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Question Number 134852 by mnjuly1970 last updated on 07/Mar/21

$$...nicecalculus...$$$$find:::$$$$\mathit{\varphi }={\int }_0^1\frac{ln\left(x\right)ln(1-x)}{x(1-x)}dx=?$$$$$$

Answered by mnjuly1970 last updated on 07/Mar/21

$$\mathit{\varphi }={\int }_0^1\frac{ln\left(x\right)ln(1-x)}{1-x}+\frac{ln\left(x\right)ln(1-x)}{x}dx$$$$={[-\frac{1}{2}{ln}^2(1-x)ln\left(x\right)]}_0^1+\frac{1}{2}{\int }_0^1\frac{{ln}^2(1-x)}{x}dx$$$$+{\left[\frac{1}{2}{ln}^2\left(x\right)ln(1-x)\right]}_0^1+\frac{1}{2}{\int }_0^1\frac{{ln}^2\left(x\right)}{1-x}dx$$$$=\zeta \left(3\right)+\frac{1}{2}{\int }_0^1\frac{{ln}^2(1-t)}{t}dt=2\zeta \left(3\right)...✓✓$$$$$$$$$$$$$$$$$$

Answered by Ñï= last updated on 08/Mar/21

$$\mathit{\varphi }={\int }_0^1\frac{ln\left(x\right)ln(1-x)}{x(1-x)}dx={\int }_0^1\frac{lnxln(1-x)}{x}+\frac{lnxln(1-x)}{1-x}dx$$$$=2{\int }_0^1\frac{ln(1-x)lnx}{x}$$$$=-2{Li}_2\left(x\right)lnx{\mid }_0^1+2{\int }_0^1\frac{{Li}_2\left(x\right)}{x}dx$$$$=2{Li}_3\left(x\right){\mid }_0^1$$$$=2{Li}_3\left(1\right)=2\zeta \left(3\right)$$

Commented by mnjuly1970 last updated on 08/Mar/21

$$thanksalot...$$

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