Question Number 109863 by mathdave last updated on 26/Aug/20
Answered by mathdave last updated on 26/Aug/20
$${solution} \\ $$$${recall}\:{that}\:\: \\ $$$$\:_{{a}} {D}_{{x}} ^{−\alpha} {f}\left({t}\right)=_{{a}} {I}_{{x}} ^{\alpha} {f}\left({t}\right)=\frac{\mathrm{1}}{\Gamma\left(\alpha\right)}\int_{{a}} ^{{x}} \left({x}−{t}\right)^{\alpha−\mathrm{1}} {f}\left({t}\right){dt}\:\:\:\:\:{for} \\ $$$${x}\succ{a} \\ $$$${let}\:\:\alpha=\frac{\mathrm{1}}{\mathrm{3}}\:\:\:{a}=\mathrm{0}\:\:\:{x}=\mathrm{1}\:\:{f}\left({t}\right)={t}^{\mathrm{4}} \\ $$$$\:\:\:\:_{\mathrm{0}} {I}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{3}}} {f}\left({t}^{\mathrm{4}} \right)=\frac{\mathrm{1}}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} {t}^{\mathrm{4}} {dt} \\ $$$$=\frac{\mathrm{1}}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{5}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} {dt}=\frac{\mathrm{1}}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}\beta\left(\mathrm{5},\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$=\frac{\mathrm{1}}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}\bullet\frac{\Gamma\left(\mathrm{5}\right).\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}{\Gamma\left(\mathrm{5}+\frac{\mathrm{1}}{\mathrm{3}}\right)}=\frac{\Gamma\left(\mathrm{5}\right)}{\Gamma\left(\frac{\mathrm{16}}{\mathrm{3}}\right)}=\frac{\mathrm{4}!}{\frac{\mathrm{13}}{\mathrm{3}}!}=\frac{\mathrm{24}}{\frac{\mathrm{13}}{\mathrm{3}}×\frac{\mathrm{10}}{\mathrm{3}}×\frac{\mathrm{7}}{\mathrm{3}}×\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{0}!} \\ $$$$\:\:\:\because\:\:_{\mathrm{0}} {I}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{3}}} {f}\left({t}^{\mathrm{3}} \right)=\frac{\mathrm{729}}{\mathrm{455}} \\ $$$${by}\:{mathdave} \\ $$