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Question Number 154058 by mnjuly1970 last updated on 13/Sep/21

Answered by qaz last updated on 14/Sep/21

$$\mathrm{I}={\int }_0^1\frac{\mathrm{xsin}\left(\mathrm{lnx}\right)}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$=-{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-2\mathrm{u}}\sin \mathrm{u}}{1+{\mathrm{e}}^{-2\mathrm{u}}}\mathrm{du}$$$$=-{\int }_0^{\mathrm{\infty }}\frac{\sin \mathrm{u}}{{\mathrm{e}}^{2\mathrm{u}}+1}\mathrm{du}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-2\mathrm{nu}}\sin \mathrm{udu}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{{4\mathrm{n}}^2+1}$$$$=\frac{\pi }{4}\mathrm{csch}\left(\frac{\pi }{2}\right)-\frac{1}{2}$$$$--------------$$$$\mathrm{J}={\int }_0^{\pi /2}{\ln }^{2021}\left(\tan \mathrm{x}\right)\mathrm{dx}$$$$={\int }_{+\mathrm{\infty }}^{-\mathrm{\infty }}\frac{{\mathrm{u}}^{2021}{\mathrm{e}}^{-\mathrm{u}}}{1+{\mathrm{e}}^{-2\mathrm{u}}}\mathrm{du}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{2021}{\mathrm{e}}^{-\mathrm{u}}}{1+{\mathrm{e}}^{-2\mathrm{u}}}\mathrm{du}$$$$=2\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{u}}^{2021}{\mathrm{e}}^{-(2\mathrm{n}+1)\mathrm{u}}\mathrm{du}$$$$=2\cdot 2021!\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{{(2\mathrm{n}+1)}^{2022}}$$$$=2\cdot \beta \left(2022\right)\cdot 2021!$$

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