I am having the following trouble:

  1. From Is $f(x)=x\sin(\frac{1}{x})$ with $f(0)=0$ of bounded variation on $[0,1]$?, $x\sin(1/x)$ has not bounded variation in $[0,1]$.

  2. $x\sin(1/x)$ has derivative $-\cos(1/x)/x + \sin(1/x)$ a.e. and $\int_0^1 (-\cos(1/x)/x + \sin(1/x)) dx=\sin(1)$, i.e., $x\sin(1/x)$ is abs. continuous.

  3. Every abs. continuous function is of bounded variation.

What is the error?

Thank you, really much.


1 Answer 1


$$g(x) = -\frac{1}{x}\cos\left(\frac{1}{x}\right) + \sin\left(\frac{1}{x}\right)$$ is not integrable on $[0,1]$. Hence the equality

$$\int_0^1 (-\cos(1/x)/x + \sin(1/x)) dx=\sin(1)$$ is wrong and $g$ is not absolutely continuous.

  • 1
    $\begingroup$ Not Lebesgue integrable, that is. $\endgroup$
    – aschepler
    Feb 19 at 14:39
  • $\begingroup$ Thank you really much! Please, how can I prove that this function is not Lebesgue integrable on $[0,1]$? $\endgroup$ Feb 19 at 15:22
  • $\begingroup$ The result I've posted is from Wolfram... What can be the error? $\endgroup$ Feb 19 at 15:30
  • 2
    $\begingroup$ Wolfram can make errors... It can make computations without checking if those computations make sense. You have to check that $\int_0^1 \vert g \vert = \infty$. $\endgroup$ Feb 19 at 15:55
  • $\begingroup$ Thank you so much, and I've forgotten to put the absolute value $\endgroup$ Feb 19 at 18:55

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