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Question Number 74503 by crystal0207 last updated on 25/Nov/19

Commented by mathmax by abdo last updated on 25/Nov/19

$$a){\int }_0^{\mathrm{\infty }}{x}^{\alpha -1}{e}^{-\lambda x}dx{=}_{\lambda x=t}{\int }_0^{\mathrm{\infty }}{\left(\frac{t}{\lambda }\right)}^{\alpha -1}{e}^{-t}\frac{dt}{\lambda }$$$$=\frac{1}{{\lambda }^{\alpha }}{\int }_0^{\mathrm{\infty }}{t}^{\alpha -1}{e}^{-t}dt=\frac{1}{{\lambda }^{\alpha }}\times \mathrm{\Gamma }\left(\alpha \right)(\lambda >0)$$$$b)\mathrm{\Gamma }(\alpha +1)={\int }_0^{\mathrm{\infty }}{x}^{\alpha }{e}^{-x}dxandbypartsu=x^{\alpha }and{v}^{{}^{\prime }}=e^{-x}$$$$\mathrm{\Gamma }(\alpha +1)={[-x^{\alpha }{e}^{-x}]}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}\alpha x^{\alpha -1}{e}^{-x}dx$$$$=\alpha \mathrm{\Gamma }\left(\alpha \right)$$$$c){\int }_0^{\mathrm{\infty }}{x}^n{e}^{-\lambda x}dx={\int }_0^{\mathrm{\infty }}{x}^{n+1-1}{e}^{-\lambda x}dx=\frac{\mathrm{\Gamma }(n+1)}{{\lambda }^{n+1}}$$$$\mathrm{\Gamma }(n+1)=n\mathrm{\Gamma }(n-1)=n(n-1)\mathrm{\Gamma }(n-2)=n!\mathrm{\Gamma }\left(1\right)$$$$\mathrm{\Gamma }\left(1\right)={\int }_0^{\mathrm{\infty }}{e}^{-x}dx={[-e^{-x}]}_0^{+\mathrm{\infty }}=1\Rightarrow {\int }_0^{\mathrm{\infty }}{x}^n{e}^{-\lambda x}dx=\frac{n!}{{\lambda }^{n+1}}$$

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