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For $n\in \mathbb{N}$, let $f_n : \mathbb{R} \rightarrow \mathbb{R}$ be given by $f_n(x)=2|x|(1-sin^{2n}(\frac{\pi}{2}x))$. Let $\beta = lim_{n\rightarrow \infty}\int_{-9}^{44}f_n(x)dx$.

The question is to find the value of $\beta$ using either Fatou's Lemma(FL), the Dominated Convergence Theorem(DCT) or Monotone Convergence Theorem(MCT).

I tried to use the DCT first, however I seem to be unable to find the pointwise limit of $f_n$ due to the $sin$ term. Then I tried to apply the MCT, however due to the same term I am unable to find a function to which $f_n$ increases. Lastly, I think FL is not suitable for this problem since it focusus on $lim_{n\rightarrow \infty}inf$.

I would appreciate a hint on finding the pointwise limit, or funtion to which $f_n$ increases such that I can use the DCT or MCT.

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The DCT is the right way to go. Some hints to help you get started:

  1. Since $-1 \le \sin(\tfrac{\pi}{2}) \le 1$ and $2n$ is an even integer, we have $0 \le \sin^{2n}(\tfrac{\pi}{2}x) \le 1$. Can you use this to bound the integrand?

  2. If $x$ is not an integer, then $-1 < \sin(\tfrac{\pi}{2}x) < 1$, and so $\displaystyle\lim_{n \to \infty}\sin^{2n}(\tfrac{\pi}{2}x) = 0$. Can you find the pointwise limit for non-integer values of $x$?

  3. The set of values of $x \in [-9,44]$ where $x$ is an integer has measure $0$.

Note that you can also use the MCT if you can show that $2|x|(1-\sin^{2n}(\tfrac{\pi}{2}x))$ increases as $n \to \infty$.

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  • $\begingroup$ Great answer including point 3. $\endgroup$ – Olba12 Dec 19 '17 at 9:46
  • $\begingroup$ Well, in that case the integrand is bounded by $2|x|$. I would also say that for non-integer x this is the pointwise limit since $f_n(x)$ goes to $2|x|$ for every $x$ as $n$ tends to infinity. So for x integer we use additivity of an integral and conclude your point 3. Then we consider non-integer x and compute the integral where we use the pointwise limit. Is this correct? Thanks $\endgroup$ – LMATH Dec 19 '17 at 10:22
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Hint:

For some $x, \sin$ will be $1$, and for some it will be $-1$, thus depending on $x$ you have to cases, either of the form $(+-1)^{2n}\to 1$, the other case is when $ |\sin|$ is not $1$, i.e some a between $-1<a<1$ and then $a^{2n} \to 0$

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