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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2$\begingroup$ And how did you try to use the hint? $\endgroup$ – J. M. isn't a mathematician May 8 '11 at 15:01
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$\begingroup$ A related question: math.stackexchange.com/questions/9867/… $\endgroup$ – Shai Covo May 8 '11 at 15:10
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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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4$\begingroup$ I think you mean that "...$f$ and $g$ are two nonnegative functions (over the domain of interest)..." $\endgroup$ – cardinal May 8 '11 at 16:07
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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.
