\frac{x}{2} e^{-x/2} \\ \end{align} $$

For $k\in\mathbb Z$ define $W_k\colon\mathbb C\to\mathbb C$ so that $z=W_k(z)e^{W_k(z)}$. Now apply the Lambert function on both sides:

$$ \begin{align} W_k(\pm\frac{1}{2\sqrt h}) &= W_k(-\frac{x}{2} e^{-x/2})=-\frac{x}{2} \\ \end{align} \Longrightarrow x=-2W_k(\pm\frac{1}{2\sqrt h}). $$

Answer 2

If $x>0$ and $y=e^x/x^2$, then $(-x/2)e^{-x/2}=-y^{-1/2}/2$, leading to an expression using Lambert's $W$. The case $x<0$ is handled similarly.