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$$ \int_1^{\infty} \left\langle t\right\rangle\dfrac{\cos\left(t\right) - \sin\left(t\right)}{t^2}\,dt $$

where $\left\langle t\right\rangle$ is the rationale part of $t$.

I would like to use the improper integral comparison test with $ \dfrac 2{t^2} $, but I have no reason to think the integral is positive. How can I evaluate if this function is integrable in the improper sense? Thank you.

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    $\begingroup$ Two hints: $\langle t \rangle < 1$ and absolute convergence implies conditional convergence! $\endgroup$ – Nigel Overmars Mar 10 '16 at 11:20
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Don't worry about $\cos t - \sin t$, because $$ \left|\langle t\rangle \frac{\cos t - \sin t}{t^2}\right| \le \frac{2}{t^2} $$ and so we can apply absolute convergence of improper integral.

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  • $\begingroup$ Makes sense, didn't think of that. Thank you! $\endgroup$ – test9995 Mar 10 '16 at 11:26

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