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Question Number 143495 by lapache last updated on 15/Jun/21

$$Ondefinitlafonction$$$$\mathcal{L}\left(f\left(t\right)\right)\left(p\right)={\int }_0^{+\mathrm{\infty }}f\left(t\right)e^{-pt}dt$$$$Calculer\mathcal{L}\left(\left(\frac{t^n}{n!}\right)\right)\left(p\right)$$

Answered by Ar Brandon last updated on 15/Jun/21

$$\mathcal{L}\left(\frac{{\mathrm{t}}^{\mathrm{n}}}{\mathrm{n}!}\right)\left(\mathrm{p}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{n}}}{\mathrm{n}!}{\mathrm{e}}^{-\mathrm{pt}}\mathrm{dt}=\frac{1}{{\mathrm{p}}^{\mathrm{n}+1}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{n}}}{\mathrm{n}!}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}$$$$=\frac{1}{{\mathrm{p}}^{\mathrm{n}+1}}\cdot \frac{\mathrm{\Gamma }(\mathrm{n}+1)}{\mathrm{n}!}=\frac{1}{{\mathrm{p}}^{\mathrm{n}+1}}$$

Commented by lapache last updated on 15/Jun/21

$$Tufaiscommentpouravoirle$$$$\frac{1}{p^{n+1}}{\int }_0^{\mathrm{\infty }}\frac{t^n}{n!}{e}^{-t}dt??????$$

Commented by Ar Brandon last updated on 15/Jun/21

$$\stackrel{`}{\mathrm{A}}{\mathrm{l}}^{\prime }\mathrm{aide}{\mathrm{d}}^{\prime }\mathrm{un}\mathrm{changement}\mathrm{de}\mathrm{variable}$$$$\mathrm{u}=\mathrm{pt}\Rightarrow \mathrm{t}=\frac{\mathrm{u}}{\mathrm{p}}\Rightarrow \mathrm{dt}=\frac{1}{\mathrm{p}}\mathrm{du}$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{n}}}{\mathrm{n}!}{\mathrm{e}}^{-\mathrm{pt}}\mathrm{dt}={\int }_0^{\mathrm{\infty }}\frac{1}{\mathrm{n}!}{\left(\frac{\mathrm{u}}{\mathrm{p}}\right)}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{u}}\left(\frac{\mathrm{du}}{\mathrm{p}}\right)$$$$={\int }_0^{\mathrm{\infty }}\frac{1}{\mathrm{n}!}{\left(\frac{1}{\mathrm{p}}\right)}^{\mathrm{n}+1}{\mathrm{u}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{u}}\mathrm{du}$$

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