Question Number 61530 by maxmathsup by imad last updated on 04/Jun/19
$${let}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{−\mathrm{2}{n}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx}\:\:\:{with}\:{n}\:{integr}\:{natural}\:{and}\:\:\:{n}\geqslant\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{U}_{{n}} \:{interms}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{n}^{\mathrm{2}} \:{U}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{study}\:{the}\:{serie}\:\Sigma\:{U}_{{n}} \\ $$
Commented by maxmathsup by imad last updated on 04/Jun/19
$$\left.\mathrm{1}\right)\:{changement}\:{x}^{\mathrm{4}} ={t}\:{give}\:{x}\:={t}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\Rightarrow{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\left({t}^{\frac{\mathrm{1}}{\mathrm{4}}} \right)^{\frac{−\mathrm{2}}{{n}}} \:\:\:}{\mathrm{1}+{t}}\:\frac{\mathrm{1}}{\mathrm{4}}{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{−\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\mathrm{1}+{t}}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}{n}}\:−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\:\:{let}\:\:{a}=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}{n}}\:\Rightarrow{a}\:=\frac{\mathrm{2}{n}−\mathrm{4}}{\mathrm{8}{n}}\:=\frac{{n}−\mathrm{2}}{\mathrm{4}{n}}>\mathrm{0} \\ $$$${for}\:{n}>\mathrm{2}\:\:{and}\:{a}−\mathrm{1}\:=\frac{{n}−\mathrm{2}}{\mathrm{4}{n}}\:−\mathrm{1}\:=\frac{{n}−\mathrm{2}−\mathrm{4}{n}}{\mathrm{4}{n}}\:=\frac{−\mathrm{3}{n}−\mathrm{2}}{\mathrm{4}{n}}\:<\mathrm{0}\:\Rightarrow\mathrm{0}<{a}<\mathrm{1}\:\Rightarrow \\ $$$${U}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\pi}{{sin}\left(\pi{a}\right)}\:=\frac{\pi}{\mathrm{4}{sin}\left(\pi\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right)\right)}\:=\frac{\pi}{\mathrm{4}\:{sin}\left(\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{2}{n}}\right)} \\ $$
Commented by maxmathsup by imad last updated on 04/Jun/19
$${forgive}\:{U}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{−\frac{\mathrm{2}}{{n}}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx}\:{with}\:{n}\geqslant\mathrm{3} \\ $$
Commented by maxmathsup by imad last updated on 04/Jun/19
$${i}\:{have}\:{used}\:\:{the}\:{result}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\pi}{{sin}\left(\pi{a}\right)}\:\:{with}\:\mathrm{0}<{a}<\mathrm{1} \\ $$
Commented by maxmathsup by imad last updated on 04/Jun/19
$$\left.\mathrm{2}\right)\:{sin}\left(\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{2}{n}}\right)\:={sin}\left(\frac{\pi}{\mathrm{4}}\right){cos}\left(\frac{\pi}{\mathrm{2}{n}}\right)−{cos}\left(\frac{\pi}{\mathrm{4}}\right){sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({cos}\left(\frac{\pi}{\mathrm{2}{n}}\right)−{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)\right)\:\Rightarrow \\ $$$${cos}\left(\frac{\pi}{\mathrm{2}{n}}\right)\sim\mathrm{1}−\frac{\left(\frac{\pi}{\mathrm{2}{n}}\right)^{\mathrm{2}} }{\mathrm{2}}\:=\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{8}{n}^{\mathrm{2}} }\:\:\:\:,\:\:{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)\sim\:\frac{\pi}{\mathrm{2}{n}}\:\Rightarrow \\ $$$${U}_{{n}} \sim\:\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}\left\{\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{8}{n}^{\mathrm{2}} }−\frac{\pi}{\mathrm{2}{n}}\right\}}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\left\{\:\frac{\mathrm{1}}{\pi}−\frac{\pi}{\mathrm{8}{n}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{\mathrm{2}{n}}\right\}}\:\Rightarrow{n}^{\mathrm{2}} \:{U}_{{n}} \sim\:\frac{{n}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}\left\{\frac{\mathrm{1}}{\pi}−\frac{\pi}{\mathrm{8}{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}{n}}\right\}} \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:\left(\frac{\mathrm{1}}{\pi}\:−\frac{\pi}{\mathrm{8}{n}^{\mathrm{2}} }\:\frac{\mathrm{1}}{\mathrm{2}{n}}\right)\:=\frac{\mathrm{1}}{\pi}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} {n}^{\mathrm{2}} \:{U}_{{n}} =+\infty \\ $$
Commented by maxmathsup by imad last updated on 04/Jun/19
$$\left.\mathrm{3}\right)\:{we}\:\:{lim}_{{n}\rightarrow\infty} {U}_{{n}} =\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\neq\mathrm{0}\:\Rightarrow\:\Sigma\:{U}_{{n}} \:{diverges}\:. \\ $$