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Question Number 145399 by Willson last updated on 04/Jul/21

$$\mathrm{Prove}\mathrm{that}$$$$\underset{\mathbf{n}\to +\mathrm{\infty }}{\mathbf{lim}}{\underset{0}{\int }}^{\mathbf{n}}\frac{{\mathbf{t}}^{\mathbf{n}}}{\mathbf{n}!}{\mathit{e}}^{-\mathbf{t}}\mathbf{dt}=\frac{1}{2}$$

Answered by ArielVyny last updated on 04/Jul/21

$${\int }_0^n\frac{t^n}{n!}{e}^{-t}dt=\frac{1}{n!}{\int }_0^{+\mathrm{\infty }}{e}^{-t}{t}^{n}dt\frac{1}{n!}\mathrm{\Gamma }(n+1)=\frac{n!}{n!}=1$$

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