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How to prove that? I try to use the comparasion test, but i don't know with that function compare.

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    $\begingroup$ how do you know it is convergent? Alternating series test might be relevant, to apply to $\sin(\frac1x)$ near $0$. $\endgroup$ – Mirko Jun 9 '19 at 23:49
  • $\begingroup$ @Mirko it's probably given $\endgroup$ – Saketh Malyala Jun 9 '19 at 23:51
  • $\begingroup$ Why not just use the definition? $\endgroup$ – Arthur Jun 9 '19 at 23:59
  • $\begingroup$ 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 $1\times1=1$ $\endgroup$ – whpowell96 Jun 10 '19 at 0:02
  • $\begingroup$ Your integral is absolutely convergent ( $\sin$ is bounded by 1). This implies that your integral converges. $\endgroup$ – Julian Mejia Jun 10 '19 at 0:20
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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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    $\begingroup$ Boundedness of the integrand does not by itself imply convergence; it can still be non-integrable. $\endgroup$ – Arthur Jun 10 '19 at 0:00
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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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