Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I think I am getting a little better at these MCT, DCT-type exercises. The issue is to show/prove the existence and finiteness (if they apply) to the following function:

$$f(x)=\sin \left(\dfrac1{x^2} \right)$$

Where applicable I want to show as rigorously as possible the justification for existence, etc. For example, it is not enough to say simply, "this exists by MCT"; rather, I need to show that the conditions of measurability, monotone-increasing, etc. are met.

Work/attempt at a solution:

First, although I have a limited experience with this type of integral, I believe the function can be re-written as follows:


If I have this correctly (I attempted to extrapolate from an integral table I found regarding $\sin\left(\dfrac1x \right)$, but please correct me if I am wrong), I can utilize Dominated Convergence to show the integral exists. That is,

If $f_1,f_2,...$ are measurable functions and $|f_n|\le g$ for $\mu$-integrable function, $g$, and if $f_n\to f$, $\mu$-a.e, then $\lim_{n\to\infty}\int f_n d\mu\to\int fd\mu$.

First, I need to show measurability. I believe that continuity on the interval in question, $(0,\infty)$, is sufficient to show measurability, correct?

Next, define

$$S_n(x):=\sum_{n=0}^{t}\frac{(-1)^n}{x^{2(2n+1)}(2n+1)!}\to f(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{x^{2(2n+1)}(2n+1)!}$$

Then, define

$g(x)=\dfrac1x\ge |f_n|,\forall x>0$

and therefore conclude by Dominated Convergence Theorem, that $\displaystyle \int_{(0,\infty)}\sin \left(\dfrac1{x^2} \right)d\mu$ exists.

Please comment, add answers, critique, tell me where I made mistakes, etc. I am learning this on my own so any and all improvements are welcome.

From here I need to show whether this integral is finite or infinite. Is there a way to show this without an explicit calculation of the integral?

share|cite|improve this question
up vote 5 down vote accepted

$\sin(1/x^2)$ is continuous in $(0,\infty)$ and hence it is also measurable.

Split your domain $(0,\infty)$ into $(0,1] \cup (1,\infty)$.

In the interval $(0,1]$, bound $\sin(1/x^2)$ by $1$ and in the interval $(1,\infty)$, bound $\sin(1/x^2)$ by $1/x^2$ and then argue out.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.