# Proving $\sin(1/x)$ has an antiderivative

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 $\mathbb{R}$?

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To type $\mathbb{R}\setminus 0$, use the \setminus command instead of just \. – Michael Dyrud Sep 24 '12 at 13:57
Thanks, I'll keep that in mind. – Johnny Westerling Sep 24 '12 at 14:00
Your last question is whether adding a constant to a function changes its derivative: is $(d/dx)f(x)$ the same as $(d/dx)(f(x)+c)$? – Michael Hardy Sep 24 '12 at 15:35
No. My last question is different. We know that $\sin(1/x)$ is undefined for $0$, that's why we defined $f(0)=0$; however, if we redefined $f$ AT point $0$ to be $f(0)=c$, then would this have implications on $f$ having an antiderivative on $\mathbb{R}$. – Johnny Westerling Sep 24 '12 at 20:00

It should be clear that the antiderivative restricted to $(0,\infty)$ has to be $$F(x) = \int_1^x \sin(1/t)\, dt + k$$ for some constant of integration $k$. This partial $F$ has to have a limit for $x\to 0^+$, or there cannot be any continuous full $F$ at all, so prove that. Now we can select $k$ such that $F(0)=0$ makes $F$ continuous and extend to negative inputs by $F(-x)=F(x)$.
The function thus defined is the only possible continuous function (upto to constant terms) that has the right derivative for $x\ne 0$, so whatever its derivative at $0$ is (if it exists at all), that has to be the value of $f(0)$. So what you need to prove is that this derivative is in fact $0$.
(See also Darboux' theorem which would immediately have told you that you cannot select an arbitrary value for $f(0)$ and still expect $f$ to have an antiderivative).
@Johnny: That's just the Fundamental Theorem of Calculus on the interval $(0,\infty)$. The lower limit $1$ is arbitrary; it could be anything inside the interval (which would shift the definite integral upwards or downwards by an amount that can be absorbed into $k$). – Henning Makholm Sep 24 '12 at 20:55
Why is the value we take for $k$ relevant for if $F$ is continuous of not? I also don't understand the last paragraph: a point has measure zero so changing a function at a single point do not alter it's integration properties. Based on this why is $f(0)=0$ relevant? – Kibble Jan 12 at 9:22
@Kibble: As for the last paragraph: Even if changing the function at a single point doesn't change its integral, it can certainly change whether it has an antiderivative. For example, there is no function whose derivative is $$f'(x)=\begin{cases}1 & \text{when }x=0 \\ 0 & \text{otherwise} \end{cases}$$ – Henning Makholm Jan 12 at 10:08