Question Number 172996 by Mathspace last updated on 04/Jul/22
$${let}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{{n}} }{ln}^{\mathrm{2}} {xdx} \\ $$$$\left.\mathrm{1}\right){lim}\:{U}_{{n}} ? \\ $$$$\left.\mathrm{2}\right){equivalent}\:{of}\:{U}_{{n}} \left({n}\rightarrow\infty\right) \\ $$
Answered by aleks041103 last updated on 05/Jul/22
$${U}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{{n}} }{ln}^{\mathrm{2}} {x}\:{dx}\Rightarrow{x}\in\left(\mathrm{0},\mathrm{1}\right) \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {{lim}x}^{{n}} =\mathrm{0} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {{lim}U}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−\mathrm{0}}{ln}^{\mathrm{2}} {xdx}=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}^{\mathrm{2}} {x}\:{dx} \\ $$$${x}={e}^{−{t}} ,{x}\in\left(\mathrm{0},\mathrm{1}\right)\Rightarrow{t}\in\left(\infty,\mathrm{0}\right)\Rightarrow{dx}=−{e}^{−{t}} {dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}^{\mathrm{2}} {x}\:{dx}=\int_{\infty} ^{\:\mathrm{0}} {ln}^{\mathrm{2}} \left({e}^{−{t}} \right)\left(−{e}^{{t}} {dt}\right)= \\ $$$$=\int_{\mathrm{0}} ^{\infty} {t}^{\mathrm{2}} {e}^{−{t}} {dt}=\mathrm{2} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {{lim}U}_{{n}} =\mathrm{2} \\ $$