Question Number 42695 by prof Abdo imad last updated on 31/Aug/18
$${calculate}\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{x}^{\mathrm{4}} }{{x}^{\mathrm{8}\:} \:+\mathrm{16}}{dx} \\ $$
Commented by maxmathsup by imad last updated on 01/Sep/18
$${let}\:{A}\:\:=\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}^{\mathrm{4}} }{{x}^{\mathrm{8}} \:+\mathrm{16}}{dx}\:\Rightarrow\:{A}\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{4}} }{{x}^{\mathrm{8}} \:\:+\left(\sqrt{\mathrm{2}}\right)^{\mathrm{8}} }\:\:\:{changement}\:{x}={t}\sqrt{\mathrm{2}}{give} \\ $$$${A}\:\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{4}{t}^{\mathrm{4}} }{\mathrm{16}{t}^{\mathrm{8}} \:+\mathrm{16}}{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}^{\mathrm{4}} }{\mathrm{1}+{t}^{\mathrm{8}} }{dt}\: \\ $$$$=_{{t}\:={u}^{\frac{\mathrm{1}}{\mathrm{8}}} } \:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{{u}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+{u}}\:\frac{\mathrm{1}}{\mathrm{8}}\:{u}^{\frac{\mathrm{1}}{\mathrm{8}}−\mathrm{1}} {du}\:=\:\frac{\mathrm{1}}{\mathrm{16}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{u}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{8}}−\mathrm{1}} }{\mathrm{1}+{u}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{u}^{\frac{\mathrm{5}}{\mathrm{8}}−\mathrm{1}} }{\mathrm{1}+{u}}{du}\:=\frac{\mathrm{1}}{\mathrm{16}}\:\:\frac{\pi}{{sin}\left(\frac{\mathrm{5}\pi}{\mathrm{8}}\right)}\:\:\:{but}\:\:{sin}\left(\frac{\mathrm{5}\pi}{\mathrm{8}}\right)\:={sin}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\right)={cos}\left(\frac{\pi}{\mathrm{8}}\right)=\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{A}\:\:=\frac{\pi}{\mathrm{16}}\:\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}\:\Rightarrow\:{A}\:\:=\frac{\pi\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{32}}\:. \\ $$