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Question Number 160734 by mnjuly1970 last updated on 05/Dec/21

$$$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\mathrm{\Phi }={\int }_0^1{\left(\frac{{ln}^{}\left(\frac{1}{1-x}\right)}{x}\right)}^{3}dx\stackrel{?}{=}3(\zeta \left(2\right)+\zeta \left(3\right))$$$$----solution----$$$$\mathrm{\Phi }\stackrel{\mathrm{I}.\mathrm{B}.\mathrm{P}}{=}{\left[\frac{1}{2x^2}{ln}^3(1-x)\right]}_0^1+\frac{3}{2}{\int }_0^1\frac{{ln}^2(1-x)}{x^2(1-x)}dx$$$$=\frac{1}{2}{lim}_{\xi \to 1^{-}}\frac{{ln}^3(1-\xi )}{{\xi }^2}$$$$+\frac{3}{2}[{\int }_0^1\frac{{ln}^2(1-x)}{x}dx=2\zeta \left(3\right)]$$$$+\frac{3}{2}[{\int }_0^1\frac{{ln}^2(1-x)}{x^2}dx=\frac{{\pi }^2}{3}=2\zeta \left(2\right)]$$$$+\frac{3}{2}{\int }_0^1\frac{{ln}^2(1-x)}{1-x}dx\}$$$$=\frac{1}{2}{lim}_{\xi \to 1^{-}}\{\frac{{ln}^3(1-\xi )}{{\xi }^2}-{ln}^3(1-\xi )\}$$$$+\frac{3}{2}\left(2\zeta \left(3\right)\right)+\frac{3}{2}\left(2\zeta \left(2\right)\right)$$$$=3(\zeta \left(3\right)+3\zeta \left(2\right))◼m.n$$$$$$

Commented by Gbenga last updated on 06/Dec/21

$$\mathit{c}\mathit{a}\mathit{n}\mathit{y}\mathit{o}\mathit{u}\mathit{t}\mathit{e}\mathit{a}\mathit{c}\mathit{h}\mathit{m}\mathit{e}\mathit{t}\mathit{h}\mathit{i}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

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