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Question Number 115055 by mnjuly1970 last updated on 23/Sep/20

$$...nicecalculus...$$$$evaluation:$$$$$$$$\chi ={\int }_0^{1}log(1-x).log(1+x)=???$$$$...m.n.july.197o...$$$$$$

Answered by mathmax by abdo last updated on 23/Sep/20

$$\mathrm{I}={\int }_0^1\ln (1-\mathrm{x}).\ln (1+\mathrm{x})\mathrm{dx}\mathrm{we}\mathrm{have}\frac{\mathrm{d}}{\mathrm{dx}}\ln (1-\mathrm{x})=\frac{-1}{1-\mathrm{x}}$$$$=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}\Rightarrow \ln (1-\mathrm{x})=-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}$$$$\mathrm{also}\ln (1+\mathrm{x})=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}\Rightarrow$$$$\ln (1-\mathrm{x}).\ln (1+\mathrm{x})=\left(\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}\right).\left(\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\right)$$$$=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\mathrm{with}{\mathrm{c}}_{\mathrm{n}}=\sum _{\mathrm{i}+\mathrm{j}=\mathrm{n}}{\mathrm{a}}_{\mathrm{i}}{\mathrm{b}}_{\mathrm{j}}=\sum _{\mathrm{i}+\mathrm{j}=\mathrm{n}}\frac{1}{\mathrm{i}}\times \frac{{(-1)}^{\mathrm{j}}}{\mathrm{j}}$$$$=\sum _{\mathrm{i}=1}^{\mathrm{n}-1}\frac{1}{\mathrm{i}}\times \frac{{(-1)}^{\mathrm{n}-\mathrm{i}}}{\mathrm{n}-\mathrm{i}}\Rightarrow \ln (1-\mathrm{x}).\ln (1+\mathrm{x})=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\left(\sum _{\mathrm{i}=1}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{n}-\mathrm{i}}}{\mathrm{i}(\mathrm{n}-\mathrm{i})}\right){\mathrm{x}}^{\mathrm{n}}$$$$\Rightarrow {\int }_0^1\ln (1-\mathrm{x})\ln (1+\mathrm{x})\mathrm{dx}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\left(\sum _{\mathrm{i}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{n}-\mathrm{i}}}{\mathrm{i}(\mathrm{n}-\mathrm{i})}\right)\times \frac{1}{\mathrm{n}+1}$$$$=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+1}\left(\sum _{\mathrm{i}=1}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{i}}}{\mathrm{i}(\mathrm{n}-\mathrm{i})}\right)\mathrm{but}$$$$\sum _{\mathrm{i}=1}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{i}}}{\mathrm{i}(\mathrm{n}-\mathrm{i})}=\frac{1}{\mathrm{n}}\sum _{\mathrm{i}=1}^{\mathrm{n}-1}{(-1)}^{\mathrm{i}}(\frac{1}{\mathrm{i}}-\frac{1}{\mathrm{n}-\mathrm{i}})=\frac{1}{\mathrm{n}}\sum _{\mathrm{i}=1}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{i}}}{\mathrm{i}}$$$$-\frac{1}{\mathrm{n}}\sum _{\mathrm{i}=1}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{i}}}{\mathrm{n}-\mathrm{i}}(\mathrm{n}-\mathrm{i}=\mathrm{p})=\frac{1}{\mathrm{n}}\sum _{\mathrm{i}=1}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{i}}}{\mathrm{i}}-\frac{1}{\mathrm{n}}\sum _1^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{n}-\mathrm{p}}}{\mathrm{p}}$$$$=\frac{1}{\mathrm{n}}\sum _{\mathrm{i}=1}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{i}}}{\mathrm{i}}-\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}\sum _{\mathrm{i}=1}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{i}}}{\mathrm{i}}$$$$=\frac{1}{\mathrm{n}}(1-{(-1)}^{\mathrm{n}})\sum _{\mathrm{i}=1}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{i}}}{\mathrm{i}}=\frac{1}{\mathrm{n}}\sum _{\mathrm{i}=1}^{2\mathrm{n}}\frac{{(-1)}^{\mathrm{i}}}{\mathrm{i}}....\mathrm{be}\mathrm{continued}....$$

