An inverse function for $-\log x/\log (1+x)$

Question

In the context of another question I asked on here a while ago, I came across the problem of inverting the function $$f(x) = -\frac{\log x}{\log (1 + x)}$$ for positive real $x$. Let $f^{-1}(x)$ denote this branch of the inverse function of $f(x)$.

Alternatively, it can also be defined implicitly as the positive real solution $y$ of the algebraic equation $$y(1 + y)^x = 1$$

This function $f^{-1}$ is somewhat related to the Lambert $W$ function; for $x\to\infty$, it has the asymptotic expansion $$f^{-1}(x) = \frac{W(x)}{x} + \frac{W(x)^2}{2 x^2(1 + \frac{1}{W(x)})} + O\left(\frac{W(x)^3}{x^3}\right)$$ but apart from that, I was not able to relate $f^{-1}(x)$ to any other special functions that I am aware of.

I was wondering whether this kind of inverse function was studied in the literature before. If not, I wonder if it may be interesting enough to publish.

Answer 1

COMMENT.-Your function $f$ is an open interval bijection, from $]0,\infty[$ on $]0,-1$[ so $f^{-1}$exists but does not have of it an easy analytical expression and you give an approximation with Lambert $W$ function yourself. You can however draw some points of its graph the following way (maybe you know already this way to draw the inverse of a function):

On the graph of $f(x)$, draw lines $y=-x + a$ for distinct values of $a$ with which you obtain points $(x_0, y_0)$ by intersection of the straight line with the graph of the function. Thus each point $(y_0, x_0)$ is a point of $f^{-1}$.

This is another thing to say that the graph of the inverse of a function is symmetric to the graph of the function with respect to the main diagonal of equation $y = x$.