Inverse Function of $(1-e^{-t})t$

Question

I'm interested if there is some hope to obtain the inverse of $$ f(t) = (1-e^{-t})t $$ for $t$ positive. If there is a formula I suspect that the Lambert W function will be involved on it. Clever approximations are also welcomed (not looking for local ones though ;).

Answer 1

For large $t$ $f(t)\sim t$, while for small t $f(t)\sim t^2$. Then for large $y$ $f^{-1}(y)\sim y$ and for small $y$ $f^{-1}(y)\sim \sqrt y$. Using this as a first approximation in Newton's method gives the following approximation: $$ f^{-1}(y)\sim h(y)=\begin{cases} \sqrt{y}-\frac{-y-e^{-\sqrt{y}} \sqrt{y}+\sqrt{y}}{e^{-\sqrt{y}} \sqrt{y}-e^{-\sqrt{y}}+1}& 0<y<1,\\ y+\frac{e^{-y} y}{e^{-y} y-e^{-y}+1}& y\ge1. \end{cases} $$ This is the graph of $f^{-1}(y)-h(y)$:

Answer 2

Maple does give us a series for it: if $(1-e^{-t})t = y$ and $t>0$, then $$ t = {y}^{1/2}+\frac{1}{4}\,y+{\frac {7}{96}}\,{y}^{3/2}+1/48\,{y}^{2}+{\frac {491} {92160}}\,{y}^{5/2}+{\frac {1}{960}}\,{y}^{3}+{\frac {983}{20643840}} \,{y}^{7/2}-{\frac {11}{120960}}\,{y}^{4}-{\frac {2455961}{39636172800 }}\,{y}^{9/2}-{\frac {79}{2903040}}\,{y}^{5}-{\frac {1131731179}{ 125567395430400}}\,{y}^{11/2}+\dots $$