Commented by mnjuly1970 last updated on 23/Sep/20

$$thankyousomuchsir$$$$max...$$

Commented by mathmax by abdo last updated on 23/Sep/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}\mathrm{sir}$$

Answered by mathdave last updated on 23/Sep/20

$$solution$$$$iflet{I}_1={\int }_0^1\ln (1-x)\ln (1+x)dx.....\left(1\right)$$$$putx=-x$$$$I_2={\int }_{-1}^0\ln (1+x)\ln (1-x)dx........\left(2\right)sincetheintegrandareunchangedthenweaddthemup$$$$whichmeanthat{I}_1=I_2$$$$I=I_1+I_2=2I_1,I_1=\frac{1}{2}I$$$$I_1={\int }_0^1\ln (1-x)\ln (1+x)dx=\frac{1}{2}{\int }_0^1\ln (1-x)\ln (1+x)dx$$$$putx=2y-1,dx=2dy$$$$I_1=\frac{1}{2}{\int }_0^1\ln (2-2y)\ln \left(2y\right)2dy={\int }_0^1\left[(\ln 2+\ln (1-y))(\ln 2+\ln y)\right]dy$$$$I={\int }_0^1({\ln }^{2}2+\ln 2(\ln (1-y)+\ln y)+\ln y\ln (1-y))dx$$$$I={\ln }^2\left(2\right){\int }_0^{1}dy+\ln 2{\int }_0^1(\ln y+\ln (1-y))dy+{\int }_0^1\ln y\ln (1-y)dy$$$$letA={\ln }^2\left(2\right){\int }_0^{1}dy={\ln }^2\left(2\right)......\left(1\right)$$$$let$$$$B=\ln 2{\int }_0^1(\ln y+\ln (1-y))dy=2{\int }_0^1\ln ydy$$$$B=2\ln 2{(y\ln y-y)}_0^1=-2\ln 2.....\left(2\right)$$$$let$$$$C={\int }_0^1\ln y\ln (1-y)dy=-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}{\int }_0^1{y}^n\ln ydy$$$$C=-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}\frac{\partial }{\partial a}{\mid }_{a=1}{\int }_0^1{y}^n.y^{a-1}dy=-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}\frac{\partial }{\partial a}{\mid }_{a=1}\left(\frac{1}{n+a}\right)$$$$C=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n{(n+1)}^2}=\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n}-\frac{1}{n+1}-\frac{1}{{(n+1)}^2})$$$$C=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{n}\right)-\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n}-1)-\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{{\pi }^2}{6}-1)$$$$C=1-\frac{{\pi }^2}{6}+1=(2-\frac{{\pi }^2}{6}).........\left(3\right)$$$$but$$$$I=A+B+C={\ln }^2\left(2\right)-2\ln 2+2-\frac{{\pi }^2}{6}=2-\frac{{\pi }^2}{6}+{\ln }^2\left(2\right)-\ln 4$$$$\because {\int }_0^1\ln (1-x)\ln (1+x)dx=2-\frac{{\pi }^2}{6}+{\ln }^2\left(2\right)-\ln 4$$$$bymathdave\left(23/09/2020\right)$$

Commented by mnjuly1970 last updated on 23/Sep/20

$$thankyoumrdave$$$$nicesolutionthatwasi$$$$wanted...$$

Commented by mnjuly1970 last updated on 23/Sep/20

$$finalanswer:$$$${ln}^2\left(2\right)-ln\left(4\right)+2-\frac{{\pi }^2}{6}$$$$thankyousomuchmr$$$$dave.$$

Commented by mnjuly1970 last updated on 23/Sep/20

$$youareverygentlman$$$$mrdave.thankyoufor$$$$youreffort.$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}$$

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