# Is $f(x)=x\sin(1/x)$ with $f(0)=0$ of bounded variation on $[0,1]$? - Problem with abs. continuous

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.

$$g(x) = -\frac{1}{x}\cos\left(\frac{1}{x}\right) + \sin\left(\frac{1}{x}\right)$$ 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.
• Thank you really much! Please, how can I prove that this function is not Lebesgue integrable on $[0,1]$? Feb 19 at 15:22
• Wolfram can make errors... It can make computations without checking if those computations make sense. You have to check that $\int_0^1 \vert g \vert = \infty$. Feb 19 at 15:55