Inverse of $f(x) = xe^x-x$

Question

I'm wondering if there is a way to obtain the inverse of the function $y=xe^x-x$. I am aware of the use of Lambert's W function in the inverse of $xe^x$ but as can be seen this is a different animal altogether. I have plotted the function and the inverse looks close to the W function.

This is related to a problem I'm working on in which I need to find the density of the random variable $Y=Xe^X-X$, where I know the density of X. I am aware that this function is not one-to-one. For my purposes, we can restrict to $x>0$. Any help would be greatly appreciated!!

Answer 1

This equation can be placed in the form:

$$e^x=\frac{x+f}{x}$$

We know that mixed exponential/bilinear equations can be solved by the extended Lambert function $W_r(x)$, which can be represented by the following Lagrange inverting series:

$$z(A,t,s)=t- (t-s) \sum_{n=1} \frac{L_n' (n(t-s))}{n} e^{-nt} A^n$$

which is the solution of:

$$e^z=A\frac{z-t}{z-s}$$

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References

On the generalization of the Lambert $W$ function with applications in theoretical physics, http://arxiv.org/abs/1408.3999

[68] C. E. Siewert and E. E. Burniston, "Solutions of the Equation $ze^z=a(z+b)$," Journal of Mathematical Analysis and Applications, 46 (1974) 329-337. http://www4.ncsu.edu/~ces/pdfversions/68.pdf

Answer 2

Beside the formal answer in terms of the generalized Lambert function, we can obtain a good approximation of the inverse at the price of a single iteration of Halley method using $x_0=W(y)$. This gives $$x_1=W(y)+\frac{2 ((y-1) W(y)+y)}{2 \left(2 (y-1) e^{W(y)}+e^{2 W(y)}+y (y-1)+1\right)+y W(y)}$$ We could even do better changing from Halley to Householder (the formula is too long to be typed here). $$\left( \begin{array}{cccc} y & x_ 1\text{ (Halley)}& x_ 1\text{ (Householder)} & \text{solution} \\ 1 & 0.79480270709 & 0.80794504316 & 0.80646599424 \\ 2 & 1.05568629122 & 1.06039805820 & 1.06009031989 \\ 3 & 1.23095181927 & 1.23340361714 & 1.23329258687 \\ 4 & 1.36596225621 & 1.36747274984 & 1.36742103137 \\ 5 & 1.47681961058 & 1.47784536126 & 1.47781737575 \\ 6 & 1.57134651217 & 1.57208888765 & 1.57207216458 \\ 7 & 1.65400279510 & 1.65456484098 & 1.65455411438 \\ 8 & 1.72759676052 & 1.72803683495 & 1.72802957840 \\ 9 & 1.79402191858 & 1.79437560652 & 1.79437048920 \\ 10 & 1.85462068870 & 1.85491094750 & 1.85490721630 \end{array} \right)$$