A doubt about an improper integral After a class about improper integrals (in high school) I was wondering about the integral $$\displaystyle\int\limits_0^1\sin{\frac{1}{x}}$$

but I found out that I'm unable to calculate the corresponding indefinite integral, is there a way to compute this integral by hand?

p.s.
after the edits this question is different from the original one, I figured out that my problem was different from the one I was originally asking
Since the function $x\mapsto\sin\frac1x$ is continuous on the interval $(0,1]$ and 
$$\left|\sin\frac1x\right|\le1$$
then the integral is convergent.
Set $y=x^{-1}$. Then

\begin{equation*}
\int_{0}^{1}dx\sin \frac{1}{x}=\int_{1}^{\infty }dy\frac{1}{y^{2}}\sin y
\end{equation*}
which is well-behaved.