How do I show that for $\alpha > 1$ the integral $\displaystyle \int_1^\infty \! \sin^\alpha(1/x) \, \mathrm{d}x$ converges?
I am given the hint:
Compare with the integral $\displaystyle \int_1^\infty \! x^{-\alpha} \, \mathrm{d}x$.
How do I show that for $\alpha > 1$ the integral $\displaystyle \int_1^\infty \! \sin^\alpha(1/x) \, \mathrm{d}x$ converges?
I am given the hint:
Compare with the integral $\displaystyle \int_1^\infty \! x^{-\alpha} \, \mathrm{d}x$.