# Integration and existence of an antiderivative

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?

• 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 $g(x) =x^2\cos(1/x), g(0)=0$. – Paramanand Singh Mar 11 '19 at 11:54
• Also see this answer for two definite results on existence of anti-derivatives : math.stackexchange.com/a/3120054/72031 – Paramanand Singh Mar 11 '19 at 12:01
• @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? – Darkmaster Mar 12 '19 at 3:58
• @ParamanandSingh Is it easy to see that the function $f$ is discontinuous for any point $c \in [-1,1]$? – Darkmaster Mar 12 '19 at 4:17
• 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 $c=0$. – 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?
• 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. – zhw. Mar 12 '19 at 16:18
• 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? – Darkmaster Mar 12 '19 at 17:00
• Note the last expression $\to 0.$ – zhw. Mar 12 '19 at 17:02
• 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? – Darkmaster Mar 12 '19 at 17:06