Let $ f: \mathbb {R} \rightarrow {\mathbb {R}} $ be given by

$$ f (x) = \left \{\begin {matrix} \sin ({\frac {1} {x}}), & \text {if} \phantom {a} x \neq 0; \\  c, & \text {if} \phantom {a} x = 0 \end {matrix} \right. $$ where $ c \in [-1,1]$. For what values ​​of $c$ is there an antiderivative of $f$?

I do not know of a theorem of the form "if and only if" that it tells me when a function $ f $ has no antiderivative.

Anyone have a suggestion?

  • $\begingroup$ See this answer math.stackexchange.com/a/2603106/72031 and then you will find that $c=0$ fits the bill here. Try to have a look at derivative of $g(x) =x^2\cos(1/x), g(0)=0$. $\endgroup$ – Paramanand Singh Mar 11 '19 at 11:54
  • $\begingroup$ Also see this answer for two definite results on existence of anti-derivatives : math.stackexchange.com/a/3120054/72031 $\endgroup$ – Paramanand Singh Mar 11 '19 at 12:01
  • $\begingroup$ @ParamanandSingh Your argument contradicts Lebesgue's theorem, right? Because this function is discontinuous for every point $ c \in [-1,1] $. Then by Lebesgue's theorem I can conclude that this function is not Riemann-integrable, right? $\endgroup$ – Darkmaster Mar 12 '19 at 3:58
  • $\begingroup$ @ParamanandSingh Is it easy to see that the function $ f $ is discontinuous for any point $ c \in [-1,1] $? $\endgroup$ – Darkmaster Mar 12 '19 at 4:17
  • $\begingroup$ Perhaps you are confused. The function in your question is discontinuous at single point $0$ no matter what value of $c$ is chosen. Thus it is Riemann integrable. A surprise is that the integral function $F$ given in zhw's answer is differentiable at $0$ making it an anti-derivative of your $f$ if $c=0$. $\endgroup$ – Paramanand Singh Mar 12 '19 at 6:33

Hint: Define $F(x) = \int_0^x \sin (1/t)\, dt.$ Then $F'(x) = \sin(1/x),$ $x\ne 0.$ Does $F'(0)$ exist?

  • $\begingroup$ Is that easy to see? I tried by definition, but I do not know what to do with that integral? $\endgroup$ – Darkmaster Mar 12 '19 at 16:05
  • $\begingroup$ Deleted by previous comment as it may be easier to write the integrand as $\sin(1/t)(-1/t^2)(-t^2)$ and integrating by parts. $\endgroup$ – zhw. Mar 12 '19 at 16:18
  • $\begingroup$ Are you sure about that?The result is $$f(0)=F'(0)=\lim_{x\rightarrow{0}}{\frac{F(x)-F(0)}{x}}=\lim_{x\rightarrow{0}}{\frac{F(x)}{x}}=\lim_{x\rightarrow{0}}{\frac{x^{2}\cos{\frac{1}{x}-\int_{0}^{x}{2t\cos{\frac{1}{t}}dt}}}{x}}=0?$$ right? $\endgroup$ – Darkmaster Mar 12 '19 at 17:00
  • $\begingroup$ Note the last expression $\to 0.$ $\endgroup$ – zhw. Mar 12 '19 at 17:02
  • $\begingroup$ It's clear that $x\cos\frac{1}{x}\rightarrow{0}$, when $x\rightarrow{0}$, but $\frac{-\int_{0}^{x}{2t\cos{\frac{1}{t}}dt}}{x}\rightarrow{0}$, when $x\rightarrow{0}$. Why? $\endgroup$ – Darkmaster Mar 12 '19 at 17:06

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