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Question Number 143296 by lapache last updated on 12/Jun/21

$$Montrerque$$$$\mathrm{\Gamma }\left(n\right)=(n-1)!$$$$$$

Answered by Olaf_Thorendsen last updated on 12/Jun/21

$$\mathrm{By}\mathrm{definition}\mathrm{\Gamma }\left(z\right)={\int }_0^{\mathrm{\infty }}{t}^{z-1}{e}^{-t}dt$$$$\mathrm{If}z=n\in \mathbb{N}:$$$$\mathrm{\Gamma }\left(n\right)={\int }_0^{\mathrm{\infty }}{t}^{n-1}{e}^{-t}dt$$$$\mathrm{\Gamma }\left(n\right)={\int }_0^{\mathrm{\infty }}1\times t^{n-1}{e}^{-t}dt$$$$\mathrm{\Gamma }\left(n\right)={[t.t^{n-1}{e}^{-t}]}_0^{\mathrm{\infty }}$$$$-{\int }_0^{\mathrm{\infty }}t[(n-1)t^{n-2}-t^{n-1}]e^{-t}dt$$$$\mathrm{\Gamma }\left(n\right)=-(n-1){\int }_0^{\mathrm{\infty }}{t}^{n-1}{e}^{-t}dt+{\int }_0^{\mathrm{\infty }}{t}^n{e}^{-t}dt$$$$\mathrm{\Gamma }\left(n\right)=-(n-1)\mathrm{\Gamma }\left(n\right)+\mathrm{\Gamma }(n+1)$$$$\mathrm{\Gamma }(n+1)=n\mathrm{\Gamma }\left(n\right)$$$$\mathrm{\Gamma }(n+1)=n(n-1)\mathrm{\Gamma }(n-1)$$$$...$$$$\mathrm{\Gamma }(n+1)=n!\mathrm{\Gamma }\left(1\right)$$$$$$$$\mathrm{\Gamma }\left(1\right)={\int }_0^{\mathrm{\infty }}{t}^{1-1}{e}^{-t}dt={\int }_0^{\mathrm{\infty }}{e}^{-t}dt$$$$\mathrm{\Gamma }\left(1\right)={[-e^{-t}]}_0^{\mathrm{\infty }}=1$$$$$$$$\mathrm{\Gamma }(n+1)=n!\mathrm{\Gamma }\left(1\right)=n!\times 1=n!$$$$$$$$\mathrm{and}\mathrm{of}\mathrm{course}\mathrm{\Gamma }\left(n\right)=(n-1)!$$

Answered by Ar Brandon last updated on 12/Jun/21

$$\mathrm{I}=\mathrm{\Gamma }(\mathrm{n}+1)=\mathrm{n}!$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt},\mathrm{u}\left(\mathrm{t}\right)={\mathrm{t}}^{\mathrm{n}},{\mathrm{v}}^{\prime }\left(\mathrm{t}\right)={\mathrm{e}}^{-\mathrm{t}}$$$$={[-{\mathrm{t}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{t}}+\mathrm{n}\int {\mathrm{t}}^{\mathrm{n}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}]}_0^{\mathrm{\infty }}=\mathrm{n}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{n}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}$$$$=\mathrm{n}{[-{\mathrm{t}}^{\mathrm{n}-1}{\mathrm{e}}^{-\mathrm{t}}+(\mathrm{n}-1)\int {\mathrm{t}}^{\mathrm{n}-2}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}]}_0^{\mathrm{\infty }}=\mathrm{n}(\mathrm{n}-1){\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{n}-2}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}$$$$=\mathrm{n}(\mathrm{n}-1){[-{\mathrm{t}}^{\mathrm{n}-2}{\mathrm{e}}^{-\mathrm{t}}+(\mathrm{n}-2)\int {\mathrm{t}}^{\mathrm{n}-3}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}]}_0^{\mathrm{\infty }}$$$$=\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2){\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{n}-3}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}$$$$=\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)...3\times 2\times {\int }_0^{\mathrm{\infty }}{\mathrm{te}}^{-\mathrm{t}}\mathrm{dt}=\mathrm{n}!$$

Answered by Dwaipayan Shikari last updated on 12/Jun/21

$${\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{n-1}dt{e}^{-t}=u\Rightarrow -e^{-t}dt=du$$$$={\int }_0^1{(-1)}^{n-1}t{log}^{n-1}\left(t\right)dt=\frac{{\partial }^{n-1}}{\partial a^{n-1}}{\mid }_{a=1}{\int }_0^1{t}^{a}dt=\frac{{\partial }^{n-1}}{\partial a^{n-1}}{\mid }_{a=1}.\frac{1}{(a+1)}$$$$=\frac{{(-1)}^{n-1}{(-1)}^{n-1}(n-1)!}{{(a+1)}^n}{\mid }_{a=0}=(n-1)!$$

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