I would want to prove that the function defined as follows: $f(x)=\sin(1/x)$ and $f(0)=0$ has an antiderivative on the entire $\mathbb{R}$ (well, I'm not sure if I haven't worded this awkwardly, but essentially I am trying to show that $f$ is a derivative of some function that is differentiable in every $x\in\mathbb{R}$).
First off, $f$ is continuous on $\mathbb{R}\setminus0$, hence it has an antiderivative for every $x\in\mathbb{R}\setminus0$. However, $f$ is not continuous in $x=0$. Is there a way we can deal with that?
Furthermore, I'm curious about the significance of $f(0)=0$. Suppose we redefine our function, and take $f(0)=c$, for some $c\in\mathbb{R}$, $c\ne{0}$. Would then $f$ have an antiderivative on $\mathbb{R}$?