Question Number 79527 by mathmax by abdo last updated on 25/Jan/20
$${find}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}^{\mathrm{3}} } {dx} \\ $$
Commented by mathmax by abdo last updated on 26/Jan/20
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}^{\mathrm{3}} } {dx}\:\:{changement}\:{x}^{\mathrm{3}} \:={t}\:{give}\:{x}={t}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} \:\frac{\mathrm{1}}{\mathrm{3}}{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} {dt}\:=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} \:{e}^{−{t}} \:{dt}\:\:{we}\:{have}\: \\ $$$$\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} \:{dt}\:\:\left({x}>\mathrm{0}\right)\:\Rightarrow{I}\:=\frac{\mathrm{1}}{\mathrm{3}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$ \\ $$