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Question Number 162877 by HongKing last updated on 01/Jan/22

$$\mathrm{Find}:$$$$\mathbf{\Omega }=\underset{0}{\stackrel{1}{\int }}\frac{{\mathrm{x}}^3}{{\ln }^2(1-\mathrm{x})}\mathrm{dx}$$

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

$$\mathrm{\Omega }={\int }_0^1\frac{x^3}{{\ln }^2(1-x)}dx={\int }_0^1\frac{{(1-x)}^3}{{\ln }^{2}x}dx$$$$f\left(\alpha \right)={\int }_0^1\frac{x^{\alpha }{(1-x)}^3}{{\ln }^{2}x}dx\Rightarrow f^{\left(2\right)}\left(\alpha \right)={\int }_0^1{x}^{\alpha }{(1-x)}^{3}dx$$$$=\beta (\alpha +1,4)=\frac{6\mathrm{\Gamma }(\alpha +1)}{\mathrm{\Gamma }(\alpha +5)}=\frac{6}{(\alpha +4)(\alpha +3)(\alpha +2)(\alpha +1)}$$$$=-\frac{1}{\alpha +4}+\frac{3}{\alpha +3}-\frac{3}{\alpha +2}+\frac{1}{\alpha +1}$$$$f^{\left(1\right)}\left(\alpha \right)=-\ln (\alpha +4)+3\ln (\alpha +3)-3\ln (\alpha +2)+\ln (\alpha +1)$$$$f\left(\alpha \right)=-(\alpha +4)[\ln (\alpha +4)-1]+3(\alpha +3)[\ln (\alpha +3)-1]$$$$-3(\alpha +2)[\ln (\alpha +2)-1]+(\alpha +1)[\ln (\alpha +1)-1]$$$$f\left(0\right)=-4(\ln 4-1)+9(\ln 3-1)-6(\ln 2-1)-1$$$$=\ln \left(\frac{3^9}{2^8\times 2^6}\right)={\int }_0^1\frac{x^3}{{\ln }^2(1-x)}dx$$

Commented by Lordose last updated on 02/Jan/22

After calculating F⁽²⁾(a), when integrating the result, There's supposed to be a constant, i. e ∫F⁽²⁾(a)da = F'(a) + C ∫F'(a) + C = F(a) + aC + D where C and D are constants

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 Ar Brandon last updated on 02/Jan/22

$$f^{\left(1\right)}\left(\alpha \right)={\int }_0^1\frac{x^{\alpha }{(1-x)}^3}{\ln x}dx$$$$=-\ln (\alpha +4)+3\ln (\alpha +3)-3\ln (\alpha +2)+\ln (\alpha +1)+C$$$$\underset{\alpha \to -\mathrm{\infty }}{\lim }{f}^{\left(1\right)}\left(\alpha \right)=0=\underset{\alpha \to -\mathrm{\infty }}{\lim ln}\left(\frac{{(\alpha +3)}^3(\alpha +1)}{(\alpha +4){(\alpha +2)}^3}\right)+\mathrm{C}$$$$=\underset{\alpha \to -\mathrm{\infty }}{\lim ln}\left(\frac{{(1+\frac{3}{\alpha })}^3(1+\frac{1}{\alpha })}{(1+\frac{4}{\alpha }){(1+\frac{2}{\alpha })}^3}\right)+\mathrm{C}=\ln \left(1\right)+\mathrm{C}=0\Rightarrow \mathrm{C}=0$$$$f^{\left(2\right)}\left(\alpha \right)={\int }_0^1\frac{x^{\alpha }{(1-x)}^3}{{\ln }^{2}x}dx$$$$=-(\alpha +4)[\ln (\alpha +4)-1]+3(\alpha +3)[\ln (\alpha +3)-1]$$$$-3(\alpha +2)[\ln (\alpha +2)-1]+(\alpha +1)[\ln (\alpha +1)-1]+\mathrm{D}$$$$\underset{\alpha \to -\mathrm{\infty }}{\lim }f\left(\alpha \right)=0$$$$=\underset{\alpha \to -\mathrm{\infty }}{\lim ln}\left(\frac{{(\alpha +3)}^9(\alpha +1)}{{(\alpha +4)}^4{(\alpha +2)}^6}\right)-\alpha \ln (\alpha +4)+3\alpha \ln (\alpha +3)$$$$-3\alpha \ln (\alpha +2)+\alpha \ln (\alpha +1)+\mathrm{D}$$$$=\mathrm{D}=0$$

Commented by Lordose last updated on 02/Jan/22

Nice one Bro.

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

Thanks bro. You too made a good remark there.

Commented by HongKing last updated on 02/Jan/22

$$\mathrm{magnificent}\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{my}\mathrm{dear}\mathrm{Sir}$$

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

$$\mathrm{\Omega }={\int }_0^1\frac{x^3}{{\ln }^2(1-x)}dx={\int }_0^1\frac{{(1-x)}^3}{{\ln }^{2}x}dx$$$$x=e^{-t}\Rightarrow dx=-e^{-t}dt$$$$={\int }_0^{\mathrm{\infty }}\frac{{(1-e^{-t})}^3}{t^2}\cdot e^{-t}dt={\int }_0^{\mathrm{\infty }}(\frac{e^{-t}}{t^2}-3\frac{e^{-2t}}{t^2}+3\frac{e^{-3t}}{t^2}-\frac{e^{-4t}}{t^2})dt$$$$={\int }_0^{\mathrm{\infty }}(\frac{t}{1+t}-\frac{3t}{2+t}+\frac{3t}{3+t}-\frac{t}{4+t})dt,\mathrm{maz}\mathrm{identity}$$$$={\int }_0^{\mathrm{\infty }}(1-\frac{1}{1+t}-3+\frac{6}{2+t}+3-\frac{9}{3+t}-1+\frac{4}{4+t})dt$$$$={[\ln (1+t)+6\ln (2+t)-9\ln (3+t)+4\ln (4+t)]}_0^{\mathrm{\infty }}$$$$={\left[\ln \left(\frac{(1+t){(2+t)}^6{(4+t)}^4}{{(3+t)}^9}\right)\right]}_0^{\mathrm{\infty }}=\ln \left(1\right)-\ln \left(\frac{2^6\times 4^4}{3^9}\right)$$$$=-\ln \left(\frac{2^6\times 2^8}{3^9}\right)=\ln \left(\frac{3^9}{2^8\times 2^6}\right)$$

Commented by Lordose last updated on 02/Jan/22

Exactly what i used��

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

I stole the method from you this morning ��

Commented by Lordose last updated on 03/Jan/22

����

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