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Question Number 74776 by chess1 last updated on 30/Nov/19

Answered by mind is power last updated on 01/Dec/19

$$\mathrm{we}\mathrm{show}$$$${\int }_1^{\mathrm{n}}{\left(\frac{{\tan }^{-}\left(\mathrm{x}\right)}{{\tan }^{-}\left(\mathrm{x}\right)-\mathrm{x}}\right)}^2\mathrm{dx}<\mathrm{\infty }$$$$\Rightarrow \lim =0$$$$\mathrm{\forall }\mathrm{x}\in {\mathbb{R}}^{+}{\tan }^{-}\left(\mathrm{x}\right)⩽\mathrm{x}$$$$\frac{1}{1+{\mathrm{x}}^2}⩽1\Rightarrow {\int }_0^{\mathrm{t}}\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}⩽{\int }_0^{\mathrm{t}}1\mathrm{dx}\Rightarrow {\tan }^{-}\left(\mathrm{t}\right)⩽\mathrm{t}$$$$\mathrm{for}\mathrm{t}>0\mathrm{we}\mathrm{have}{\tan }^{-}\left(\mathrm{t}\right)<\mathrm{t}\Rightarrow {\int }_1^{\mathrm{n}}\frac{{\left({\tan }^{-}\left(\mathrm{x}\right)\right)}^2}{{({\tan }^{-}\left(\mathrm{x}\right)-\mathrm{x})}^2}\mathrm{dx}\mathrm{is}\mathrm{well}\mathrm{definde}$$$${\left(\frac{{\tan }^{-}\left(\mathrm{x}\right)}{({\tan }^{-\mathrm{x}}-\mathrm{x})}\right)}^2\sim \frac{{\pi }^2}{{4\mathrm{x}}^2}\mathrm{as}\mathrm{x}\to +\mathrm{\infty }\mathrm{wich}\mathrm{is}\mathrm{integrable}$$$$\Rightarrow {\int }_1^{+\mathrm{\infty }}{\left(\frac{{\tan }^{-}\left(\mathrm{x}\right)}{({\tan }^{-}\left(\mathrm{x}\right)-\mathrm{x})}\right)}^2\mathrm{dx}<\mathrm{\infty }$$$$\lim =0$$

Commented by chess1 last updated on 01/Dec/19

$$\mathrm{thank}\mathrm{you}$$

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