Inverse of $x\cdot\sin(x)$ In the function 
$$f(x) = x\cdot\sin(x)$$
every codomain value $y$ occurrs infinitely many times. So, in principle, there are "inverse" functions $f^{-1}$ such that 
$$f^{-1}(y)=x \mbox{ and } f(x)=y$$
for every $y \in \mathbb{R}$. 
(I know that there is no ONE inverse function in the traditional sense (therefore, the "quotes"), but I'm looking for a function that has the above and below properties which are fairly close to an inverse.)
I'm particularly interested in inverse values $f^{-1}(y)$ that are closest to zero. That is, for values of $y$ below the maximum of the smallest "sine hill", I'd like to get the respective $x$ value on the zero-facing side of that hill, for values larger than that, but smaller than the maximum of the second hill, the function should jump to the zero-facing side of the second hill and so on. (Only looking for positive $x$ values.)
There are probably no closed forms for that (but maybe there are partial closed forms that can be defined interval-wise in some way)? Numerical solutions would also suffice - or any other ideas.
 A: $$x\sin(x)$$
$$-\frac{1}{2}ix(e^{ix}-e^{-ix})$$
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.
$$x\sin(x)=y$$
$$-\frac{1}{2}ix(e^{ix}-e^{-ix})=y$$
$x\to \frac{t}{i}$:
$$-\frac{1}{2}t(e^t)^2-ye^t+\frac{1}{2}t=0$$
This is a polynomial equation of both $t$ and $e^t$ which is quadratic for $e^t$. This kind of equations is therefore not in a form to apply Lambert W or Generalized Lambert W.
How can we show that $A(z,e^z)$ and $A(\ln (z),z)$ have no elementary inverse?
A: Strictly speaking, an 'inverse' trigonometric function like $\arcsin$ is not a proper inverse of $\sin:\mathbb{R}\rightarrow[-1,1]$. This is because $\sin$ is not injective over the domain $\mathbb{R}$. In order to define $\arcsin$, you have to decide on a 'principal value', such as the range $[-\pi/2,\pi/2]$.
The same is true of your function $f$, which is also not injective. Indeed you have specified a principal value, but it is good to remember that $f^{-1}$ is not a true inverse, despite its notation.
The function $f^{-1}$, defined using these principal values, does exist (is well-defined) and will be piecewise smooth. You can justify this using the inverse function theorem. There is no closed form for $f^{-1}$, but you can work out numerical values for it using a scheme such as interval bisection or Newton's method.
