Question Number 63662 by mathmax by abdo last updated on 06/Jul/19
$$\:{let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{{a}−\mathrm{1}} }{\mathrm{1}+{x}^{{n}} }{dx}\:\:{with}\:{n}\:{integr}\:{and}\:{n}\geqslant\mathrm{2}\:\:{and}\:\mathrm{0}<{a}<\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{values}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{a}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{a}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{3}} }{dx} \\ $$$$\left.\mathrm{3}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\:\sqrt{{x}}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)}\:\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left(^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} }\right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)} \\ $$
Commented by mathmax by abdo last updated on 10/Jul/19
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{{a}−\mathrm{1}} }{\mathrm{1}+{x}^{{n}} }{dx}\:\:{changement}\:{x}^{{n}} ={t}\:{give}\:{x}={t}^{\frac{\mathrm{1}}{{n}}} \:\Rightarrow \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left({t}^{\frac{\mathrm{1}}{{n}}} \right)^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}\:\frac{\mathrm{1}}{{n}}{t}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} {dt}\:=\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{{a}−\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt} \\ $$$$=\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{{a}}{{n}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\mathrm{1}}{{n}}\:\frac{\pi}{{sin}\left(\frac{\pi{a}}{{n}}\right)}\:{by}\:{use}\:{of}\:{result}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{\alpha−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\pi}{{sin}\left(\pi\alpha\right)} \\ $$$$\left.\mathrm{2}\right)\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{a}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:={A}_{\mathrm{2}} =\frac{\pi}{\mathrm{2}{sin}\left(\frac{\pi{a}}{\mathrm{2}}\right)} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{{a}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{3}} }\:{dx}\:={A}_{\mathrm{3}} =\frac{\pi}{\mathrm{3}{sin}\left(\frac{\pi{a}}{\mathrm{3}}\right)} \\ $$$$\left.\mathrm{3}\right)\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\:\sqrt{{x}}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:\:\left({a}=\frac{\mathrm{1}}{\mathrm{2}}\:\:{and}\:{n}=\mathrm{4}\right) \\ $$$$=\:\frac{\pi}{\mathrm{4}{sin}\left(\frac{\pi}{\mathrm{4}}\right)}\:=\frac{\pi}{\mathrm{4}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:. \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left(^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} }\right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{−\frac{\mathrm{2}}{\mathrm{3}}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{4}} }\left(\rightarrow{n}=\mathrm{4}\:{and}\:{a}=\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$=\frac{\pi}{\mathrm{4}{sin}\left(\frac{\pi}{\mathrm{6}}\right)}\:=\frac{\pi}{\mathrm{4}.\frac{\mathrm{1}}{\mathrm{2}}}\:=\frac{\pi}{\mathrm{2}} \\ $$