# Show that $\lim_{n \rightarrow \infty} \int_0^{2\pi} \sin(x)^n \, dx=0$

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.

• Note that $\lim\sin^n(x) = 0$ almost everywhere. – amsmath Oct 14 '19 at 17:37
• Is it $(\sin x)^n$ or $\sin(x^n)$. Sorry but I am confused. – Rishi Oct 14 '19 at 17:48

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 $$\epsilon \downarrow 0$$.

$$(\sin{x})^n \to 0$$ 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.

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