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Question Number 109863 by mathdave last updated on 26/Aug/20

Answered by mathdave last updated on 26/Aug/20

$$solution$$$$recallthat$$$${}_a{D}_x^{-\alpha }f\left(t\right){=}_a{I}_x^{\alpha }f\left(t\right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_a^x{(x-t)}^{\alpha -1}f\left(t\right)dtfor$$$$x\succ a$$$$let\alpha =\frac{1}{3}a=0x=1f\left(t\right)=t^4$$$${}_0{I}_1^{\frac{1}{3}}f\left(t^4\right)=\frac{1}{\mathrm{\Gamma }\left(\frac{1}{3}\right)}{\int }_0^1{(1-t)}^{\frac{1}{3}-1}{t}^{4}dt$$$$=\frac{1}{\mathrm{\Gamma }\left(\frac{1}{3}\right)}{\int }_0^1{t}^{5-1}{(1-t)}^{\frac{1}{3}-1}dt=\frac{1}{\mathrm{\Gamma }\left(\frac{1}{3}\right)}\beta (5,\frac{1}{3})$$$$=\frac{1}{\mathrm{\Gamma }\left(\frac{1}{3}\right)}\bullet \frac{\mathrm{\Gamma }\left(5\right).\mathrm{\Gamma }\left(\frac{1}{3}\right)}{\mathrm{\Gamma }(5+\frac{1}{3})}=\frac{\mathrm{\Gamma }\left(5\right)}{\mathrm{\Gamma }\left(\frac{16}{3}\right)}=\frac{4!}{\frac{13}{3}!}=\frac{24}{\frac{13}{3}\times \frac{10}{3}\times \frac{7}{3}\times \frac{4}{3}\times \frac{1}{3}\times 0!}$$$$\because {}_0{I}_1^{\frac{1}{3}}f\left(t^3\right)=\frac{729}{455}$$$$bymathdave$$

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