A doubt about an improper integral

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

One possibility, as suggested in another answer, is to subsitute $x=1/y$ in the integral. Next, you can split the integral in two pieces: $$ \int_1^\infty \frac{\sin y}{y^2}\,dy =\int_1^a \frac{\sin y}{y^2}\,dy+\int_a^\infty \frac{\sin y}{y^2}\,dy $$ where $a$ is to be determined later. You can find a numerical approximation of the first integral on the right by Simpson's method; then it remains to estimate the last integral. If you let $a=n\pi$ then the last integral can be written as $$ \int_{n\pi}^\infty \frac{\sin y}{y^2}\,dy=\sum_{k=n}^\infty \int_{k\pi}^{(k+1)\pi} \frac{\sin y}{y^2}\,dy, $$ which is an alternating series whose terms decreas in absolute value. Incidentally, this proves the existence of the limit. But you can also use the alternating series error estimate to conclude that the value is between $0$ and the first term of the series. (No need to go further, as then you might as well just pick a bigger $n$ to begin with.)

You can speed things up a bit by partial integration, integrating the trigonometric part and differentiating the inverse power. Two partial integrations, for example, yield $$ \int_a^\infty \frac{\sin y}{y^2}\,dy=\frac{\cos a}{a^2}+2\frac{\sin a}{a^3} +6\int_a^\infty \frac{\sin y}{y^4}\,dy $$ (modulo any sign errors I may have committed). The integral on the right converges faster than the original, so you should be able to get away with a smaller $a$ for the same accuracy.

(Too many partial integrations may be counter-productive, since there will be a factor $k!$ in front of the integral after $k-1$ partial integrations. The resulting series is known as an asymptotic series. It will not converge, but it is still useful for computation.