Question Number 28262 by abdo imad last updated on 22/Jan/18
$${find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{5}} }\:. \\ $$
Commented by abdo imad last updated on 24/Jan/18
$${let}\:{put}\:{x}^{\mathrm{5}} ={t}\:\:\Leftrightarrow\:{x}=\:{t}^{\frac{\mathrm{1}}{\mathrm{5}}} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{5}} }\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\frac{\mathrm{1}}{\mathrm{5}}{t}^{\frac{\mathrm{1}}{\mathrm{5}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\:\frac{\mathrm{1}}{\mathrm{5}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{5}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:{but}\:{we} \\ $$$${know}\:{that}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:^{} =\:\frac{\pi}{{sin}\left(\pi{a}\right)}\:{with}\:\mathrm{0}<{a}<\mathrm{1} \\ $$$${so}\:\:{I}=\:\:\frac{\frac{\pi}{\mathrm{5}}}{{sin}\left(\frac{\pi}{\mathrm{5}}\right)}\:\:\:{but}\:{cos}\left(\frac{\pi}{\mathrm{5}}\right)=\:\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\:\:{and} \\ $$$${sin}\left(\frac{\pi}{\mathrm{5}}\right)\:=\sqrt{\mathrm{1}−\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{16}}}\:\:=\frac{\sqrt{\mathrm{16}−\left(\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\right)}}{\mathrm{4}} \\ $$$$=\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}\:\:{so}\:{the}\:{value}\:{of}\:{I}\:{is}\:{known}. \\ $$