Inverse of $f(x)=x/\log(1+x)$ for $x>1$

Question

Is there an expression of $f^{-1}$ in terms of product-log or other special functions which holds for $x>1$?

WolframAlpha finds the solution: $$g(y)=-y\,W\big(-\mathrm e^{-1/y}/y\big)-1,$$ which however only holds for $-1<x<1$, $0<y<1$.

Edit: Corrected $g(y)$ as per @jjagmath's comment.

The Lambert-W function is multivalued. Try using the other branch, $W_{-1}$ instead. — user1337, Jun 09 '21 at 12:49
The formula you typed for $g$ is not the answer from Wolfram, you missed a minus one term and a sing in the $W$ argument. — jjagmath, Jun 09 '21 at 14:01

score 4 · Accepted Answer · answered Jun 09 '21 at 13:55

4

As user1337 said in his comment, you just need to consider the other branch of the Lambert function:

answered Jun 09 '21 at 13:55

jjagmath

1 Answers1