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Question Number 154143 by mnjuly1970 last updated on 14/Sep/21

Answered by phanphuoc last updated on 14/Sep/21

$$u=-lnx\to e^{-u}=x\to -e^{-u}du=dx$$$$\mathrm{\Omega }={\int }_{\mathrm{\infty }}^0\frac{e^{-x}sinx}{1+e^{-2x}}(-e^{-x}dx)={\int }_0^{\mathrm{\infty }}\frac{e^{-2x}sinxdx}{1+e^{-2x}}$$$$={\int }_0^{\mathrm{\infty }}{e}^{-2x}sinx\mathrm{\Sigma }{(-1)}^{\mathit{n}}{\mathit{e}}^{-2\mathit{n}\mathit{x}}\mathit{d}\mathit{x}=$$$$\mathrm{\Sigma }{(-1)}^n{\int }_0^{\mathrm{\infty }}{e}^{-2x(n+1)}{sinxdx}_{=}$$

Answered by puissant last updated on 14/Sep/21

$$\mathrm{\Omega }={\int }_0^1\frac{xsin\left(lnx\right)}{1+x^2}dx$$$$u=-lnx\to x=e^{-u}\to dx=-e^{-u}du$$$$\mathrm{\Omega }=-{\int }_{\mathrm{\infty }}^0\frac{e^{-u}sinu}{1+e^{-2u}}(-e^{-u}du)=-{\int }_0^{\mathrm{\infty }}\frac{e^{-2u}sinu}{1+e^{-2u}}du$$$$=-{\int }_0^{\mathrm{\infty }}\frac{sinu}{1+e^{2u}}du={\int }_0^{\mathrm{\infty }}\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^n{e}^{-2nu}sinudu$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^n{\int }_0^{\mathrm{\infty }}{e}^{-2nu}sinudu=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^n\times \frac{1}{4n^2+1}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{4n^2+1}=\frac{\pi }{4}csch\left(\frac{\pi }{2}\right)-\frac{1}{2}..$$$$$$$$\therefore \because \mathrm{\Omega }=\frac{\pi csch\left(\frac{\pi }{2}\right)-2}{4}...$$$$$$$$................\mathcal{L}epuissant..............$$

Commented by mnjuly1970 last updated on 15/Sep/21

$$grate..$$

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