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Question Number 142469 by mnjuly1970 last updated on 01/Jun/21

$$......Calculus.....$$$$Evaluate:{\int }_0^1{\left(\frac{log\left(\frac{1}{x}\right)}{1-x}\right)}^{3}dx=??$$

Answered by mindispower last updated on 01/Jun/21

$$=-{\int }_0^1{\left(\frac{ln\left(x\right)}{1-x}\right)}^3$$$$\frac{1}{{(1-x)}^3}=\frac{d^2}{2{dx}^2}.\frac{1}{1-x}=\sum _{k⩾2}k(k-1)x^{k-2}$$$$=\sum _{k⩾0}(k+1)(k+2)x^k$$$$=\sum _{k⩾0}(k+1)(k+2){\int }_0^1-{\left(ln\left(x\right)\right)}^3{x}^k$$$$=\sum _{k⩾0}(k+1)(k+2){\int }_0^{\mathrm{\infty }}{t}^3{e}^{-(k+1)t}dt$$$$=\frac{1}{2}\sum _{k⩾0}\frac{k+2}{{(k+1)}^3}{\int }_0^{\mathrm{\infty }}{t}^3{e}^{-t}dt$$$$=3\sum _{k⩾0}(\frac{1}{{(k+1)}^2}+\frac{1}{{(k+1)}^3})=3(\zeta \left(2\right)+\zeta \left(3\right))$$

Commented by mnjuly1970 last updated on 02/Jun/21

$$thankyousomuchmrpower$$$$excellentasalways...$$

Commented by mindispower last updated on 02/Jun/21

$$pleasursirthankyou$$

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