Inverse of $\frac{\sin(x)}{x}$ How would one find the inverse of the function $y=\frac{\sin(x)}{x}$? Here are my steps:
$y=\frac{\sin(x)}{x}$,
$x=\frac{\sin(y)}{y}$,
$xy=\sin(y)$,
$\arcsin(xy)=y$,
After that step, I can’t find a way to isolate $y$. 
 A: Elementary functions:
$$f(x)=\frac{\sin(x)}{x}$$
$$f(x)=-\frac{1}{2x}i(e^{ix}-e^{-ix})$$
We see, this function is an algebraic function in dependence of both $x$ and $e^x$. Liouville proved that such kind of functions (over a complex domain without isolated points) don't have (partial) inverses that are elementary functions: How can we show that $A(z,e^z)$ and $A(\ln (z),z)$ have no elementary inverse?
Lambert W, Generalized Lambert W:
The defining equation for the inverse $\frac{\sin(x)}{x}$ can be rearranged to a polynomial equation of both $x$ and $e^x$ which is quadratic for $e^x$. This equation is therefore not in a form to apply Lambert W or Generalized Lambert W.
"Leal-functions":
The partial inverses of the function mentioned in the question can be represented in terms of the function $\text{Lcsc}$ presented in [Vazquez-Leal et al. 2020].
$$\frac{\sin(x)}{x}=y$$
$\sin(x)=\frac{1}{\csc(x)}$:
$$\frac{1}{x\csc(x)}=y$$
$$x\csc(x)=\frac{1}{y}$$
$$x=\text{Lcsc}\left(\frac{1}{y}\right)$$
We can take the "Leal functions" as closed-form functions because some of their algebraic properties and their applicability for some other kinds of equations are presented in the cited article.
[Vazquez-Leal et al. 2020] Vazquez-Leal, H.; Sandoval-Hernandez, M. A.; Filobello-Ninoa, U.: The novel family of transcendental Leal-functions with applications to science and engineering. Heliyon 6 (2020) (11) e05418
