Inverse of $\frac{1-e^{-x}}{x}$ on $(0,1)$

Question

I am trying to invert (or to estimate the inverse of) $$y=\frac{1-e^{-x}}{x}$$ for $y\in(0,1)$. The function 'looks' monotonically decreasing between $x=0$ and $x=\infty$, but I have not been able to show this.

I have been able to compute the inverse function numerically, but I am wondering if there is an analytical solution or approximation that would help speed things up.

Mathematica tells me that the inverse is $$x=\frac{1+y\cdot\text{ProductLog}[-e^{-1/y}/y]}{y}$$ where $\text{ProductLog}[z]$ is the solution to $z=we^w$. I have tried re-arranging the latter expression but I cannot arrive at the original function. Plotting the latter function on $y\in(0,1)$, it looks plausible, but I don't want to use this formula without understanding where it comes from.

Can anyone show me how to invert the original function or help me estimate the inverse to some degree of precision?

Answer 1

Derivation to obtain the Lambert $W$ function : \begin{align} y&=\frac{1-e^{-x}}x\\ x&=\frac{1-e^{-x}}y\\ x\,e^x&=\frac{e^{x}-1}y\\ \left(x-\frac 1y\right)\,e^x&=-\frac 1y\\ \left(x-\frac 1y\right)\,e^{\large{x-\frac 1y}}&=-\frac {\large{e^{-\frac 1y}}}y\\ x-\frac 1y&=W\left(-\frac {\large{e^{-\frac 1y}}}y\right)\\ \end{align} and the wished formula : $\quad\boxed{\displaystyle x=\frac 1y+W\left(-\frac {\large{e^{-\frac 1y}}}y\right)}$

At this point (as indicated by robjohn) we may use the fact that $\;y\in(0,1)\;$ and observe that the parameter of the Lambert-$W$ function $\,-\dfrac 1y\;e^{-\large{\frac 1y}}\;$ will belong to $\;\left(-\dfrac 1e,\;0\right)$.

The implications are :

• for any $\,y\in(0,1)\;$ we have two real solutions from the two branches of the Lambert-$W$ function (see the picture in the Wikipedia link and the discussion about the image of $W$ under or above $-1$ corresponding to the parameter $-\dfrac 1e$) :$$x_1=\frac 1y+W\left(-\frac {\large{e^{-\frac 1y}}}y\right),\;x_2=\frac 1y+W_{-1}\left(-\frac {\large{e^{-\frac 1y}}}y\right)$$
• the parameter of $W$ may be written as $\;u\,e^u\,$ for $u=-\dfrac 1y\;$ but $\;W_{-1}(u\,e^u)=u\;$ in the second case so that $\;x_2=\dfrac 1y-\dfrac 1y\;$ with the solutions becoming simply : $$x_1=\frac 1y+W\left(-\frac {\large{e^{-\frac 1y}}}y\right),\ x_2=0$$ ($x_2=0$ is rather a solution of $\:x\;y=1-e^{-x}\;$ than of the initial equation)