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Question Number 205237 by universe last updated on 13/Mar/24

Answered by Berbere last updated on 13/Mar/24

$$n^2+x^2⩾n^2$$$$\frac{x}{1+x}⩽1\Rightarrow \frac{nx{\tan }^{-1}\left(x\right)}{(1+x)(n^2+x^2)}⩽n.1.\frac{{\tan }^{-1}\left(x\right)}{n^2+x^2}=n\frac{{\tan }^{-1}\left(x\right)}{n^2+x^2}$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{nx{\tan }^{-1}\left(x\right)}{(1+x)(n^2+x^2)}dx⩽{\int }_0^{\mathrm{\infty }}n\frac{{\tan }^{-1}\left(x\right)}{n^2+x^2}⩽\frac{n\pi }{2}{\int }_0^{\mathrm{\infty }}\frac{dx}{n^2+x^2}$$$$=\frac{n\pi }{2}{\left[\frac{1}{n}{\tan }^{-1}\left(\frac{x}{n}\right)\right]}_0^{\mathrm{\infty }}=\frac{{\pi }^2}{4}$$$$wecanexchange\int andlim$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }{\int }_0^{\mathrm{\infty }}\frac{nx{\tan }^{-1}\left(x\right)}{(1+x)(n^2+x^2)}dx={\int }_0^{\mathrm{\infty }}\underset{n\to \mathrm{\infty }}{\lim }\frac{x{\tan }^{-1}\left(x\right)}{1+x}\frac{n}{n^2+x^2}dx=0$$

Commented by universe last updated on 13/Mar/24

$$thankssir$$

Commented by Berbere last updated on 13/Mar/24

$$withePleasur$$

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