Proving that the limit of the integral of the following version of the topologist's sine curve exists, but that of the absolute value doesn't

Question

Consider the following variation of topologist's sine curve:

$$f(x) = \frac{1}{x} \cdot \sin \left(\frac{1}{x}\right)$$

I want to show that

$$\lim_{c \to 0} \int_c^1 \frac{1}{x} \cdot \sin \left(\frac{1}{x}\right) \, dx$$ exists, but

$$\lim_{c \to 0} \int_c^1 \left|\frac{1}{x} \cdot \sin \left(\frac{1}{x}\right)\right| \, dx$$

does not exist.

I am not sure how this can be shown. For instance, I am not sure how to establish that the limit above exists without knowing what the limiting value is. My ideas so far are that I will need to take the substitution $u = \frac{1}{x}$. If I do this, I get

$$\lim_{c \to 0} \int_c^1 \frac{1}{x} \cdot \sin \left(\frac{1}{x}\right) \, dx = \lim_{c \to 0} \int_{\frac{1}{c}}^1 \frac{-\sin u}{u} \, du$$

and similarly

$$\lim_{c \to 0} \int_c^1 \left|\frac{1}{x} \cdot \sin \left(\frac{1}{x}\right)\right| \, dx = \lim_{c \to 0} \int_{\frac{1}{c}}^1 \left|\frac{-\sin u}{u} \, du\right|$$

How can I develop this idea further to show what I need to show? Thanks for any help.

Answer 1

Sorry if hints should be reserved for comments. I'm new to this and wasn't sure I should post this as a comment. Hope it helps!

Hint: focus on the fact that the original integrand gives a well-defined limit while the absolute value of that integrand doesn't give a limit. More specifically, notice that while the absolute value of the integrand is certainly monotonically decreasing on your interval, you can bound it below by a harmonic series in this case. On the other hand, when you don't take the absolute value, you have a cyclic function which $\textit{alternates}$ sign. Try to use this along with the fact that the absolute value of the integranded is monotonically decreasing to prove the desired result.

Answer 2

Try a substitution: $$ \lim_{c\to 0^+} \int_c^1 \frac{1}{x}\sin\left(\frac{1}{x}\right)\,dx $$ $$ \stackrel{z=1/x}{=}\lim_{r\to +\infty} \int_r^1 z\sin\left(z\right)(-z^{-2})\,dz $$ $$ =\lim_{r\to +\infty} \int_1^r\frac{\sin(z)}{z}\,dz $$This is improperly integrable, but not absolutely so. If we extend the range of integration to $[0,\infty)$, the result is well-known to equal $\pi/2$; on the other hand, if we take the absolute value of the integrand, the result diverges (a nice exercise: try comparing it to the harmonic series by integrating over intervals of length $\pi$, maybe over $[\pi,\infty)$).

Answer 3

To finish off the other answers: to show that $$ \lim_{r\to \infty} \int_1^r \frac{\sin z}{z} \, dz $$ exists, use integration by parts: $$ \int_1^r \frac{\sin z}{z} \, dz = - \left[\frac{\cos z}{z}\right]_1^r - \int_1^r \frac{\cos z}{z^2} \, dz .$$ You know $$ \left|\int_r^\infty \frac{\cos z}{z^2} \, dz \right| \le \int_r^\infty \frac{1}{z^2} \, dz = \frac1r \to 0. $$