Let $F:[0,1]\rightarrow \mathbb{R}$ defined by $$F(x)= \left\{ \begin{array} .x\cdot \cos\left(\frac{\pi}{x}\right), &\textrm{if } x\in[0,1] \\ 0, &\textrm{ if $x=0$} \end{array} \right. $$ How can I prove that $f=F'$ is Henstock-Kurzweil integrable in $[0,1]?$ But $|f|$ is not? My biggest problem is that $F'$ is not defined at $0$, for $\lim_{x\rightarrow\infty}f(x)=\infty.$


1 Answer 1



For all $0<a<1$

$$\int_a^1 f(t) \ dt =F(1)-F(a)$$ as $f$ is continuous on $[a,1]$ and $\lim\limits_{a \to 0^+} F(a) =0$. Hence, according to Hake’s theorem, the HK-integral of $f$ on $[0,1]$ exists and is equal to $F(1)-F(0)=F(1)=-1$.

To prove that the HK-integral of $\vert f \vert$ on $[0,1]$ doesn’t exist, notice that $$f(x)= \cos\left(\frac{\pi}{x}\right)+\frac{\pi}{x}\sin\left(\frac{\pi}{x}\right)$$ and

$$\vert f(x)\vert \ge \left\vert \frac{\pi}{x}\sin\left(\frac{\pi}{x}\right)\right\vert -1$$

And then mimic the standard way used to prove that $$\int_1^\infty \frac{\vert \sin x \vert}{x} \ dx$$ diverges.

  • $\begingroup$ Sorry, but to prove that $f$ is not integrable in $[0,1]$ using the fact that $|f(x)|\geq \left|\frac{\pi}{x}sin\left(\frac{\pi}{x}\right) \right|-1$, don't we need to integrate $\frac{|sinx|}{x}$ from 0 to 1? $\endgroup$ Jun 22, 2018 at 3:13
  • $\begingroup$ No. You just need to prove that it is unbounded. You can minorite the integral with a series proportional to the harmonic series which is diverging. $\endgroup$ Jun 22, 2018 at 5:50

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