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Question Number 172564 by Mathspace last updated on 28/Jun/22

$$find{\int }_0^1\sqrt{x}\sqrt{1-\sqrt{x}}lnxdx$$

Answered by Ar Brandon last updated on 28/Jun/22

$$I={\int }_0^1\sqrt{x}\sqrt{1-\sqrt{x}}\ln xdx,x=t^2$$$$=4{\int }_0^1{t}^2\sqrt{1-t}\ln tdt=4\frac{\partial }{\partial \alpha }{\mid }_{\alpha =3}{\int }_0^1{t}^{\alpha -1}\sqrt{1-t}dt$$$$=4\frac{\partial }{\partial \alpha }{\mid }_{\alpha =3}\beta (\alpha ,\frac{3}{2})=4\frac{\partial }{\partial \alpha }{\mid }_{\alpha =3}\frac{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }(\alpha +\frac{3}{2})}$$$$=2\sqrt{\pi }\frac{\partial }{\partial \alpha }{\mid }_{\alpha =3}\frac{\mathrm{\Gamma }\left(\alpha \right)}{\mathrm{\Gamma }(\alpha +\frac{3}{2})}=2\sqrt{\pi }{(\frac{{\mathrm{\Gamma }}^{\prime }\left(\alpha \right)}{\mathrm{\Gamma }(\alpha +\frac{3}{2})}-\frac{\mathrm{\Gamma }\left(\alpha \right){\mathrm{\Gamma }}^{\prime }(\alpha +\frac{3}{2})}{{\mathrm{\Gamma }}^2(\alpha +\frac{3}{2})})}_{\alpha =3}$$$$=2\sqrt{\pi }(\frac{\mathrm{\Gamma }\left(3\right)\psi \left(3\right)}{\mathrm{\Gamma }\left(\frac{9}{2}\right)}-\frac{\mathrm{\Gamma }\left(3\right)\mathrm{\Gamma }\left(\frac{9}{2}\right)\psi \left(\frac{9}{2}\right)}{{\mathrm{\Gamma }}^2\left(\frac{9}{2}\right)})$$$$=2\sqrt{\pi }(\frac{2(\frac{1}{2}+1-\gamma )}{\frac{7}{2}\cdot \frac{5}{2}\cdot \frac{3}{2}\cdot \frac{1}{2}\cdot \sqrt{\pi }}-\frac{2(\frac{2}{7}+\frac{2}{5}+\frac{2}{3}+2-\gamma -2\ln 2)}{\frac{7}{2}\cdot \frac{5}{2}\cdot \frac{3}{2}\cdot \frac{1}{2}\cdot \sqrt{\pi }})$$$$=\frac{32}{105}(3-2\gamma -\frac{704}{105}+2\gamma +4\ln 2)=\frac{32}{105}(4\ln 2-\frac{389}{105})$$

Answered by Mathspace last updated on 29/Jun/22

$$changement\sqrt{x}=tgivex=t^2$$$$andI={\int }_0^{1}t\sqrt{1-t}\left(2lnt\right)2tdt$$$$=4{\int }_0^1{t}^2{(1-t)}^{\frac{1}{2}}lntdt$$$$letf\left(a\right)={\int }_0^1{t}^{a+2}{(1-t)}^{\frac{1}{2}}dt$$$$wehave{f}^{{}^{\prime }}\left(a\right)={\int }_0^1{t}^{a+2}{(1-t)}^{\frac{1}{2}}lntdt$$$$\Rightarrow f^{{}^{\prime }}\left(0\right)={\int }_0^1{t}^2{(1-t)}^{\frac{1}{2}}lntdt\Rightarrow$$$$I=4f^{{}^{\prime }}\left(0\right)$$$$wehavef\left(a\right)={\int }_0^1{t}^{a+3-1}{(1-t)}^{\frac{3}{2}-1}dt$$$$=B(a+3,\frac{3}{2})=\frac{\mathrm{\Gamma }(a+3)\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }(a+3+\frac{3}{2})}$$$$=\frac{\sqrt{\pi }}{2}\times \frac{\mathrm{\Gamma }(a+3)}{\mathrm{\Gamma }(a+\frac{9}{2})}\Rightarrow$$$$f^{{}^{\prime }}\left(a\right)=\frac{\sqrt{\pi }}{2}\times \frac{{\mathrm{\Gamma }}^{{}^{\prime }}(a+3)\mathrm{\Gamma }(a+\frac{9}{2})-\mathrm{\Gamma }(a+3){\mathrm{\Gamma }}^{{}^{\prime }}(a+\frac{9}{2})}{{\mathrm{\Gamma }}^2(a+\frac{9}{2})}$$$$I=4f^{{}^{\prime }}\left(0\right)=2\sqrt{\pi }\times \frac{{\mathrm{\Gamma }}^{{}^{\prime }}\left(3\right)\mathrm{\Gamma }\left(\frac{9}{2}\right)-\mathrm{\Gamma }\left(3\right){\mathrm{\Gamma }}^{{}^{\prime }}\left(\frac{9}{2}\right)}{{\mathrm{\Gamma }}^2\left(\frac{9}{2}\right)}$$$$afterweuse\psi \left(x\right)=\frac{{\mathrm{\Gamma }}^{{}^{\prime }}\left(x\right)}{\mathrm{\Gamma }\left(x\right)}\Rightarrow$$$${\mathrm{\Gamma }}^{{}^{\prime }}\left(3\right)=\mathrm{\Gamma }\left(3\right).\psi \left(3\right)$$$$\mathrm{\Gamma }\left(3\right)=2!=2$$$$\psi \left(s\right)=-\gamma +{\int }_0^1\frac{1-x^{s-1}}{1-x}dx\Rightarrow$$$$\psi \left(3\right)=-\gamma +{\int }_0^1\frac{1-x^2}{1-x}dx$$$$=-\gamma +{\int }_0^1(1+x)dx=-\gamma +\frac{3}{2}$$$${\mathrm{\Gamma }}^{{}^{\prime }}\left(\frac{9}{2}\right)=\mathrm{\Gamma }\left(\frac{9}{2}\right)\psi \left(\frac{9}{2}\right)$$$$\psi \left(\frac{9}{2}\right)=-\gamma +{\int }_0^1\frac{1-x^{\frac{7}{2}}}{1-x}dx$$$${\int }_0^1\frac{1-x^{\frac{7}{2}}}{1-x}dx$$$$={\int }_0^1\frac{1-x^3\sqrt{x}}{1-x}dx(x=t^2)$$$$={\int }_0^1\frac{1-t^7}{1-t^2}\left(2t\right)dt$$$$=2{\int }_0^1\frac{t(1+t+t^2+t^3+t^4+t^5+t^6)}{1+t}dt=...$$

Commented by Ar Brandon last updated on 29/Jun/22

$$\psi (1+s)=\frac{1}{s}+\psi \left(s\right)$$$$\psi \left(3\right)=\frac{1}{2}+1+\psi \left(1\right)=\frac{3}{2}-\gamma$$$$\psi \left(\frac{9}{2}\right)=\frac{2}{7}+\frac{2}{5}+\frac{2}{3}+2+\psi \left(\frac{1}{2}\right)$$$$=\frac{352}{105}-\gamma -2\ln 2$$

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