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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.

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  • $\begingroup$ 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. $\endgroup$ – Alexey Burdin Jul 10 '20 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 '20 at 19:19
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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.

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  • $\begingroup$ shouldn't be $\int_{0}^{\infty} \frac{sin(u)}{u^{\alpha}}$ ? $\endgroup$ – Karl Jul 10 '20 at 19:34
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    $\begingroup$ @Karl $dx=-\frac{dt}{t^2}$ $\endgroup$ – Alexey Burdin Jul 10 '20 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 '20 at 20:07
  • $\begingroup$ @Karl: This may be of help. math.stackexchange.com/questions/390809/…. $\endgroup$ – Oliver Diaz Jul 10 '20 at 20:24
  • $\begingroup$ @Karl: I just added more generous hints for you. $\endgroup$ – Oliver Diaz Jul 10 '20 at 20:39

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