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Question Number 151144 by mnjuly1970 last updated on 18/Aug/21

Answered by qaz last updated on 18/Aug/21

$$\frac{1}{\mathrm{n}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\tan }^{-1}\left({\mathrm{e}}^{\mathrm{x}}\right)}{\cosh \left(\frac{\mathrm{x}}{\mathrm{n}}\right)}\mathrm{dx}$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\tan }^{-1}\left({\mathrm{e}}^{\mathrm{nx}}\right)}{\cosh \mathrm{x}}\mathrm{dx}$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\tan }^{-1}\left({\mathrm{e}}^{-\mathrm{nx}}\right)}{\cosh \mathrm{x}}\mathrm{dx}$$$$=\frac{\pi }{4}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{\cosh \mathrm{x}}$$$$=\pi {\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{x}}}{1+{\mathrm{e}}^{-2\mathrm{x}}}\mathrm{dx}$$$$=-\pi {\tan }^{-1}{\mathrm{e}}^{-\mathrm{x}}{\mid }_0^{\mathrm{\infty }}$$$$=\frac{{\pi }^2}{4}$$

Commented by mnjuly1970 last updated on 18/Aug/21

$$excellentmasterbravo...$$

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