Q1) How do I prove that the improper Riemann Integral $$\int_0^1\frac{\sin(\frac 1x)}x\,dx$$ converges?

It does, according to WolframAlpha (http://www.wolframalpha.com/input/?i=integrate+sin(1%2Fx)%2Fx+from+0+to+1).

The estimates $\sin(\frac 1x)\leq\frac 1x$ or $\sin(\frac 1x)\leq 1$ both do not seem to work here since $\frac 1x$ and $\frac{1}{x^2}$ are both not integrable on $(0,1]$.

Q2) How do we prove that as a Lebesgue integral, $$\int_0^1\frac{\sin(\frac 1x)}x\,dx$$ does not exist?

Roughly I know it is because of the $\infty-\infty$ reason because $f^+=\infty$, $f^-=\infty$, but how do we show that? Or alternatively, we could show $|f|$ is not Lebesgue integrable?

Thanks for any help.


Hint: Setting $u = 1/x$, we have that \begin{align} \int^1_{0} \frac{\sin \frac{1}{x}}{x}\ dx = \int^\infty_1 \frac{\sin u}{u}\ du. \end{align} There are many related post.

  • $\begingroup$ This doesn't prove anything. How do you even know you can make this substitution? When you are first learning about the integrals, you should be very precise. $\endgroup$ – Euler....IS_ALIVE Oct 1 '16 at 4:50
  • $\begingroup$ @Euler....IS_ALIVE It's not meant as a solution. I just want the reader to look at the problem for $\int^\infty_1 \frac{\sin u}{u}\ du$ since it has been looked at in a few MSE post. $\endgroup$ – Jacky Chong Oct 1 '16 at 4:52
  • $\begingroup$ This is quite helpful to me though. Thanks. $\endgroup$ – yoyostein Oct 1 '16 at 4:52
  • 1
    $\begingroup$ @Euler....IS_ALIVE That substitution is standard, and since the word "Lebesgue" has appeared I think we can assume the OP didn't start learning about integrals yesterday. $\endgroup$ – zhw. Oct 1 '16 at 15:14
  • $\begingroup$ @zhw. Hmmmmmmm do you think he's learning about Lebesgue integrals? This is a standard question when doing so. $\endgroup$ – Euler....IS_ALIVE Oct 1 '16 at 15:16

For Q2), hint: Let $a_n = 1/(3\pi/4+n\pi), b_n = 1/(\pi/4+n\pi).$ Observe

$$\int_{a_n}^{b_n} \frac{|\sin (1/x)|}{x}\, dx \ge \frac{\sqrt2 /2}{b_n}(b_n-a_n).$$


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