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Question Number 133369 by mnjuly1970 last updated on 21/Feb/21

$$$$$$calculus\left(2\right).....$$$$\mathit{\varphi }={\int }_1^2\left(\frac{ln\left(\frac{x+2}{x+1}\right)}{x}\right)dx=???$$$$$$

Commented by Dwaipayan Shikari last updated on 21/Feb/21

$$\frac{{log}^2\left(2\right)}{2}$$

Commented by mnjuly1970 last updated on 21/Feb/21

$$correct...$$

Answered by mathmax by abdo last updated on 21/Feb/21

$$\mathrm{\Phi }={\int }_1^2\frac{1}{\mathrm{x}}\ln \left(\frac{\mathrm{x}+2}{\mathrm{x}+1}\right)\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\frac{1}{\mathrm{x}}=\mathrm{t}\Rightarrow$$$$\mathrm{\Phi }={\int }_1^{\frac{1}{2}}\mathrm{tln}\left(\frac{\frac{1}{\mathrm{t}}+2}{\frac{1}{\mathrm{t}}+1}\right)(-\frac{\mathrm{dt}}{{\mathrm{t}}^2})={\int }_{\frac{1}{2}}^1\frac{1}{\mathrm{t}}\ln \left(\frac{1+2\mathrm{t}}{1+\mathrm{t}}\right)\mathrm{dt}$$$$={\int }_{\frac{1}{2}}^1\frac{\ln (1+2\mathrm{t})}{\mathrm{t}}\mathrm{dt}-{\int }_{\frac{1}{2}}^1\frac{\ln (1+\mathrm{t})}{\mathrm{t}}\mathrm{dt}=\mathrm{H}-\mathrm{K}$$$$\mathrm{K}={[\mathrm{lnt}.\ln (1+\mathrm{t})]}_{\frac{1}{2}}^1-{\int }_{\frac{1}{2}}^1\frac{\ln \left(\mathrm{t}\right)}{\mathrm{t}+1}\mathrm{dt}$$$$=-\ln \left(\frac{1}{2}\right)\ln \left(\frac{3}{2}\right)-{\int }_{\frac{1}{2}}^1\frac{\ln \left(\mathrm{t}\right)}{1+\mathrm{t}}\mathrm{dt}\mathrm{and}$$$${\int }_{\frac{1}{2}}^1\frac{\ln \left(\mathrm{t}\right)}{1+\mathrm{t}}\mathrm{dt}={\int }_{\frac{1}{2}}^1\ln \left(\mathrm{t}\right)\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}}\mathrm{dt}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_{\frac{1}{2}}^1{\mathrm{t}}^{\mathrm{n}}\ln \left(\mathrm{t}\right)\mathrm{dt}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{u}}_{\mathrm{n}}$$$${\mathrm{u}}_{\mathrm{n}}={\left[\frac{{\mathrm{t}}^{\mathrm{n}+1}}{\mathrm{n}+1}\mathrm{lnt}\right]}_{\frac{1}{2}}^1-{\int }_{\frac{1}{2}}^1\frac{{\mathrm{t}}^{\mathrm{n}+1}}{\mathrm{n}+1}\frac{\mathrm{dt}}{\mathrm{t}}=-\ln \left(\frac{1}{2}\right)\frac{1}{(\mathrm{n}+1)2^{\mathrm{n}+1}}-\frac{1}{\mathrm{n}+1}{\int }_{\frac{1}{2}}^1{\mathrm{t}}^{\mathrm{n}}\mathrm{dt}$$$$=\frac{\ln \left(2\right)}{(\mathrm{n}+1)2^{\mathrm{n}}}-\frac{1}{{(\mathrm{n}+1)}^2}(1-\frac{1}{2^{\mathrm{n}+1}})\Rightarrow$$$${\int }_{\frac{1}{2}}^1\frac{\mathrm{lnt}}{1+\mathrm{t}}\mathrm{dt}=\ln 2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{(\mathrm{n}+1)2^{\mathrm{n}}}-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{(\mathrm{n}+1)}^2}+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{(\mathrm{n}+1)}^2.2^{\mathrm{n}+1}}$$$$\mathrm{rest}\mathrm{calculus}\mathrm{of}\mathrm{those}\mathrm{series}...$$$$\mathrm{H}={\int }_{\frac{1}{2}}^1\frac{\ln (1+2\mathrm{t})}{\mathrm{t}}\mathrm{dt}{=}_{2\mathrm{t}=\mathrm{y}}2{\int }_1^2\frac{\ln (1+\mathrm{y})}{\mathrm{y}}\frac{\mathrm{dy}}{2}$$$${=}_{\mathrm{y}=\mathrm{u}+1}{\int }_{\mathrm{o}}^1\frac{\ln (\mathrm{u}+2)}{\mathrm{u}+1}\mathrm{du}={\int }_0^1\ln (2+\mathrm{u})\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{u}}^{\mathrm{n}}\mathrm{du}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_0^1{\mathrm{u}}^{\mathrm{n}}\ln (2+\mathrm{u})\mathrm{du}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{v}}_{\mathrm{n}}$$$${\mathrm{v}}_{\mathrm{n}}={\left[\frac{{\mathrm{u}}^{\mathrm{n}+1}}{\mathrm{n}+1}\ln (2+\mathrm{u})\right]}_0^1-{\int }_0^1\frac{{\mathrm{u}}^{\mathrm{n}+1}}{\mathrm{n}+1}\frac{\mathrm{du}}{2+\mathrm{u}}$$$$=\frac{\ln \left(3\right)}{\mathrm{n}+1}-\frac{1}{\mathrm{n}+1}{\int }_0^1\frac{{\mathrm{u}}^{\mathrm{n}+1}}{\mathrm{u}+2}\mathrm{du}=....\mathrm{be}\mathrm{continued}....$$

Answered by mnjuly1970 last updated on 22/Feb/21

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