# How to prove $\int_0^1\sin(x+\frac{1}{x}) \, dx$ is convergent?

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

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

$$-1\le \sin\left(x+\frac{1}{x}\right)\le1$$ And, therefore $$-1\le \int_0^1\sin\left(x+\frac{1}{x}\right)\mathrm{d}x\le1$$
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.