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Question Number 123462 by mnjuly1970 last updated on 25/Nov/20

$$...nicecalculus...$$$$provethat::$$$$\mathrm{\Omega }={\int }_0^1\frac{ln(1-x)ln(1-x^2)}{x}\stackrel{??}{=}\frac{11\zeta \left(3\right)}{8}$$$$.................$$

Answered by Lordose last updated on 25/Nov/20

$$$$$$\mathrm{\Omega }={\int }_0^1\frac{\ln (1-\mathrm{x})(\ln (1-\mathrm{x})+\ln (1+\mathrm{x}))}{\mathrm{x}}\mathrm{dx}$$$$\mathrm{\Omega }={\int }_0^1\frac{{\ln }^2(1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}+{\int }_0^1\frac{\ln (1-\mathrm{x})\ln (1+\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$\mathrm{\Omega }=2\zeta \left(3\right)-\frac{5\zeta \left(3\right)}{8}$$$$\mathrm{\Omega }=\frac{11\zeta \left(3\right)}{8}$$

Commented by Dwaipayan Shikari last updated on 25/Nov/20

$${\int }_0^1\frac{{log}^2(1-x)}{x}dx$$$$={\int }_0^1\frac{{log}^{2}x}{1-x}dx={\int }_0^1{log}^{2}x\underset{n=0}{\sum ^{\mathrm{\infty }}}{x}^n=\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{x}^n{log}^2\left(x\right)logx=t$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_{-\mathrm{\infty }}^0{t}^2{e}^{(n+1)x}dx(n+1)x=-\mathrm{\Phi }$$$$=-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+1)}^3}{\int }_{\mathrm{\infty }}^0{\mathrm{\Phi }}^2{e}^{-\mathrm{\Phi }}d\mathrm{\Phi }=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+1)}^3}.\mathrm{\Gamma }\left(3\right)=2\zeta \left(3\right)$$

Commented by mnjuly1970 last updated on 26/Nov/20

$$thankyou$$

Commented by mnjuly1970 last updated on 26/Nov/20

$$thankyou..$$

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