Inverse function of $x \mapsto \coth x - 1/x$

Question

Consider the function $f$ defined over $\mathbb{R}$ as $$f(x) = \coth x - \frac{1}{x}$$ if $x \neq 0$ and $f(0)=0$.

Since the function $\coth$ can be developed in series as $\coth x = \frac{1}{x} + \frac{x}{3} - \frac{x^3}{45} + \cdots$ around the origin, the function $f$ can easily be shown to be infinitely smooth over $\mathbb{R}$, strictly increasing, with limits $\pm 1$ in $\pm \infty$. The function $f$ is therefore a smooth bijection from $\mathbb{R} \rightarrow (-1,1)$.

My question is the following: Is there an analytic expression for the inverse $f^{-1} : (-1,1) \rightarrow \mathbb{R}$ of $f$ such that $f^{-1}(y) = x$ if and only if $f(x) = y$?

Context: The function $f$ naturally appears in an optimization problem I am considering. It would be useful for me to understand and possibly compute what the inverse is.

Answer 1

The “Langevin function” is invertible via Lagrange reversion:

$$L(x)=\coth(x)-\frac1x=a\mathop\iff^{x=\frac{y-1}{a+1}} y=e^\frac{2(y-1)}{a+1} \left(\frac{a+1}{a-1}(y-1)+1\right)\implies x=\frac1{a+1}\left(-1+\sum_{n=1}^\infty\frac{e^{-\frac{2n}{a+1}}}{n!}\left.\frac{d^{n-1}}{dt^{n-1}}e^\frac{2nt}{a+1}\left(\frac{a-1}{a+1}(t-1)+1\right)^n\right|_0\right)$$

Using general Leibniz rule:

$$\left.\frac{d^{n-1}}{dt^{n-1}}e^\frac{2nt}{a+1}\left(\frac{a-1}{a+1}(t-1)+1\right)^n\right|_0=\sum_{k=0}^{n-1}\binom{n-1}k\left.\frac{d^{n-1}}{dt^{n-1}}e^\frac{2n t}{a+1}\right|_0 \left.\frac{d^{n-1}}{dt^{n-1}} \left(\frac{a-1}{a+1}(t-1)+1\right)^n\right|_0$$

and summing over $k$ uses hypergeometric $\operatorname U(a,b,x)$. Therefore:

$$\bbox[2px, border: 2px solid silver]{L^{-1}(x)=-\frac1{x+1}\left(1+2\sum_{n=1}^\infty\frac{(-1)^n(x-1)^{n-1}}{e^\frac{2n}{x+1}(x+1)^nn!}\operatorname U\left(1-n,2,-\frac{4n}{x^2-1}\right)\right)}$$

shown here. The series converges more quickly as $x\to-1$ and symmetry extends the convergence radius to $x>0$

Answer 2

Use th Lagrange inversion formula: if $$f(z) = \sum_{n=1}^\infty a_n z^n$$ with $a_1 \neq 0$ (interpreted either as an analytic function or as a formal power series), then \begin{equation} f^{-1}(w) \;=\; \sum_{m=1}^\infty {w^m \over m} \, [\zeta^{m-1}] \biggl( {\zeta \over f(\zeta)}\biggr)^m \;, \end{equation} where $[\zeta^n] g(\zeta) = g^{(n)}(0)/n!$ denotes the coefficient of $\zeta^n$ in the power series $g(\zeta)$. More generally, if $h(z) = \sum_{n=0}^\infty b_n z^n$, we have \begin{equation} h(f^{-1}(w)) \;=\; h(0) \,+\, \sum_{m=1}^\infty {w^m \over m} \, [\zeta^{m-1}] \, h'(\zeta) \biggl( {\zeta \over f(\zeta)}\biggr)^m \;. \end{equation} Proofs of these formulae can be found in many books on analytic function theory.