Integral of $\sin(1/x)$ from $0$ to $\infty$ I am interested in proving that the integral $$\int_0^{\infty} \sin \left(\frac{1}{x}\right) dx$$ does not converge. In particular, I would like to show this by appealing to the small angle property of sine. We know that for sufficiently small $\theta$, $\sin(\theta) \approx \theta.$ Said another way, $$\sin\left(\frac{1}{x}\right) \approx \frac{1}{x}\ \text{ as } x \rightarrow \infty.$$ We know that the integral $$\int_0^{\infty}\frac{1}{x} dx$$ does not converge. With these two ideas in mind, I am trying to find a way to bound $\int \sin(1/x) dx$ from below by $\int 1/x dx$ for sufficiently large $x$ in order to show that $\int \sin(1/x) dx$ does not converge. 
Note: one can appeal to the sum $$\sum_{n=1}^{\infty} \sin\left(\frac{1}{n}\right)$$ and show this does not converge by observing that $$\lim_{n \rightarrow \infty} \frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}} = \lim_{x \rightarrow 0} \frac{\sin(x)}{x} = \lim_{x \rightarrow 0} \frac{\cos(x)}{1} = 1$$ via L'Hopital's rule and since $\sum_{n=1}^{\infty} 1/n$ does not converge, $\sum_{n=1}^{\infty} \sin(1/n)$ does not converge. I would simply like to see this notion applied to its integral analogue! 
 A: For every $x\in[0,\pi/2]$, we have
$$\sin(x)\ge\frac{\sin(\pi/2)-\sin(0)}{\pi/2-0}x=\frac2\pi x$$
Indeed, all I did was draw a secant line against the $\sin$ function.  Thus, we have
$$\int_{2/\pi}^\infty\sin(1/x)\ dx\ge\int_{2/\pi}^\infty\frac2\pi\frac1x\ dx$$
which diverges.

A visuallization of the inequality:

A: As you said, $\frac{\sin(1/x)}{1/x} \stackrel{x \to \infty}{\to} 1.$ In particular, for $x$ sufficiently large,
$$\frac{\sin(1/x)}{1/x}>1/2,$$
and therefore
$$\sin(1/x)>\frac{1}{2x}.$$
A: The other answers gave you nice direct solutions, but there is a general result lurking here, an analogue of the reasoning in your infinite series example. What you're using there is basically the limit comparison test for infinite series; there is also a limit comparison test for improper integrals. The proof technique is just a generalization of Aloizio's answer.
A: The convergence can be verified simply by the Integral Convergence Test.
OR we can also consider the function as a series and check if it is Uniformly Convergent.
