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I want to show that: $$ \lim_{n \rightarrow \infty} \int_0^{2\pi} \sin(x)^n \, dx=0 $$

and my idea was to use DCT (dominated convergenece theorem).

However, my textbook has the requirement that $u(x)=\lim_{n \rightarrow \infty} u_n(x)$ for all x. And this is not true for e.g. $x=-\frac{\pi}{2}$ which alternates between $1$ and $-1$.

Am I overseeing something using DCT or should another theorem be applied? I also thought about Fatou's lemma but the problem is that $\sin(x)^n \notin \mathcal{M}^+$.

Any hint would be appreciated.

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  • $\begingroup$ Note that $\lim\sin^n(x) = 0$ almost everywhere. $\endgroup$ – amsmath Oct 14 '19 at 17:37
  • $\begingroup$ Is it $(\sin x)^n$ or $\sin(x^n)$. Sorry but I am confused. $\endgroup$ – Rishi Oct 14 '19 at 17:48
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DCT only requires almost-everywhere convergence, which is the case in your limit. If the version in your textbook requires everywhere convergence, then here is a simple trick: $$\int_{0}^{2\pi}\sin^n(x)\,\mathrm{d}x=\int_{0}^{2\pi}u_n(x)\,\mathrm{d}x$$ where $$ u_n(x) = \begin{cases} \sin^n(x), & \text{if $|\sin(x)|<1$}\\0, & \text{otherwise} \end{cases}. $$ Now $u_n(x) \to 0$ everywhere and you are good to go.


For a more elementary trick, for any $\epsilon\in(0,\frac{\pi}{2})$ notice that

$$\left|\int_{0}^{2\pi}\sin(x)^n\,\mathrm{d}x\right|\leq 4\left(\epsilon + (\tfrac{\pi}{2}-\epsilon)\cos^n(\epsilon) \right). $$

Taking limsup as $n\to\infty$, this reduces to

$$\limsup_{n\to\infty} \left|\int_{0}^{2\pi}\sin(x)^n\,\mathrm{d}x\right|\leq 4\epsilon,$$

and then the desired claim follows by letting $\epsilon \downarrow 0$.

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$(\sin{x})^n \to 0$ a.e on $[0,2\pi]$

And $|\sin{x}|^n \leq 1 \in L^1([0,2\pi])$

So from Dominated Convergence theorem you have the conclusion.

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You could also use the DCT to show $\int_0^\pi\sin^nxdx\to0$, then use the squeeze theorem with $\left|\int_0^{2\pi}\sin^nxdx\right|\le2\int_0^\pi\sin^nxdx$.

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