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Question Number 102089 by Ar Brandon last updated on 06/Jul/20

$$$$$$\mathrm{The}\mathrm{Gamma}\mathrm{function}\mathrm{\Gamma }\left(\alpha \right)\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{follows};$$$$\mathrm{\Gamma }\left(\alpha \right)={\int }_0^{\mathrm{\infty }}{\mathrm{y}}^{\alpha -1}{\mathrm{e}}^{-\mathrm{y}}\mathrm{dy},\alpha >0$$$$\mathrm{a}\mathrm{\setminus }\mathrm{Show}\mathrm{that}\mathrm{\Gamma }(\alpha +1)=\alpha \mathrm{\Gamma }\left(\alpha \right).$$$$\mathrm{b}\mathrm{\setminus }\mathrm{Conclude}\mathrm{that}\mathrm{\Gamma }\left(\mathrm{n}\right)=(\mathrm{n}-1)!,\mathrm{n}=1,2,3,...$$$$\mathrm{c}\mathrm{\setminus }\mathrm{Determine}\mathrm{\Gamma }\left(55\right).$$

Answered by Ar Brandon last updated on 06/Jul/20

$$\mathrm{a}\mathrm{\setminus }\mathrm{\Gamma }(\alpha +1)={\int }_0^{\mathrm{\infty }}{\mathrm{y}}^{\alpha }{\mathrm{e}}^{-\mathrm{y}}\mathrm{dy}={[{\mathrm{y}}^{\alpha }\int {\mathrm{e}}^{-\mathrm{y}}\mathrm{dy}]}_0^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}\{\frac{{\mathrm{dy}}^{\alpha }}{\mathrm{dy}}\cdot \int {\mathrm{e}}^{-\mathrm{y}}\mathrm{dy}\}\mathrm{dy}$$$$={[-{\mathrm{y}}^{\alpha }{\mathrm{e}}^{-\mathrm{y}}]}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}\alpha {\mathrm{y}}^{\alpha -1}{\mathrm{e}}^{-\mathrm{y}}\mathrm{dy}=\alpha {\int }_0^{\mathrm{\infty }}{\mathrm{y}}^{\alpha -1}{\mathrm{e}}^{-\mathrm{y}}\mathrm{dy}$$$$=\alpha \mathrm{\Gamma }\left(\alpha \right)$$

Answered by floor(10²Eta[1]) last updated on 06/Jul/20

$$\mathrm{\Gamma }\left(55\right)=54!:)$$

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