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Question Number 137093 by mnjuly1970 last updated on 29/Mar/21

$$.....advancedcalculus....$$$$pleaseevaluate::$$$$1:\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{H_n}{{(2n+1)}^2}=?$$$$2:{\int }_0^1\frac{xln\left(x\right)ln(1-x)}{1-x}dx=?$$$$note:H_n=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}={\int }_0^1\frac{1-x^n}{1-x}dx$$

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

$$\mathrm{\Lambda }={\int }_0^1\frac{\mathrm{xln}\left(\mathrm{x}\right)\ln (1-\mathrm{x})}{1-\mathrm{x}}\mathrm{dx}$$$$={\{\mathrm{xln}\left(\mathrm{x}\right)\cdot \frac{{\ln }^2(1-\mathrm{x})}{-2}+\frac{1}{2}\int (1+\mathrm{lnx}){\ln }^2(1-\mathrm{x})\mathrm{dx}\}}_0^1$$$$=\frac{1}{2}{\int }_0^1(1+\mathrm{lnx}){\ln }^2(1-\mathrm{x})\mathrm{dx}=\frac{1}{2}{\int }_0^1(1+\ln (1-\mathrm{x})){\ln }^2\mathrm{xdx}$$$$=\frac{1}{2}\{{\int }_0^1{\ln }^2\mathrm{xdx}+{\int }_0^1\ln (1-\mathrm{x}){\ln }^2\mathrm{xdx}\}$$$$=\frac{1}{2}\{\frac{{\partial }^2}{\partial {\alpha }^2}{\mid }_{\alpha =0}{\int }_0^1{\mathrm{x}}^{\alpha }\mathrm{dx}-\frac{{\partial }^2}{\partial {\rho }^2}{\mid }_{\rho =0}{\int }_0^1\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{x}}^{\mathrm{n}+\rho }}{\mathrm{n}}\}$$$$=\frac{1}{2}\{\frac{{\partial }^2}{\partial {\alpha }^2}{\mid }_{\alpha =0}\frac{1}{\alpha +1}-\frac{{\partial }^2}{\partial {\rho }^2}{\mid }_{\rho =0}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{n}(\mathrm{n}+\rho +1)}\}$$$$=\frac{1}{2}\{2-\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{2}{\mathrm{n}{(\mathrm{n}+1)}^3}\}=1-\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{n}{(\mathrm{n}+1)}^3}$$$$=1-\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\{\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1}-\frac{1}{{(\mathrm{n}+1)}^2}-\frac{1}{{(\mathrm{n}+1)}^3}\}$$$$=1-1+\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(\mathrm{n}+1)}^2}+\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(\mathrm{n}+1)}^3}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathrm{n}}^2}-1+\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathrm{n}}^3}-1=\zeta \left(2\right)+\zeta \left(3\right)-2=\frac{{\pi }^2}{6}+\zeta \left(3\right)-2$$

Answered by Dwaipayan Shikari last updated on 29/Mar/21

$$\mathrm{ð}(a,b)={\int }_0^1{x}^{\alpha -1}{(1-x)}^{\beta -1}dx$$$$\frac{\partial }{\partial a\partial b}\mathrm{ð}(a,b)={\int }_0^{1}log\left(x\right)log(1-x)x^{\alpha -1}{(1-x)}^{\beta -1}dx$$$$\frac{\partial \mathrm{ð}(a,b)}{\partial a\partial b}=\frac{\partial \left(\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }(a+b)}\right)}{\partial a\partial b}=\frac{\partial }{\partial a}\left(\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }(a+b){\mathrm{\Gamma }}^{\prime }\left(b\right)-\mathrm{\Gamma }\left(b\right)\mathrm{\Gamma }\left(a\right){\mathrm{\Gamma }}^{\prime }(a+b)}{{\mathrm{\Gamma }}^2(a+b)}\right)$$$$\frac{\partial }{\partial a}(\frac{\mathrm{\Gamma }\left(a\right){\mathrm{\Gamma }}^{\prime }\left(b\right)}{\mathrm{\Gamma }(a+b)}-\frac{\mathrm{\Gamma }\left(b\right)\mathrm{\Gamma }\left(a\right)\psi (a+b)}{\mathrm{\Gamma }(a+b)})$$$$=\frac{\mathrm{\Gamma }(a+b){\mathrm{\Gamma }}^{\prime }\left(a\right){\mathrm{\Gamma }}^{\prime }\left(b\right)-{\mathrm{\Gamma }}^{\prime }(a+b)\mathrm{\Gamma }\left(a\right){\mathrm{\Gamma }}^{\prime }\left(b\right)}{{\mathrm{\Gamma }}^2(a+b)}-\frac{{\mathrm{\Gamma }}^{\prime }\left(b\right)\psi (a+b)\mathrm{\Gamma }\left(a\right)+{\psi }^{\prime }(a+b)\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)-{\psi }^2(a+b)\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)\mathrm{\Gamma }(b+a)}{{\mathrm{\Gamma }}^2(a+b)}$$$$a=1b=0$$$$$$

Answered by Dwaipayan Shikari last updated on 29/Mar/21

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{H_n}{{(2n+1)}^2}={\int }_0^1\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{H_n}{2n+1}{x}^{2n+1}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{H}_n{x}^{2n+\alpha }=-x^{\alpha }\frac{log(1-x^2)}{1-x^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{H_n}{2n+\alpha +1}=-{\int }_0^1\frac{x^{\alpha }log(1-x^2)}{1-x^2}dx$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{H_n}{{(2n+\alpha +1)}^2}=\frac{\partial }{\partial \alpha }{\int }_0^1\frac{x^{\alpha }log(1-x^2)}{1-x^2}dx$$$$=\frac{\partial }{\partial \alpha }{\int }_0^1\frac{u^{\frac{\alpha -1}{2}}log(1-u)}{1-u}du=\frac{\partial }{\partial \alpha }.\frac{\partial }{\partial \vartheta }{\mid }_{\vartheta =0}\left(\frac{\mathrm{\Gamma }\left(\frac{\alpha +1}{2}\right)\mathrm{\Gamma }\left(\vartheta \right)}{\mathrm{\Gamma }(\frac{\alpha }{2}+\frac{1}{2}+\vartheta )}\right)\mid$$

Commented by mnjuly1970 last updated on 29/Mar/21

$$thanksalot...$$

Answered by Ñï= last updated on 29/Mar/21

$${\int }_0^1\frac{xln\left(x\right)ln(1-x)}{1-x}dx={\int }_0^1\frac{(1-x)ln\left(x\right)ln(1-x)}{x}dx$$$$={\int }_0^1\frac{ln\left(x\right)ln(1-x)}{x}dx-{\int }_0^{1}ln\left(x\right)ln(1-x)dx$$$$={\int }_0^1\frac{{Li}_2\left(x\right)}{x}dx+{\int }_0^{1}x(\frac{ln(1-x)}{x}-\frac{lnx}{1-x})dx$$$$={Li}_3\left(1\right)+{\int }_0^{1}ln(1-x)dx-{\int }_0^1(1-x)\frac{ln(1-x)}{x}dx$$$$={Li}_3\left(1\right)+{\int }_0^{1}ln(1-x)dx-{\int }_0^1\frac{ln(1-x)}{x}-ln(1-x)dx$$$$={Li}_3\left(1\right)+2{\int }_0^{1}lnxdx+{Li}_2\left(1\right)$$$$={Li}_3\left(1\right)+{Li}_2\left(1\right)-2$$

Commented by mnjuly1970 last updated on 29/Mar/21

$$thanksalot...$$

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