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Question Number 154927 by physicstutes last updated on 23/Sep/21

$$\mathrm{Let}{I}_n=\int x^n{e}^{-x}dx$$$$\mathrm{show}\mathrm{that}$$$${\int }_0^{\mathrm{\infty }}{x}^n{e}^{-x}dx=n!$$

Answered by ArielVyny last updated on 23/Sep/21

$$usegammafunction.$$$$\mathrm{\Gamma }\left(m\right)={\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{m-1}dxandweknowthat$$$$\mathrm{\Gamma }(m+1)=m!$$$$then{\int }_0^{\mathrm{\infty }}{x}^n{e}^{-x}dx=\mathrm{\Gamma }(n+1)=n!$$

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