How to prove that? I try to use the comparasion test, but i don't know with that function compare.
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2$\begingroup$ how do you know it is convergent? Alternating series test might be relevant, to apply to $\sin(\frac1x)$ near $0$. $\endgroup$– MirkoJun 9, 2019 at 23:49
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$\begingroup$ @Mirko it's probably given $\endgroup$– Saketh MalyalaJun 9, 2019 at 23:51
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$\begingroup$ Why not just use the definition? $\endgroup$– ArthurJun 9, 2019 at 23:59
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$\begingroup$ You are integrating a bounded function (bounded by 1) over a bounded interval (of length 1). Therefore, the absolute value of the integral must be less than or equal to $1\times1=1$ $\endgroup$– whpowell96Jun 10, 2019 at 0:02
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$\begingroup$ Your integral is absolutely convergent ( $\sin$ is bounded by 1). This implies that your integral converges. $\endgroup$– Julian MejiaJun 10, 2019 at 0:20
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2 Answers
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As Azif said this function is bounded. It's continuous and consequently integrable on $[δ,1]$ and |$\int_o^{\delta} \sin(x+\frac{1}{x}) \mathrm{d}x| \le \delta$ which converges to $0$ as $\delta \rightarrow 0$ so it is intégrable.
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$$-1\le \sin\left(x+\frac{1}{x}\right)\le1$$ And, therefore $$-1\le \int_0^1\sin\left(x+\frac{1}{x}\right)\mathrm{d}x\le1$$
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8$\begingroup$ Boundedness of the integrand does not by itself imply convergence; it can still be non-integrable. $\endgroup$– ArthurJun 10, 2019 at 0:00