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

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.

• 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. Jul 10, 2020 at 19:06
• 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)
– E-A
Jul 10, 2020 at 19:19

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.

• shouldn't be $\int_{0}^{\infty} \frac{sin(u)}{u^{\alpha}}$ ?
– Karl
Jul 10, 2020 at 19:34
• @Karl $dx=-\frac{dt}{t^2}$ Jul 10, 2020 at 19:35
• 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 $\alpha +1 \gt 0$ ?
– Karl
Jul 10, 2020 at 20:07
• @Karl: This may be of help. math.stackexchange.com/questions/390809/…. Jul 10, 2020 at 20:24
• @Karl: I just added more generous hints for you. Jul 10, 2020 at 20:39