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Question Number 98537 by student work last updated on 14/Jun/20

Commented by student work last updated on 14/Jun/20

$$\mathrm{helpe}\mathrm{me}$$

Answered by mathmax by abdo last updated on 14/Jun/20

$$\mathrm{at}\mathrm{form}\mathrm{of}\mathrm{serie}\mathrm{I}=\int \frac{\mathrm{lnx}}{1+\mathrm{x}}\mathrm{dx}\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}\frac{\ln \left(\mathrm{t}\right)}{1+\mathrm{t}}\mathrm{dt}(\mathrm{x}>0)$$$$\mathrm{case}1\mathrm{o}<\mathrm{x}<1\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}\ln \left(\mathrm{t}\right)\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}}\mathrm{dt}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_0^{\mathrm{x}}{\mathrm{t}}^{\mathrm{n}}\ln \left(\mathrm{t}\right)\mathrm{dt}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{A}}_{\mathrm{n}}\mathrm{by}\mathrm{parts}$$$${\mathrm{A}}_{\mathrm{n}}={\left[\frac{{\mathrm{t}}^{\mathrm{n}+1}}{\mathrm{n}+1}\ln \left(\mathrm{t}\right)\right]}_0^{\mathrm{x}}-{\int }_0^{\mathrm{x}}\frac{{\mathrm{t}}^{\mathrm{n}}}{\mathrm{n}+1}\mathrm{dt}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}\mathrm{lnx}-\frac{{\mathrm{x}}^{\mathrm{n}+1}}{{(\mathrm{n}+1)}^2}\Rightarrow$$$$\mathrm{I}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+1}{\mathrm{x}}^{\mathrm{n}+1}\ln \left(\mathrm{x}\right)-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{(\mathrm{n}+1)}^2}{\mathrm{x}}^{\mathrm{n}+1}$$$$\mathrm{case}2\mathrm{x}>1\Rightarrow \frac{1}{\mathrm{x}}<1\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{t}=\frac{1}{\mathrm{u}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)={\int }_0^1\frac{\ln \left(\mathrm{t}\right)}{1+\mathrm{t}}\mathrm{dt}+{\int }_1^{\mathrm{x}}\frac{\ln \left(\mathrm{t}\right)}{1+\mathrm{t}}\mathrm{dt}(\to \mathrm{t}=\frac{1}{\mathrm{u}})$$$$={\int }_0^1\frac{\ln \left(\mathrm{t}\right)}{1+\mathrm{t}}\mathrm{dt}-{\int }_{\frac{1}{\mathrm{x}}}^1\frac{-\mathrm{lnu}}{1+\frac{1}{\mathrm{u}}}\times \frac{-\mathrm{du}}{{\mathrm{u}}^2}$$$$={\int }_0^1\frac{\ln \left(\mathrm{t}\right)}{1+\mathrm{t}}\mathrm{dt}-{\int }_{\frac{1}{\mathrm{x}}}^1\frac{\mathrm{lnu}}{1+{\mathrm{u}}^2}\mathrm{du}={\int }_0^1\ln \left(\mathrm{t}\right)\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}}\mathrm{dt}-{\int }_{\frac{1}{\mathrm{x}}}^1\mathrm{lnt}\left(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{t}}^{2\mathrm{n}}\right)\mathrm{dt}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_0^1{\mathrm{t}}^{\mathrm{n}}\mathrm{lnt}\mathrm{dt}-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_{\frac{1}{\mathrm{x}}}^1{\mathrm{t}}^{2\mathrm{n}}\mathrm{lnt}\mathrm{dt}$$$${\int }_0^1{\mathrm{t}}^{\mathrm{n}}\ln \left(\mathrm{t}\right)\mathrm{dt}={\left[\frac{{\mathrm{t}}^{\mathrm{n}+1}}{\mathrm{n}+1}\mathrm{lnt}\right]}_0^1-{\int }_0^1\frac{{\mathrm{t}}^{\mathrm{n}}}{\mathrm{n}+1}\mathrm{dt}=-\frac{1}{{(\mathrm{n}+1)}^2}\Rightarrow$$$${\int }_0^1\frac{\mathrm{lnt}}{1+\mathrm{t}}\mathrm{dt}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}+1}}{{(\mathrm{n}+1)}^2}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^2}=(2^{1-2}-1)\xi \left(2\right)=-\frac{{\pi }^2}{12}$$$${\int }_{\frac{1}{\mathrm{x}}}^1{\mathrm{t}}^{2\mathrm{n}}\ln \left(\mathrm{t}\right)\mathrm{dt}={\left[\frac{{\mathrm{t}}^{2\mathrm{n}+1}}{2\mathrm{n}+1}\ln \left(\mathrm{t}\right)\right]}_{\frac{1}{\mathrm{x}}}^1-{\int }_{\frac{1}{\mathrm{x}}}^1\frac{{\mathrm{t}}^{2\mathrm{n}}}{2\mathrm{n}+1}\mathrm{dt}$$$$=\frac{\ln \left(\mathrm{x}\right)}{(2\mathrm{n}+1){\mathrm{x}}^{2\mathrm{n}+1}}-\frac{1}{{(2\mathrm{n}+1)}^2}(1-\frac{1}{{\mathrm{x}}^{2\mathrm{n}+1}})\Rightarrow$$$${\int }_{\frac{1}{\mathrm{x}}}^1\frac{\mathrm{lnu}}{1+{\mathrm{u}}^2}\mathrm{du}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}\ln \left(\mathrm{x}\right)}{(2\mathrm{n}+1){\mathrm{x}}^{2\mathrm{n}+1}}-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{(2\mathrm{n}+1)}^2}(\to \mathrm{k}\mathrm{catalan}\mathrm{constant})$$$$+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{(2\mathrm{n}+1)}^2{\mathrm{x}}^{2\mathrm{n}+1}}$$$$$$

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