Question Number 63664 by mathmax by abdo last updated on 07/Jul/19
$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{{x}+{t}^{{n}} }\:{dt}\:\:\:{with}\:\mathrm{0}<{a}<\mathrm{1}\:\:{and}\:\:{x}>\mathrm{0}\:{and}\:{n}\geqslant\mathrm{2} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\left({x}+{t}^{{n}} \right)^{\mathrm{2}} }\:{dt} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{f}^{\left({k}\right)} \left({x}\right)\:{at}\:{form}\:{of}\:{integrals} \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{9}+{t}^{\mathrm{2}} }\:{dt}\:\:\:\:{and}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\left(\mathrm{9}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{5}\right)\:{calculate}\:\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} }{\mathrm{2}^{{n}} \:+{t}^{{n}} }\:{dt}\:\:{and}\:{study}\:{the}\:{convergence}\:{of}\:\Sigma\:{U}_{{n}} \\ $$
Commented by mathmax by abdo last updated on 09/Jul/19
$$\left.\mathrm{1}\right){we}\:{have}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{{x}\:+{t}^{{n}} }{dt}\:\:{changement}\:{t}\:=^{{n}} \sqrt{{x}}{u}\:{give} \\ $$$${f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\left(^{{n}} \sqrt{{x}}{u}\right)^{{a}−\mathrm{1}} }{{x}\:+{xu}^{{n}} }\:^{{n}} \sqrt{{x}}{du} \\ $$$$=\frac{{x}^{\frac{{a}−\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}}} }{{x}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{u}^{{a}−\mathrm{1}} }{\mathrm{1}+{u}^{{n}} }\:{du}\:={x}^{\frac{{a}}{{n}}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{u}^{{a}−\mathrm{1}} }{\mathrm{1}+{u}^{{n}} } \\ $$$$=_{{u}=\alpha^{\frac{\mathrm{1}}{{n}}} } \:\:\:\:{x}^{\frac{{a}}{{n}}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\alpha^{\frac{{a}−\mathrm{1}}{{n}}} }{\mathrm{1}+\alpha}\:\frac{\mathrm{1}}{{n}}\:\alpha^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} \:{d}\alpha \\ $$$$=\:{x}^{\frac{{a}}{{n}}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\alpha^{\frac{{a}−\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}}−\mathrm{1}} }{\mathrm{1}+\alpha}\:{d}\alpha\:=\:{x}^{\frac{{a}}{{n}}−\mathrm{1}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\alpha^{\frac{{a}}{{n}}−\mathrm{1}} }{\mathrm{1}+\alpha}\:{d}\alpha\:\Rightarrow \\ $$$${f}\left({x}\right)={x}^{\frac{{a}}{{n}}−\mathrm{1}} \:\:\frac{\pi}{{sin}\left(\frac{\pi{a}}{{n}}\right)} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 09/Jul/19
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{by}\:{derivation}\:{f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \frac{\partial}{\partial{x}}\left(\frac{{t}^{{a}−\mathrm{1}} }{{x}+{t}^{{n}} }\right){dt} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\left({x}+{t}^{{n}} \right)^{\mathrm{2}} }{dt}\:=−{g}\left({x}\right)\:\Rightarrow{g}\left({x}\right)=−{f}^{'} \left({x}\right) \\ $$$${f}\left({x}\right)=\frac{\pi}{{sin}\left(\frac{\pi{a}}{{n}}\right)}\:{x}^{\frac{{a}}{{n}}−\mathrm{1}} \:\Rightarrow{f}^{'} \left({x}\right)\:=\frac{\pi\left(\frac{{a}}{{n}}−\mathrm{1}\right)}{{sin}\left(\frac{\pi{a}}{{n}}\right)}\:{x}^{\frac{{a}}{{n}}−\mathrm{2}} \:\Rightarrow \\ $$$${g}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\left({x}+{t}^{{n}} \right)^{\mathrm{2}} }{dt}\:=\frac{\pi\left(\mathrm{1}−\frac{{a}}{{n}}\right)}{{sin}\left(\frac{\pi{a}}{{n}}\right)}\:{x}^{\frac{{a}}{{n}}−\mathrm{2}} \\ $$
Commented by mathmax by abdo last updated on 09/Jul/19
$$\left.\mathrm{3}\right)\:{we}\:{have}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{{x}\:+{t}^{{n}} }\:{dt}\:\Rightarrow{f}^{\left({k}\right)} \left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\partial^{{k}} }{\partial{x}^{{k}} }\left(\frac{{t}^{{a}−\mathrm{1}} }{{x}+{t}^{{n}} }\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} {k}!}{\left({x}+{t}^{{n}} \right)^{{k}+\mathrm{1}} }\:{t}^{{a}−\mathrm{1}} \:{dt}\:\:\:\Rightarrow{f}^{\left({k}\right)} \left({x}\right)\:=\left(−\mathrm{1}\right)^{{k}} {k}!\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\left({x}+{t}^{{n}} \right)^{{k}+\mathrm{1}} }{dt} \\ $$
Commented by mathmax by abdo last updated on 09/Jul/19
$$\left.\mathrm{4}\right)\:{we}\:{have}\:{proved}\:{that}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{{x}+{t}^{{n}} }\:{dt}\:\:=\frac{\pi}{{sin}\left(\frac{\pi{a}}{{n}}\right)}\:{x}^{\frac{{a}}{{n}}−\mathrm{1}} \:{and} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\left({x}+{t}^{{n}} \right)^{\mathrm{2}} }{dt}\:=\frac{\pi\left(\mathrm{1}−\frac{{a}}{{n}}\right)}{{sin}\left(\frac{\pi{a}}{{n}}\right)}\:{x}^{\frac{{a}}{{n}}−\mathrm{2}} \\ $$$${x}=\mathrm{9}\:{and}\:{n}=\mathrm{2}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{9}+{t}^{\mathrm{2}} }{dt}\:=\frac{\pi}{{sin}\left(\frac{\pi{a}}{\mathrm{2}}\right)}×\mathrm{9}^{\frac{{a}}{\mathrm{2}}−\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\left(\mathrm{9}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:=\frac{\pi\left(\mathrm{1}−\frac{{a}}{\mathrm{2}}\right)}{{sin}\left(\frac{\pi{a}}{\mathrm{2}}\right)}×\mathrm{9}^{\frac{{a}}{\mathrm{2}}−\mathrm{2}} \\ $$
Commented by mathmax by abdo last updated on 09/Jul/19
$$\left.\mathrm{5}\right)\:{we}\:{have}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{{x}+{t}^{{n}} }{dt}\:=\frac{\pi}{{sin}\left(\frac{\pi{a}}{{n}}\right)}\:{x}^{\frac{{a}}{{n}}−\mathrm{1}} \\ $$$${a}=\frac{\mathrm{1}}{{n}}\:\:{and}\:{x}\:=\mathrm{2}^{{n}} \:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} }{\mathrm{2}^{{n}} \:+{t}^{{n}} }{dt}\:=\:\frac{\pi}{{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)}\left(\mathrm{2}^{{n}} \right)^{\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\mathrm{1}} \\ $$$$=\frac{\pi}{{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)}×\mathrm{2}^{\frac{\mathrm{1}}{{n}}−{n}} \:=\frac{\pi}{\mathrm{2}^{{n}−\frac{\mathrm{1}}{{n}}} \:\:\:{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)}\:. \\ $$
Commented by mathmax by abdo last updated on 09/Jul/19
$$\Rightarrow\:{U}_{{n}} =\frac{\pi}{\mathrm{2}^{{n}−\frac{\mathrm{1}}{{n}}} \:{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)} \\ $$