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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 \.2012-09-24
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    Thanks, I'll keep that in mind.2012-09-24
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    Your last question is whether adding a constant to a function changes its derivative: is $(d/dx)f(x)$ the same as $(d/dx)(f(x)+c)$?2012-09-24
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    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 $\mathbb{R}$.2012-09-24

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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).

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    Thank you for the answer. To be honest, I do not understand your first step - could you please make it clear from what does it follow?2012-09-24
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    @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 $k$).2012-09-24
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    OK, now this looks more clear...2012-09-24
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    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?2016-01-12
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    @Kibble: In order for $F$ to be continuous at $0$ you need to choose a value for $k$ that is consistent with the (otherwise arbitrary) value you choose for $F(0)$. As long as these choices match each other you'll be good; I just suggesting to use $F(0)=0$ because that will bring out the symmetry of the situation most clearly.2016-01-12
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    @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 $$ f'(x)=\begin{cases}1 & \text{when }x=0 \\ 0 & \text{otherwise} \end{cases} $$2016-01-12