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Question Number 143889 by mnjuly1970 last updated on 19/Jun/21

$$$$$$AChallangingIntegral:$$$$$$$$\mathrm{\Phi }={\int }_0^1\frac{log\left(x\right).log(1+x)}{1-x}dx$$$$$$$$$$

Answered by Olaf_Thorendsen last updated on 19/Jun/21

$$$$$$\mathrm{\Phi }={\int }_0^1\frac{\ln \left(x\right).\ln (1+x)}{1-x}dx$$$$\mathrm{Let}u=1-x$$$$\mathrm{\Phi }={\int }_0^1\frac{\ln (1-u).\ln (2-u)}{u}du$$$$\mathrm{\Phi }={\int }_0^1\frac{\ln (2-u)}{u}(-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{u^{n+1}}{n+1})du$$$$\mathrm{\Phi }=-{\int }_0^1\ln (2-u)\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{u^n}{n+1}du$$$$\mathrm{\Phi }=-{\left[\ln (2-u)\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{u^{n+1}}{{(n+1)}^2}\right]}_0^1$$$$-{\int }_0^1\frac{1}{2-u}.\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{u^{n+1}}{{(n+1)}^2}du$$$$\mathrm{\Phi }=\underset{n=0(}{\sum ^{\mathrm{\infty }}}\frac{1}{{n+1)}^2}{\int }_0^1\frac{u^{n+1}}{2-u}du$$$$\mathrm{Let}{\mathrm{I}}_n={\int }_0^1\frac{u^n}{2-u}du$$$${2\mathrm{I}}_n-{\mathrm{I}}_{n+1}={\int }_0^1\frac{2u^n-u^{n+1}}{2-u}du$$$${2\mathrm{I}}_n-{\mathrm{I}}_{n+1}={\int }_0^1{u}^{n}du$$$${2\mathrm{I}}_n-{\mathrm{I}}_{n+1}=\frac{1}{n+1}$$$${\mathrm{I}}_{n+1}={2\mathrm{I}}_n-\frac{1}{n+1}$$$${\mathrm{I}}_{n+1}=2({2\mathrm{I}}_{n-1}-\frac{1}{n})-\frac{1}{n+1}$$$${\mathrm{I}}_{n+1}=2^2{\mathrm{I}}_{n-1}-\frac{2}{n}-\frac{1}{n+1}$$$${\mathrm{I}}_{n+1}=2^2{\mathrm{I}}_{n-1}-\frac{2}{n}-\frac{1}{n+1}$$$${\mathrm{I}}_{n+1}=2^3{\mathrm{I}}_{n-2}-\frac{2^2}{n-1}-\frac{2}{n}-\frac{1}{n+1}$$$$...$$$${\mathrm{I}}_{n+1}=2^{n+1}{\mathrm{I}}_0-\underset{k=1}{\sum ^{n+1}}\frac{2^{n+1-k}}{k}$$$${\mathrm{I}}_0={\int }_0^1\frac{u^0}{2-u}du={[-\ln \mid 2-u\mid ]}_0^1=\ln 2$$$${\mathrm{I}}_{n+1}=2^{n+1}\ln 2-\underset{k=1}{\sum ^{n+1}}\frac{2^{n+1-k}}{k}$$$$\mathrm{\Phi }=\underset{n=0(}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{I}}_{n+1}}{{n+1)}^2}$$$$\mathrm{\Phi }=\underset{n=0(}{\sum ^{\mathrm{\infty }}}\frac{1}{{n+1)}^2}(2^{n+1}\ln 2-\underset{k=1}{\sum ^{n+1}}\frac{2^{n+1-k}}{k})$$$$\mathrm{\Phi }=\underset{n=0(}{\sum ^{\mathrm{\infty }}}\frac{2^{n+1}}{{n+1)}^2}(\ln 2-\underset{k=1}{\sum ^{n+1}}\frac{2^{-k}}{k})$$$$\underset{k=1}{\sum ^n}\frac{1}{2^{k}k}=\ln 2-\frac{n-1}{n}.\frac{\mathrm{LerchPhi}(\frac{1}{2},1,n)}{2^{n+1}}$$$$\mathrm{\Phi }=\frac{1}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}(n.\frac{\mathrm{LerchPhi}(\frac{1}{2},1,n+1)}{{(n+1)}^3})$$$$...$$$$\mathrm{Sorry},\mathrm{I}\mathrm{tried}\mathrm{to}\mathrm{solve}\mathrm{but}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{lost}...$$

Commented by mnjuly1970 last updated on 20/Jun/21

$$thanksalotmrolaf..$$

Answered by mnjuly1970 last updated on 20/Jun/21

$$\mathrm{\Phi }:={\int }_0^1\frac{log(1-x)log(2-x)}{x}dx$$$$:={\int }_0^1\frac{log(1-x)\{log\left(2\right)+log(1-\frac{x}{2})\}}{x}dx$$$$:=-log\left(2\right)\frac{{\pi }^2}{6}-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2^n}{\int }_0^1{x}^{n-1}log(1-x)dx$$$$:=-log\left(2\right).\frac{{\pi }^2}{2}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{H_n}{n^2{2}^n}$$$$:=-log\left(2\right).\frac{{\pi }^2}{6}+\zeta \left(3\right)-\frac{{\pi }^2}{12}log\left(2\right)$$$$:=\frac{-{\pi }^2}{4}log\left(2\right)+\zeta \left(3\right)...$$$$$$

Commented by Dwaipayan Shikari last updated on 20/Jun/21

$$Nicesir!$$

Commented by mnjuly1970 last updated on 20/Jun/21

$$thankyousomuch...$$

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