Question Number 37360 by math khazana by abdo last updated on 12/Jun/18
$${calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{3}} } \\ $$
Commented by math khazana by abdo last updated on 13/Jun/18
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{3}} }\:\:{changement}\:{x}^{\mathrm{3}} \:={t}\:{give} \\ $$$${I}\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{t}}\:\frac{\mathrm{1}}{\mathrm{3}}\:{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} \:{dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} }{\mathrm{1}+{t}}\:{dt}\:\:=\frac{\mathrm{1}}{\mathrm{3}}\:\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{3}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\:\frac{\pi}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:\Rightarrow\:{I}\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:. \\ $$$${I}\:{have}\:{used}\:{tbe}\:{formula}\:{for}\:\mathrm{0}<{a}<\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}\:{dt}\:\:=\:\frac{\pi}{{sin}\left(\pi{a}\right)}\:. \\ $$$$ \\ $$