Conditions for the inverse of a bijection to be "almost explicit" The following map
$$f:\left[-1,+\infty\right[\to\left[-\frac1e,+\infty\right[,\,x\mapsto x\,e^x$$is a bijection and it can be shown that its inverse (known as the Lambert function W) is representable as a power series on $\left[-\frac1e,\frac1e\right]$ :
$$\forall x\in\left[-\frac1e,\frac1e\right],f^{-1}(x)=\sum_{n=1}^\infty\frac{(-n)^{n-1}}{n!}x^n$$
My question (which is rather a fuzzy one) :
What kind of sufficient conditions on a bijection $u:I\to J$ (where $I,J$ are intervals) imply the possibility for its inverse $u^{-1}$ to be representable (at least on some sub-interval of $J$) as a series (or an integral) involving "elementary functions" ?
For example, is it possible to do something similar with the following bijections ?

*

*$\mathbb{R}\to\mathbb{R},x\mapsto x+e^x$

*$\left]0,+\infty\right[\to\mathbb{R},x\mapsto x+\ln(x)$

*$\left]0,+\infty\right[\to\mathbb{R},x\mapsto e^x\ln(x)$
 A: The best generalization that I know is essentially this theorem:
Theorem. If $f(x) = a_0 + a_1 x + a_2 x^2 + \cdots$ is a formal power series with $a_0 = 0$ and $a_1 \ne 0$, then $f$ has an inverse formal power series $g$ (i.e. $g(f(x)) = f(g(x)) = x$).
I've stated it in terms of formal power series rather than convergent power series, but that's no obstacle. If $f$ is analytic near $0$, then by the inverse function theorem, locally it has an analytic inverse $g$.
To find the coefficients of $g(y) = b_0 + b_1 y + b_2 y^2 + \cdots $, you can simply write
$$
  b_0 + b_1(a_1 x + a_2 x^2 + \cdots) + b_2(a_1 x + \cdots)^2 + \cdots = x
$$
and solve for $b_0$, $b_1$, $b_2$, etc. Each coefficient of $g$ only depends on finitely many of the coefficients of $f$. The computations involved are merely algebra: The algorithm to find any $b_j$ does not require us to sum any infinite series or take any limits.
So, for example, this is exactly how you can invert $f(x) =x e^x$.
As for your other examples, the theorem does not directly apply because it has the condition that $a_0 = 0$, but we can get around that. For example, with $f(x) = x + e^x$, $f(0)$ is $1$ instead of $0$. But our method basically still works: the difference is that we will get a $g$ of the form $\sum_j b_j (y-1)^j$ instead of $g(y) = \sum_j b_j y^j$. This is still a series of suitably "elementary" terms which are easy to compute.
(For some references, see e.g. Wikipedia on formal power series, or Wilf, generatingfunctionology, Chapter 2, available on the author's website.)
