# 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 ?

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.

• Yes correct you can use this argument. – user63181 Jan 31 '14 at 17:20
• That's right. The function is bounded, and it is continuous except for a single point, therefore it is Riemann integrable. The argument works even if you allow a at most a countable number of points of discontinuity. – Eric Auld Jan 31 '14 at 17:24
• Thanks a lot to you both ! – homogenity Jan 31 '14 at 17:28
• It should be emphasized that bounded is crucial here. $1/x$ is continuous on $[0,1]$ except at one point, but $\int_0^1 \frac{1}{x}\;dx$ does not converge. – wckronholm Jan 31 '14 at 17:49
• When I had this exact question on an analysis hand-in set of problems a number of years back, the intended solution was to show from the definition that it is integrable. That is, given an $\epsilon > 0$, find a partition of $[0, 1]$ so that the upper and lower sums are within $\epsilon$ of one another. The trick is to let one piece be $[0, \epsilon/4]$, and exploit that on $[\epsilon/4, 1]$, the function is (finitely) piecewise injective, and thus integrable. – Arthur Jan 31 '14 at 18:47

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) $$\left| \int_s^t \sin \frac 1x \, dx \right| < \epsilon.$$

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.