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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 $\displaystyle \int_1^\infty \! x^{-\alpha} \, \mathrm{d}x$.

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And how did you try to use the hint? – J. M. May 8 '11 at 15:01
A related question:… – Shai Covo May 8 '11 at 15:10

What basically your hint means that if $f$ and $g$ are two non-negative functions such that $f \leq g$ and $\int g$ converges then $\int f$ converges. This is basically the comparison test.

Just evaluate your integral: \begin{align*} \lim_{t \to \infty} \int\limits_{1}^{t} \frac{1}{x^{\alpha} } \ \textrm{dx} &=\lim_{t \to \infty} \Biggl[\frac{x^{-\alpha +1}}{-\alpha+1}\Biggr]_{1}^{t} \end{align*}

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I think you mean that "...$f$ and $g$ are two nonnegative functions (over the domain of interest)..." – cardinal May 8 '11 at 16:07

The integral \begin{equation} \int _{1}^{\infty} \frac{1}{x^\alpha} \ \textrm{dx} \end{equation} converges, as show @Chandru1. Moreover, $$ \lim_{x\to \infty} \left( \frac{\sin\left( \frac{1}{x} \right)}{\frac{1}{x}} \right)^\alpha = 1, $$ therefore your integral and $\int _{1}^{\infty} \frac{1}{x^\alpha} \ \textrm{dx}$ have the same behavior, i.e., your integral converges.

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