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Question Number 137251 by mnjuly1970 last updated on 31/Mar/21

$$......\mathcal{A}dvanced...calculus......$$$$\mathit{\varphi }={\int }_0^1{x}^{2}ln\left(x\right)ln(1-x)dx=???$$$$$$

Answered by Ar Brandon last updated on 31/Mar/21

$$\varphi ={\int }_0^1{\mathrm{x}}^2\ln \left(\mathrm{x}\right)\ln (1-\mathrm{x})\mathrm{dx}=\frac{\partial }{\partial \alpha }{\mid }_{\alpha =0}-{\int }_0^1{\mathrm{x}}^{2+\alpha }\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}\mathrm{dx}$$$$=\frac{\partial }{\partial \alpha }{\mid }_{\alpha =0}-\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{n}}\int {\mathrm{x}}^{2+\alpha +\mathrm{n}}\mathrm{dx}=\frac{\partial }{\partial \alpha }{\mid }_{\alpha =0}-\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{n}(3+\alpha +\mathrm{n})}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{n}{(\mathrm{n}+3)}^2}=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{9\mathrm{n}}-\frac{1}{9(\mathrm{n}+3)}-\frac{1}{3{(\mathrm{n}+3)}^2})$$$$=\frac{1}{9}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+3})-\frac{1}{3}(\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathrm{n}}^2}-(1+\frac{1}{2^2}+\frac{1}{3^2}))$$$$=\frac{1}{9}(1+\frac{1}{2}+\frac{1}{3})-\frac{1}{3}(\zeta \left(2\right)-(1+\frac{1}{2^2}+\frac{1}{3^2}))$$$$=\frac{11}{54}-\frac{{\pi }^2}{18}+\frac{49}{108}=\frac{71}{108}-\frac{{\pi }^2}{18}$$

Commented by Ar Brandon last updated on 31/Mar/21

$$:)$$

Commented by Dwaipayan Shikari last updated on 31/Mar/21

$$Evenshorter:)$$

Commented by mnjuly1970 last updated on 31/Mar/21

$$thanksalotmrbrandon...$$

Commented by mnjuly1970 last updated on 31/Mar/21

$$grateful...$$

Commented by Ar Brandon last updated on 31/Mar/21

$${\mathrm{You}}^{\prime }\mathrm{re}\mathrm{welcome},\mathrm{Sir}!$$

Commented by Dwaipayan Shikari last updated on 31/Mar/21

$$But,wheneveryouarecalculating$$$${\int }_0^1{x}^{\pi }{(1-x)}^{e}log\left(x\right)log(1-x)dx$$$$youneedthis{\nu }^{\prime }(e+1,\pi +1);-)$$

Commented by Ar Brandon last updated on 31/Mar/21

OK, bro. Thanks 😊😃

Answered by Dwaipayan Shikari last updated on 31/Mar/21

$${\int }_0^1{x}^{2}log\left(x\right)log(1-x)dx={\nu }^{\prime }(3,1)$$$${\int }_0^1{x}^{a-1}{(1-x)}^{b-1}dx=\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }(a+b)}=\nu (a,b)$$$${\int }_0^1{x}^{a-1}{(1-x)}^{b-1}log(1-x)log\left(x\right)={\nu }^{\prime }(a,b)$$$$=\frac{\partial }{\partial a}\left(\frac{\mathrm{\Gamma }\left(a\right){\mathrm{\Gamma }}^{\prime }\left(1\right)-\psi (a+1)\mathrm{\Gamma }\left(1\right)\mathrm{\Gamma }\left(a\right)}{\mathrm{\Gamma }(a+1)}\right)$$$$=\frac{\partial }{\partial a}(\frac{-\gamma }{a}-\frac{\psi (a+1)}{a})=\frac{\gamma }{a^2}+\frac{\psi (1+a)}{a^2}-\frac{{\psi }^{\prime }(a+1)}{a}{\mid }_{a=3}$$$$=\frac{\gamma }{9}+\frac{\psi \left(4\right)}{9}-\frac{{\psi }^{\prime }\left(4\right)}{a}=\frac{\gamma -\gamma +1+\frac{1}{2}+\frac{1}{3}}{9}-\frac{1}{3}(\frac{{\pi }^2}{6}-1-\frac{1}{2^2}-\frac{1}{3^2})$$$$=\frac{11}{54}+\frac{1}{3}\left(\frac{49}{36}\right)-\frac{{\pi }^2}{18}=\frac{71}{108}-\frac{{\pi }^2}{18}$$$$$$

Commented by mnjuly1970 last updated on 31/Mar/21

$$thankyousomuchmrpayan..$$

Commented by mnjuly1970 last updated on 31/Mar/21

$$verynice..$$

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