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Question Number 165271 by mnjuly1970 last updated on 28/Jan/22

$$$$$$\mathrm{L}et,f:[0,1]\to \mathbb{R}isacontinuous$$$$function,provethat:$$$${lim}_{n\to \mathrm{\infty }}{\int }_0^1\frac{nf\left(x\right)}{1+n^2{x}^2}dx=\frac{\pi }{2}f\left(0\right)$$$$---proof---$$$${\mathrm{S}}_n=[{\int }_0^{\frac{1}{\sqrt{n}}}\frac{n.f\left(x\right)}{1+n^2{x}^2}dx={\mathrm{\Omega }}_n]+[{\int }_{\frac{1}{\sqrt{n}}}^1\frac{n.f\left(x\right)}{1+n^2{x}^2}dx={\mathrm{\Phi }}_n]$$$${\mathrm{\Omega }}_n\underset{\mathrm{\exists }{t}_n\in (0,\frac{1}{\sqrt{n}})}{\stackrel{\mathrm{M}ean\mathrm{V}alue\mathrm{T}heorem\left(first\right)}{=}}f\left(t_n\right){\int }_0^{\frac{1}{\sqrt{n}}}\frac{n}{1+n^2{x}^2}dx$$$$=f\left(t_n\right)\left({tan}^{-1}\left(\sqrt{n}\right)\right)$$$${lim}_{n\to \mathrm{\infty }}\left({\mathrm{\Omega }}_n\right)=\frac{\pi }{2}f\left({lim}_{n\to \mathrm{\infty }}\left(t_n\right)\right)=\frac{\pi }{2}f\left(0\right)$$$${\mathrm{\Phi }}_n={\int }_{\frac{1}{\sqrt{n}}}^1\frac{n.f\left(x\right)}{1+n^2{x}^2}dx\underset{\mathrm{\exists }\mathrm{M}>0}{\stackrel{fisbounded}{\Rightarrow }}\mid {\mathrm{\Phi }}_n\mid ⩽\mathrm{M}.{\int }_{\frac{1}{\sqrt{n}}}^1\frac{n}{1+n^2{x}^2}dx$$$$\Rightarrow \mid {\mathrm{\Phi }}_n\mid ⩽\mathrm{M}.({tan}^{-1}\left(n\right)-{tan}^{-1}\left(\sqrt{n}\right))$$$${lim}_{n\to \mathrm{\infty }}\mid {\mathrm{\Phi }}_n\mid =0\Rightarrow {lim}_{n\to \mathrm{\infty }}{\mathrm{\Phi }}_n=0$$$$\therefore {lim}_{n\to \mathrm{\infty }}\left({\mathrm{S}}_n\right)=\frac{\pi }{2}f\left(0\right)◼m.n$$$$$$

Answered by mindispower last updated on 28/Jan/22

$$niceResultThanxsir$$$$$$

Commented by mnjuly1970 last updated on 28/Jan/22

$$thankyousomuchsir...$$$$$$

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