Question Number 143495 by lapache last updated on 15/Jun/21
$${On}\:{definit}\:{la}\:{fonction}\: \\ $$$$\mathscr{L}\left({f}\left({t}\right)\right)\left({p}\right)=\int_{\mathrm{0}} ^{+\infty} {f}\left({t}\right){e}^{−{pt}} {dt} \\ $$$${Calculer}\:\mathscr{L}\left(\left(\frac{{t}^{{n}} }{{n}!}\right)\right)\left({p}\right) \\ $$
Answered by Ar Brandon last updated on 15/Jun/21
$$\mathscr{L}\left(\frac{\mathrm{t}^{\mathrm{n}} }{\mathrm{n}!}\right)\left(\mathrm{p}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{n}} }{\mathrm{n}!}\mathrm{e}^{−\mathrm{pt}} \mathrm{dt}=\frac{\mathrm{1}}{\mathrm{p}^{\mathrm{n}+\mathrm{1}} }\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{n}} }{\mathrm{n}!}\mathrm{e}^{−\mathrm{t}} \mathrm{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{p}^{\mathrm{n}+\mathrm{1}} }\centerdot\frac{\Gamma\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{n}!}=\frac{\mathrm{1}}{\mathrm{p}^{\mathrm{n}+\mathrm{1}} }\: \\ $$
Commented by lapache last updated on 15/Jun/21
$${Tu}\:{fais}\:{comment}\:{pour}\:{avoir}\:{le} \\ $$$$\frac{\mathrm{1}}{{p}^{{n}+\mathrm{1}} }\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{n}} }{{n}!}{e}^{−{t}} {dt}\:\:\:\:??????\:\:\:\: \\ $$
Commented by Ar Brandon last updated on 15/Jun/21
$$\grave {\mathrm{A}}\:\mathrm{l}'\mathrm{aide}\:\mathrm{d}'\mathrm{un}\:\mathrm{changement}\:\mathrm{de}\:\mathrm{variable} \\ $$$$\mathrm{u}=\mathrm{pt}\Rightarrow\mathrm{t}=\frac{\mathrm{u}}{\mathrm{p}}\:\Rightarrow\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{p}}\mathrm{du} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{n}} }{\mathrm{n}!}\mathrm{e}^{−\mathrm{pt}} \mathrm{dt}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{n}!}\left(\frac{\mathrm{u}}{\mathrm{p}}\right)^{\mathrm{n}} \mathrm{e}^{−\mathrm{u}} \left(\frac{\mathrm{du}}{\mathrm{p}}\right) \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{n}!}\left(\frac{\mathrm{1}}{\mathrm{p}}\right)^{\mathrm{n}+\mathrm{1}} \mathrm{u}^{\mathrm{n}} \mathrm{e}^{−\mathrm{u}} \mathrm{du} \\ $$