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Greetings I want to evaluate $\displaystyle\lim_{n\rightarrow\infty} \int_{0}^{\pi} \frac {\sin x}{1+ \cos^2 (nx)} dx$.

Here is my try: We have that $x\in[0,\pi]$ so $$\cos(n\pi)\le \cos(nx) \le 1.$$ Here I am not sure, but if it's correct then it gives: $$\frac{1}{2}\le\frac{1}{1+\cos^2(nx)}\le \frac{1}{1+\cos^2(n\pi)},$$ giving $$\lim_{n\rightarrow\infty} \frac{1}{2}\int_0^{\pi} \sin x dx \le \lim_{n\rightarrow\infty} \int_{0}^{\pi} \frac {\sin x}{1+ \cos^2 (nx)} dx \le \lim_{n\rightarrow\infty} \int_{0}^{\pi} \frac {\sin x}{1+ \cos^2 (n\pi)} dx.$$ Since $$\int_0^{\pi} \sin x dx =2$$ By squeeze theorem we may conclude that $\displaystyle \lim_{n\rightarrow\infty} \int_{0}^{\pi} \frac {\sin x}{1+ \cos^2 (nx)} dx=1$. Could you help me evaluate this, if it's wrong?

GNUSupporter 8964民主女神 地下教會
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Zacky
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1 Answers1

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One approach can be as follows: \begin{align} \int_0^{\pi}\frac{\sin x}{1+\cos^2nx}\,dx&\stackrel{(1)}{=} \frac{1}{n}\int_0^{n\pi}\frac{\sin(y/n)}{1+\cos^2y}\,dy= \frac{1}{n}\sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi}\frac{\sin(y/n)}{1+\cos^2y}\,dy\stackrel{(2)}{=}\\ &\stackrel{(2)}{=}\frac{1}{n}\sum_{k=0}^{n-1}\int_{0}^{\pi}\frac{\sin\big(\frac{t+k\pi}{n}\big)}{1+\cos^2t}\,dt= \frac{1}{\pi}\int_0^{\pi}\frac{1}{1+\cos^2t}\Big[\frac{\pi}{n}\sum_{k=0}^{n-1}\sin\Big(\frac{t+k\pi}{n}\Big)\Big]\,dt\stackrel{(3)}{\to}\\ &\stackrel{(3)}{\to}\frac{1}{\pi}\int_0^{\pi}\frac{1}{1+\cos^2t}\,dt\cdot\int_0^{\pi}\sin t\,dt=\sqrt{2}. \end{align} Explanations:

(1) change $y=nx$,

(2) change $t=y-k\pi$ and use that $\cos^2t$ is $\pi$-periodic,

(3) Riemann summa limit and dominated convergence theorem.

For the last integration: rewrite $$ \cos^2x+1=\cos^2x+\cos^2x+\sin^2x=\cos^2x(2+\tan^2x), $$ that gives \begin{align} \int_0^\pi\frac{1}{1+\cos^2x}\,dx&=2\int_0^{\pi/2}\frac{1}{1+\cos^2x}\,dx=2\int_0^{\pi/2}\frac{1}{2+\tan^2x}\,d\tan x=2\int_0^\infty\frac{1}{2+t^2}\,dt=\\ &=[t=s\sqrt{2}]=\sqrt{2}\int_0^\infty\frac{1}{1+s^2}\,ds=\sqrt{2}\frac{\pi}{2}=\frac{\pi}{\sqrt{2}}. \end{align}

A.Γ.
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