Does a function have an indefinite integral?

Question

Does this function have an indefinite integral?

$f(x)=\left\{\begin{array}{ll} \cos\left( \frac{1}{x}\right) & ,x \neq 0 \\ 0 & ,x=0\end{array}\right.$

Answer 1

Yes. Sketch: Define $F(x) = \int_0^x \cos (1/t)\ dt.$ Then $F'(x) = f(x)$ for $x\ne 0$ by the FTC. To show $F'(0)= f(0)=0,$ we need to go back to the definition. So consider

$$\frac{F(x) - F(0)}{x-0} = (1/x)\int_0^x \cos (1/t)\ dt.$$

as $x\to 0^+.$ Let $t = 1/y$ to see the last integral equals

$$(1/x)\int_{1/x}^\infty \cos (y)/y^2\ dy.$$

OK, you want to show the last expression $\to 0$ as $x\to 0^+.$ I suggest integrating by parts here.