Proving sin(1/x) has an antiderivativeI 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 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 ∈ R ).First off, f is continuous on R ∖ 0, hence it has an antiderivative for every x ∈ R ∖ 0. However, f is not continuous in x = 0. x = 0Furthermore, I'm curious about the significance of f ( 0 ) = 0. Suppose we redefine our function, and take f ( 0 ) = c, for some c ∈ R , c ≠ 0 . Would then f have an antiderivative on R ?

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Proving sin(1/x) has an antiderivativeI 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       R   (well, I&#039;m not sure if I haven&#039;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  ∈      R  ).First off, f is continuous on       R    ∖  0, hence it has an antiderivative for every   x  ∈      R    ∖  0. However, f is not continuous in   x  =  0.   x  =  0Furthermore, I&#039;m curious about the significance of   f  (  0  )  =  0. Suppose we redefine our function, and take   f  (  0  )  =  c, for some   c  ∈      R  ,   c  ≠      0  . Would then f have an antiderivative on       R  ?

Tristan Ward · Accepted Answer

Step 1It should be clear that the antiderivative restricted to   (  0  ,  ∞  ) has to be   F  (  x  )  =      ∫    1    x    sin  ⁡  (  1      /    t  )    d  t  +  k for some constant of integration k. This partial F has to have a limit for   x  →      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  ).Step 2The function thus defined is the only possible continuous function (upto to constant terms) that has the right derivative for   x  ≠  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&#039; 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).

Proving sin(1/x) has an antiderivative I would want to prove that the function defined as follows:

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