Convergence of $\int_0^1 \sin\frac{1}{x} \;\mathrm{dx}$ Given the following improper integral:
$$\int_0^1 \sin\frac{1}{x} \;\mathrm{dx}$$
I know it converges, after substituting $u=\frac{1}{x}$ and then comparing to $\frac{1}{x^2}$ . 
But, is it also legitimate to say that $| \sin\frac{1}{x} | \leq 1$ always, and since $1$ is integrable in the region $[0,1]$ , we have that also our integral of $\sin\frac{1}{x}$ converges ?
I am not sure about this argument and will be glad if you will be able to verify my thoughts.
Thanks ! 
p.s.- I am not sure about this argument, because the textbook only mentioned the substitution argument and not my comparison so it might be wrong somewhere.
 A: Here is a different way to look at it: as with most limiting processes there is a corresponding Cauchy criterion---here $$\lim_{t \to 0^+} \int_t^1 \sin \frac 1x \, dx$$  exists because for any $\epsilon > 0$ there exists $0 < t_\epsilon < 1$ with the property that $0 < s < t < t_\epsilon$ implies (taking into account $\sin$ is bounded) $$\left| \int_s^t \sin \frac 1x \, dx \right| < \epsilon.$$
A: One could also make the comparison of the integral to the sum of areas of  rectangles of "height" +1 or -1 on each of the intervals between $ \ x-$ intercepts.  It is sufficient, though, to just sum the positive areas: 
$$ \int_0^1 \ \sin \left( \frac{1}{x} \right) \ \ dx \ \ < $$
$$(+1) \cdot ( 1 - \frac{1}{\pi} ) \   + \ (+1) \cdot (\frac{1}{2\pi}  -  \frac{1}{3 \pi}) \ + \ \ (+1) \cdot (\frac{1}{4 \pi}  -  \frac{1}{5 \pi}) \ + \ \ldots$$
$$ = \ ( 1 - \frac{1}{\pi} ) \ + \  \frac{1}{(2 \cdot 3) \pi} \ + \ \frac{1}{(4 \cdot 5) \pi} \ + \   \frac{1}{(6 \cdot 7) \pi} \ + \ \ldots $$
$$ < \ \ ( 1 - \frac{1}{\pi} ) \ + \  \frac{1}{\pi} \cdot \sum_{n=2}^{\infty} \ \frac{1}{n(n+1)} \ \ , $$
the infinite series being convergent.
