How can I solve the integral $\lim_{n\rightarrow \infty} \int_0^{2022\pi} \sin^n(x) dx$?

Asked Aug 29 '22 at 08:54

Active Aug 29 '22 at 08:54

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I want to compute the integral $$\lim_{n\rightarrow \infty} \int_0^{2022\pi} \sin^n(x) dx$$

Somehow I struggle a bit. At this point now I have the following attempt.

Let me denote $I_n= \int_0^{2022\pi} \sin^n(x) dx$. Then after a small computation I get a reduction formula: $$I_n=-\frac{\sin^{n-1}(x)\cos(x)}{n}+\frac{n-1}{n}I_{n-2}$$ hence $$\lim_{n\rightarrow \infty}I_n=\lim_{n\rightarrow \infty} I_{n-2}$$ by the same reduction formula I get that $$\lim_{n\rightarrow \infty} I_n=\lim_{n\rightarrow \infty} I_{n-4}$$ But now I don't see why this could be useful for my computations.

Can maybe someone help me?

asked Aug 29 '22 at 08:54

user123234

2,885

This was just asked last week, let me find it – FShrike Aug 29 '22 at 08:57
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Do you know Lebesgue dominated convergence theorem ? – mathcounterexamples.net Aug 29 '22 at 08:57
On that linked question there is an answer which explains how to approach the problem using simple technology – FShrike Aug 29 '22 at 08:58
Wave. It is better to instead use the recurrences to bring $I_n$ as a product of many recurrence factors * $I_1$ or $I_2$. – FShrike Aug 29 '22 at 08:59
@mathcounterexamples.net yes I know this theorem in order to swap limit and integration. But I mean sinus does not converge? – user123234 Aug 29 '22 at 09:16
@FShrike I tool a look at your tagged post, so you would first try to find a product of factors and then take the limit as you wrote in the tagged post? – user123234 Aug 29 '22 at 09:17
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@Wave The sequence ${\sin^n x}$ converges to zero for almost any $x \in [0,2022 \pi]$. To be specific for any $x \notin \frac{\pi}{2} + \mathbb Z \pi$. Hence, you can easily apply Lebesgue dominated convergence theorem. – mathcounterexamples.net Aug 29 '22 at 09:20
The product method is one way of doing it. It’s a little harder, but it’s pretty and it involves the Wallis products which are related to some very nice identities for $\pi$. – FShrike Aug 29 '22 at 09:23
@mathcounterexamples.net ah I see, so I will try it! Thanks – user123234 Aug 29 '22 at 09:23
@FShrike perfect thanks a lot! – user123234 Aug 29 '22 at 09:23
This answer might help. Note that over a complete period, $\int_0^{2\pi}\cos^n(x),\mathrm{d}x=\int_0^{2\pi}\sin^n(x),\mathrm{d}x$ – robjohn Aug 29 '22 at 15:09

How can I solve the integral $\lim_{n\rightarrow \infty} \int_0^{2022\pi} \sin^n(x) dx$?

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