Question Number 145399 by Willson last updated on 04/Jul/21
$$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow+\infty} {\boldsymbol{\mathrm{lim}}}\:\:\underset{\:\mathrm{0}} {\int}^{\:\boldsymbol{\mathrm{n}}} \:\frac{\boldsymbol{\mathrm{t}}^{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}!}\:\boldsymbol{{e}}^{−\boldsymbol{\mathrm{t}}} \:\boldsymbol{\mathrm{dt}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by ArielVyny last updated on 04/Jul/21
$$\int_{\mathrm{0}} ^{{n}} \frac{{t}^{{n}} }{{n}!}{e}^{−{t}} {dt}=\frac{\mathrm{1}}{{n}!}\int_{\mathrm{0}} ^{+\infty} {e}^{−{t}} {t}^{{n}} {dt}\frac{\mathrm{1}}{{n}!}\Gamma\left({n}+\mathrm{1}\right)=\frac{{n}!}{{n}!}=\mathrm{1} \\ $$