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Question Number 125336 by mnjuly1970 last updated on 10/Dec/20

$$...nicecalculus...$$$$evaluate:$$$$\varphi ={\int }_0^1{x}^{2}ln\left(x\right).ln(1-x)dx=?$$$$$$

Answered by Dwaipayan Shikari last updated on 10/Dec/20

$$-\underset{n⩾1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}{\int }_0^1{x}^{2+n}log\left(x\right)dxlog\left(x\right)=t$$$$=-\underset{n⩾1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}{\int }_{-\mathrm{\infty }}^0{te}^{(n+3)t}dt(n+3)t=-u$$$$=-\underset{n⩾1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}{\int }_0^{\mathrm{\infty }}\frac{u}{{(n+3)}^2}{e}^{-u}du$$$$=\sum ^{\mathrm{\infty }}\frac{1}{n{(n+3)}^2}\mathrm{\Gamma }\left(2\right)=\frac{1}{3}\sum ^{\mathrm{\infty }}\frac{1}{(n+3)}(\frac{1}{n}-\frac{1}{(n+3)})$$$$=\frac{1}{9}\sum ^{\mathrm{\infty }}\frac{1}{n}-\frac{1}{n+3}-\frac{1}{3}\sum ^{\mathrm{\infty }}\frac{1}{{(n+3)}^2}$$$$=\frac{1}{9}(1+\frac{1}{2}+\frac{1}{3})-\frac{1}{3}(\frac{{\pi }^2}{6}-1-\frac{1}{4}-\frac{1}{9})$$$$=\frac{11}{54}+\frac{49}{108}-\frac{{\pi }^2}{18}=\frac{71}{108}-\frac{{\pi }^2}{18}$$

Commented by mnjuly1970 last updated on 10/Dec/20

$$verynicesolution.$$$$thanksalot...$$

Commented by Dwaipayan Shikari last updated on 10/Dec/20

$$Withpleasuresir!$$

Answered by Olaf last updated on 10/Dec/20

$$\mathrm{Let}{\mathrm{I}}_n={\int }_0^1{x}^n\ln xdx,n\in \mathbb{N}$$$${\mathrm{I}}_n={\left[\frac{x^{n+1}}{n+1}\ln x\right]}_0^1-{\int }_0^1\frac{x^{n+1}}{n+1}.\frac{dx}{x}$$$${\mathrm{I}}_n=-\frac{1}{n+1}{\int }_0^1{x}^{n}dx$$$${\mathrm{I}}_n=-\frac{1}{n+1}{\left[\frac{x^{n+1}}{n+1}\right]}_0^1=-\frac{1}{{(n+1)}^2}$$$$\varphi ={\int }_0^1{x}^2\ln x\ln (1-x)dx$$$$\varphi ={\int }_0^1{x}^2\ln x\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n}dx$$$$\varphi =\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}{\int }_0^1{x}^{n+2}\ln xdx$$$$\varphi =\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{I}}_{n+2}}{n}$$$$\varphi =-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n{(n+3)}^2}$$$$\varphi =-\frac{1}{9}\underset{n=1}{\sum ^{\mathrm{\infty }}}[\frac{1}{n}-\frac{1}{n+3}-\frac{3}{{(n+3)}^2}]$$$$\varphi =-\frac{1}{9}\underset{n=1}{\sum ^{\mathrm{\infty }}}[\frac{1}{n}-\frac{1}{n+3}]+\frac{1}{3}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+3)}^2}$$$$\varphi =-\frac{1}{9}[1+\frac{1}{2}+\frac{1}{3}]+\frac{1}{3}[\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}-(1+\frac{1}{4}+\frac{1}{9})]$$$$\varphi =-\frac{1}{9}\left[\frac{11}{6}\right]+\frac{1}{3}[\frac{{\pi }^2}{6}-\frac{49}{36}]$$$$\varphi =\frac{{\pi }^2}{18}-\frac{71}{108}$$

Commented by mnjuly1970 last updated on 10/Dec/20

$$gratefulmrolaf...$$

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