Find $\alpha$ so that the integral $\int_{0}^{\infty} x^{\alpha}\sin(\frac{1}{x})$ converges.

What I did first is to separte the integral into $\int_{0}^{1} x^{\alpha}\sin(\frac{1}{x}) dx+ \int_{1}^{\infty} x^{\alpha}\sin(\frac{1}{x})dx$ since $f(x)$ is not defined in $0$ nor $\infty$

Secondly, the only way I know to compare is ether by using $\sin(\frac{1}{x}) \lt \frac{1}{x}$ or that $|\sin(\frac{1}{x})| \lt 1$ but non of those two work for this excersice. Any hints ? Thanks in advance.

  • $\begingroup$ Somewhy this heavily reminds me of Flint-Hill Series. The solution of them involves irretionality measure of $\pi$. I'd do the variable change $t=1/x$ first. $\endgroup$ Jul 10, 2020 at 19:06
  • $\begingroup$ Use one bound for one integral, and the other for the other. (Hint: 1/x < 1 when x> 1) I think it will come down to $\alpha \in (-1,0]$ (did not check the boundaries) $\endgroup$
    – E-A
    Jul 10, 2020 at 19:19

1 Answer 1


Hint: The change of variables $u=x^{-1}$ transform your integral into $$ \int^\infty_0 \frac{\sin u}{u^{\alpha+2}}du$$

this type integral has been studied and discussed several times in this forum. for instance here they discuss something similar.

  • Case $\alpha+2\leq1$: (a) The integral converges as an improper Riemann but (b) not as a Lebesgue integral since $\int^{N\pi}_{\pi}\frac{|\sin u|}{u^{\alpha+2}} \geq C_\alpha\sum^{N-1}_{k=1}\frac{1}{k^{2+\alpha}}$ for some constant $C_\alpha>0$.

To all that, divide the integral in pieces $\int^{N\pi}_0=\sum^{N-1}_{k=1}\int^{(k+1)\pi}_{k\pi}$.

(a) This partition of the integral strategy also helps to show that $\lim_{T\rightarrow\infty}\int^T_\pi\frac{\sin u}{u^{\alpha +2}}\,du$ converges, since thee sum you get is an alternating series of the type one studies in freshmen calculus.

(b) Using that $\frac{1}{\pi (k+1)}\leq \frac{1}{t}\leq \frac{1}{\pi k}$ for $k\pi\leq t\leq (k+1)\pi)$ one gets that $\int^\infty_0\frac{|\sin u|}{u^{\alpha+2}}\,du=\infty$.

Finally, on the interval $[0,\pi]$ there are no problems since $\int^1_0\frac{\sin u}{u^{\alpha+2}}\,du\leq \int^1_0\frac{1}{u^{\alpha+2}}\,du$ converges when $\alpha+2<1$, and $\int^1_0\frac{\sin u}{u}\,du$ exists a a genuine Riemann integral (the function can be defined at zero to produce a nice continuous function)

  • Case $\alpha+2>1$: The integral diverges to $\infty$ (as both Lebesgue and roper Riemann integral) since $\int^{\pi/4}_0\frac{\sin u}{u^{\alpha+2}}\,dt\geq \sin1\int^{\pi/4}_0\frac{du}{u^{\alpha+1}}=\infty$ and $\int^{\infty}_{\pi/4}\frac{|\sin u|}{u^{\alpha+2}}\,du<\infty$. Similar arguments as above.

  • $\begingroup$ shouldn't be $\int_{0}^{\infty} \frac{sin(u)}{u^{\alpha}}$ ? $\endgroup$
    – Karl
    Jul 10, 2020 at 19:34
  • 1
    $\begingroup$ @Karl $dx=-\frac{dt}{t^2}$ $\endgroup$ Jul 10, 2020 at 19:35
  • $\begingroup$ I'm not familiar with the improper Riemann integral, is there another way to work with the integral. What I don't understand is this. If $sin(x) \lt 1$ then $\int_{1}^{\infty} \frac{sin(t)}{x^{\alpha + 2}} \lt \int_{1}^{\infty} \frac{1}{x^{\alpha +2}}$ witch should converge for $\alpha +1 \gt 0$ ? $\endgroup$
    – Karl
    Jul 10, 2020 at 20:07
  • $\begingroup$ @Karl: This may be of help. math.stackexchange.com/questions/390809/…. $\endgroup$ Jul 10, 2020 at 20:24
  • $\begingroup$ @Karl: I just added more generous hints for you. $\endgroup$ Jul 10, 2020 at 20:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.