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Question Number 63641 by aliesam last updated on 06/Jul/19

Commented by mathmax by abdo last updated on 06/Jul/19

$$let{A}_n={\int }_0^1\frac{\mid sin\left(n\pi x\right)\mid }{x^2+1}dx\Rightarrow {lim}_{n\to +\mathrm{\infty }}{A}_n={lim}_{n\to +\mathrm{\infty }}\sum _{k=0}^{n-1}{\int }_{\frac{k}{n}}^{\frac{k+1}{n}}\frac{\mid sin\left(n\pi x\right)\mid }{1+x^2}dx$$$$changementnx=tgive$$$${\int }_{\frac{k}{n}}^{\frac{k+1}{n}}\frac{\mid sin\left(n\pi x\right)\mid }{1+x^2}dx={\int }_k^{k+1}\frac{\mid sin\left(\pi t\right)\mid }{1+\frac{t^2}{n^2}}\frac{dt}{n}=n{\int }_k^{k+1}\frac{\mid sin\left(\pi t\right)\mid }{n^2+t^2}dt\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{A}_n={lim}_{n\to +\mathrm{\infty }}n{\int }_0^n\frac{\mid sin\left(\pi t\right)\mid }{n^2+t^2}dt={lim}_{n\to +\mathrm{\infty }}{\int }_R\frac{n\mid sin\left(\pi t\right)\mid }{n^2+t^2}{\chi }_{[0,n[}\left(t\right)dt$$$$let{f}_n\left(t\right)=\frac{n\mid sin\left(\pi t\right)\mid }{n^2+t^2}{\chi }_{[0,n[}\left(t\right)wehavefortinside[0,n[$$$$f_n\left(t\right)=\frac{n\mid sin\left(\pi t\right)\mid }{n^2(1+\frac{t^2}{n^2})}=\frac{1}{n}\times n^2\frac{\mid sin\left(\pi t\right)\mid }{1+\frac{t^2}{n^2}}\sim n(1-\frac{t^2}{n^2}+o\left(\frac{1}{n^2}\right))\mid sin\left(\pi t\right)\mid \Rightarrow$$$$n{\int }_0^n\frac{\mid sin\left(\pi t\right)\mid }{n^2+t^2}dt\sim {\int }_0^{n}n\mid sin\left(\pi t\right)\mid dt-\frac{1}{n}{\int }_0^n{t}^2\mid sin\left(\pi t\right)\mid dt+o\left(\frac{1}{n}\right)...becontinued...$$

Commented by mathmax by abdo last updated on 06/Jul/19

$$remsrklet{W}_n={\int }_{\frac{1}{n}}^1\frac{\mid sin\left(n\pi x\right)\mid }{x^2+1}dxwehave{lim}_{n\to +\mathrm{\infty }}{W}_n={lim}_{n\to +\mathrm{\infty }}{\int }_0^1\frac{\mid sin\left(n\pi x\right)\mid }{1+x^2}dx$$$$fromanotherside{W}_n⩽{\int }_{\frac{1}{n}}^1\frac{dx}{1+x^2}=arctan\left(1\right)-arctan\left(\frac{1}{n}\right)\Rightarrow {lim}_{n\to +\mathrm{\infty }}{W}_n⩽\frac{\pi }{4}$$

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