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

Question

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.

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.

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• 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)

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• 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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