proving that an improper integral diverges

Question

I was wondering if I could get a hint on checking the convergence of the following integral: $$\int_1^\infty x^a\sin^3x \, dx $$ For a>0 I don't know how to show it diverges(or converges). Thanks in advance!

Answer 1

just for nonexistence of the limit

Let $f (x)=x^a\sin^3 (x), $

$$u_n=\frac {\pi}{2}+2n\pi $$ and $$v_n=n\pi $$

then

$$\lim_{n\to+\infty}f (u_n)=+\infty $$ $$\lim_{n\to+\infty}f (v_n)=0$$

thus $\lim_{x\to+\infty}f (x) $ does not exist.