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Question Number 107202 by hgrocks last updated on 09/Aug/20

Answered by EmericGent last updated on 09/Aug/20

$${\int }_0^{1/2}\frac{ln1-t}{t}dt=-{\int }_0^{1/2}\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{t^{k-1}}{k}dt$$$${=-\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{t^k}{k^2}]}_0^{1/2}=-\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^2{2}^k}=-S$$$$u=ln(1-t)\Rightarrow u^{\prime }=\frac{1}{t-1}$$$$v^{\prime }=\frac{1}{t}\Rightarrow v=lnt$$$${{\int }_0^{1/2}\frac{ln1-t}{t}dt=ln(1-t)ln\left(t\right)]}_0^{1/2}+{\int }_0^{1/2}\frac{lnt}{1-t}dt$$$$u=1-t\Rightarrow -du=dt$$$${\int }_0^{1/2}\frac{lnt}{1-t}dt={\int }_{1/2}^1\frac{ln(1-u)}{u}du$$$$={\int }_0^1\frac{ln(1-t)}{t}dt-{\int }_0^{1/2}\frac{ln1-t}{t}dt$$$$\iff {\int }_0^{1/2}\frac{ln1-t}{t}dt={ln}^{2}2+{\int }_0^1\frac{ln(1-t)}{t}dt-{\int }_0^{1/2}\frac{ln1-t}{t}dt$$$$\iff 2{\int }_0^{1/2}\frac{ln1-t}{t}dt={ln}^{2}2+{\int }_0^1\frac{ln(1-t)}{t}dt$$$${\int }_0^1\frac{ln(1-t)}{t}dt=-{\int }_0^1\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{t^{k-1}}{k}dt=-\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^2}=\frac{-{\pi }^2}{6}$$$$\iff 2{\int }_0^{1/2}\frac{ln1-t}{t}dt={ln}^{2}2-\frac{{\pi }^2}{6}$$$$\iff -2S={ln}^{2}2-\frac{{\pi }^2}{6}$$$$\iff S=\frac{{\pi }^2}{12}-\frac{{ln}^{2}2}{2}$$

Commented by hgrocks last updated on 09/Aug/20

GREAT !! THANKS A LOT ����

Answered by mathmax by abdo last updated on 10/Aug/20

$$\mathrm{S}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2}{\left(\frac{1}{2}\right)}^{\mathrm{n}}\Rightarrow \mathrm{S}=\mathrm{f}\left(\frac{1}{2}\right)\mathrm{with}\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{n}}^2}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}-1}}{\mathrm{n}}=\frac{1}{\mathrm{x}}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}=-\frac{\ln (1-\mathrm{x})}{\mathrm{x}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=-{\int }_0^{\mathrm{x}}\frac{\ln (1-\mathrm{t})}{\mathrm{t}}\mathrm{dt}+\mathrm{c}\mathrm{we}\mathrm{have}\mathrm{c}=\mathrm{f}\left(0\right)=0\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=-{\int }_0^{\mathrm{x}}\frac{\ln (1-\mathrm{t})}{\mathrm{t}}\mathrm{dt}\Rightarrow \mathrm{f}\left(\frac{1}{2}\right)=-{\int }_0^{\frac{1}{2}}\frac{\ln (1-\mathrm{t})}{\mathrm{t}}\mathrm{dt}\mathrm{we}\mathrm{have}$$$$\mathrm{by}\mathrm{parts}{\int }_0^{\frac{1}{2}}\frac{\ln (1-\mathrm{t})}{\mathrm{t}}\mathrm{dt}={\left[\mathrm{lntln}(1-\mathrm{t})\right]}_0^{\frac{1}{2}}-{\int }_0^{\frac{1}{2}}\mathrm{lnt}\times \frac{-1}{1-\mathrm{t}}\mathrm{dt}$$$$={\ln }^2\left(2\right)+{\int }_0^{\frac{1}{2}}\frac{\ln \left(\mathrm{t}\right)}{1-\mathrm{t}}\mathrm{dt}\mathrm{we}\mathrm{have}$$$${\int }_0^1\frac{\ln (1-\mathrm{t})}{\mathrm{t}}\mathrm{dt}={\int }_0^{\frac{1}{2}}\frac{\ln (1-\mathrm{t})}{\mathrm{t}}\mathrm{dt}+{\int }_{\frac{1}{2}}^1\frac{\ln (1-\mathrm{t})}{\mathrm{t}}\mathrm{dt}(\to 1-\mathrm{t}=\mathrm{u})$$$$={\int }_0^{\frac{1}{2}}\frac{\ln (1-\mathrm{t})}{\mathrm{t}}\mathrm{dt}+{\int }_0^{\frac{1}{2}}\frac{\ln \left(\mathrm{u}\right)}{1-\mathrm{u}}\mathrm{du}=-\mathrm{f}\left(\frac{1}{2}\right)-\mathrm{f}\left(\frac{1}{2}\right)-{\ln }^2\left(2\right)$$$$=-2\mathrm{f}\left(\frac{1}{2}\right)-{\ln }^2\left(2\right)\Rightarrow 2\mathrm{f}\left(\frac{1}{2}\right)=-{\int }_0^1\frac{\ln (1-\mathrm{t})}{\mathrm{t}}\mathrm{dt}-{\ln }^2\left(2\right)$$$$\mathrm{we}\mathrm{have}{\ln }^{{}^{\prime }}(1-\mathrm{t})=\frac{-1}{1-\mathrm{t}}=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{n}}\Rightarrow$$$$\ln (1-\mathrm{t})=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{n}+1}{\mathrm{t}}^{\mathrm{n}+1}+\mathrm{c}(\mathrm{c}=0)=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{n}}}{\mathrm{n}}\Rightarrow$$$$\frac{-\ln (1-\mathrm{t})}{\mathrm{t}}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{n}-1}}{\mathrm{n}}\Rightarrow {\int }_0^1\frac{-\ln (1-\mathrm{t})}{\mathrm{t}}\mathrm{dt}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2}=\frac{{\pi }^2}{6}\Rightarrow$$$$2\mathrm{f}\left(\frac{1}{2}\right)=\frac{{\pi }^2}{6}-{\ln }^2\left(2\right)\Rightarrow \mathrm{S}=\mathrm{f}\left(\frac{1}{2}\right)=\frac{{\pi }^2}{12}-\frac{{\ln }^{2}2}{2}$$

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