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Question Number 162893 by mnjuly1970 last updated on 02/Jan/22

$$$$$$\mathrm{\Omega }={\int }_0^1{\left(\frac{x^{}}{{\ln }^{}(1-x)}\right)}^{2}dx\stackrel{?}{=}\ln \left(\frac{27}{16}\right)$$$$----$$$$$$

Answered by Ar Brandon last updated on 02/Jan/22

$$\mathrm{\Omega }={\int }_0^1{\left(\frac{x}{\ln (1-x)}\right)}^{2}dx={\int }_0^1\frac{{(1-x)}^2}{{\ln }^{2}x}dx$$$$f\left(\alpha \right)={\int }_0^1\frac{x^{\alpha }{(1-x)}^2}{{\ln }^{2}x}dx\Rightarrow f^{\left(2\right)}\left(\alpha \right)={\int }_0^1{x}^{\alpha }{(1-x)}^{2}dx$$$$=\beta (\alpha +1,3)=\frac{2\mathrm{\Gamma }(\alpha +1)}{\mathrm{\Gamma }(\alpha +4)}=\frac{2}{(\alpha +3)(\alpha +2)(\alpha +1)}$$$$=\frac{1}{\alpha +3}-\frac{2}{\alpha +2}+\frac{1}{\alpha +1}\Rightarrow f^{\left(1\right)}\left(\alpha \right)=\ln (\alpha +3)-2\ln (\alpha +2)+\ln (\alpha +1)$$$$f\left(\alpha \right)=(\alpha +3)[\ln (\alpha +3)-1]-2(\alpha +2)[\ln (\alpha +2)-1]+(\alpha +1)[\ln (\alpha +1)-1]$$$$f\left(0\right)={\int }_0^1\frac{{(1-x)}^2}{{\ln }^{2}x}dx=3(\ln 3-1)-4(\ln 2-1)-1$$$$\mathrm{\Omega }={\int }_0^1{\left(\frac{x}{\ln (1-x)}\right)}^{2}dx=\ln \left(\frac{27}{16}\right)$$

Commented by Lordose last updated on 02/Jan/22

$$\mathrm{No}\mathrm{intial}\mathrm{conditions}..$$

Commented by Ar Brandon last updated on 02/Jan/22

$$\mathrm{We}\mathrm{study}\mathrm{limits}\mathrm{as}\alpha \to -\mathrm{\infty }\mathrm{after}$$$$\mathrm{integrating}\mathrm{to}\mathrm{get}\mathrm{C}=0$$

Commented by mnjuly1970 last updated on 02/Jan/22

$$verynicesolutionsirbrandon$$

Answered by Lordose last updated on 02/Jan/22

$$$$$$\mathrm{\Omega }={\int }_0^1\frac{{\mathrm{x}}^2}{{\ln }^2(1-\mathrm{x})}\mathrm{dx}\stackrel{\mathrm{x}=1-\mathrm{x}}{=}{\int }_0^1\frac{{(1-\mathrm{x})}^2}{{\ln }^2\left(\mathrm{x}\right)}\mathrm{dx}$$$$\mathrm{\Omega }\stackrel{\mathrm{t}=-\ln \left(\mathrm{x}\right)}{=}{\int }_0^{\mathrm{\infty }}\frac{(1-{2\mathrm{e}}^{-\mathrm{t}}+{\mathrm{e}}^{-2\mathrm{t}}){\mathrm{e}}^{-\mathrm{t}}}{{\mathrm{t}}^2}\mathrm{dt}$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{t}}}{{\mathrm{t}}^2}\mathrm{dt}-2{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{{\mathrm{t}}^2}+{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-3\mathrm{t}}}{{\mathrm{t}}^2}\mathrm{dt}$$$$\mathbf{Maz}\mathbf{Identity}:\{\begin{array}{l}{\int }_0^{\mathrm{\infty }}\mathrm{F}\left(\mathrm{u}\right)\mathrm{g}\left(\mathrm{u}\right)\mathrm{du}={\int }_0^{\mathrm{\infty }}\mathrm{f}\left(\mathrm{u}\right)\mathrm{G}\left(\mathrm{u}\right)\mathrm{du}\\ \mathfrak{L}\left(\mathrm{f}\left(\mathrm{t}\right)\right)=\mathrm{F}\left(\mathrm{t}\right)\end{array}$$$$\mathrm{F}\left(\mathrm{u}\right)=\frac{1}{{\mathrm{u}}^2}\Rightarrow \mathrm{f}\left(\mathrm{u}\right)=\mathrm{u}$$$$\mathrm{g}\left(\mathrm{u}\right)={\mathrm{e}}^{-\mathrm{at}}\Rightarrow \mathrm{G}\left(\mathrm{u}\right)=\frac{1}{\mathrm{a}+\mathrm{u}}$$$$\mathrm{\Omega }\stackrel{\mathbf{Maz}\mathbf{Identity}}{=}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{t}}{1+\mathrm{t}}\mathrm{dt}-2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{t}}{2+\mathrm{t}}\mathrm{dt}+{\int }_0^{\mathrm{\infty }}\frac{\mathrm{t}}{3+\mathrm{t}}\mathrm{dt}$$$$\mathrm{\Omega }=\mathrm{t}-\ln (1+\mathrm{t})-2(\mathrm{t}-2\ln (2+\mathrm{t}))+(\mathrm{t}-3\ln (3+\mathrm{t})){\mid }_0^{\mathrm{\infty }}$$$$\mathrm{\Omega }=\ln \left(\frac{{(2+\mathrm{t})}^4}{(1+\mathrm{t}){(3+\mathrm{t})}^3}\right){\mid }_0^{\mathrm{\infty }}$$$$\mathrm{\Omega }=-\ln \left(\frac{16}{27}\right)=\ln \left(\frac{27}{16}\right)$$

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