On-definit-la-fonction-L-f-t-p-0-f-t-e-pt-dt-Calculer-L-t-n-n-p-

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

O n d e f i n i t l a f o n c t i o n

L (f (t)) (p) = \int_{0}^{+ \infty} f (t) e^{- p t} d t

C a l c u l e r L ((\frac{t^{n}}{n!})) (p)

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

L (\frac{t^{n}}{n!}) (p) = \int_{0}^{\infty} \frac{t^{n}}{n!} e^{- pt} dt = \frac{1}{p^{n + 1}} \int_{0}^{\infty} \frac{t^{n}}{n!} e^{- t} dt

= \frac{1}{p^{n + 1}} \cdot \frac{Γ (n + 1)}{n!} = \frac{1}{p^{n + 1}}

Commented by lapache last updated on 15/Jun/21

T u f a i s c o m m e n t p o u r a v o i r l e

\frac{1}{p^{n + 1}} \int_{0}^{\infty} \frac{t^{n}}{n!} e^{- t} d t ? ? ? ? ? ?

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

\overset{`}{A} l^{'} aide d^{'} un changement de variable

u = pt \Rightarrow t = \frac{u}{p} \Rightarrow dt = \frac{1}{p} du

\int_{0}^{\infty} \frac{t^{n}}{n!} e^{- pt} dt = \int_{0}^{\infty} \frac{1}{n!} {(\frac{u}{p})}^{n} e^{- u} (\frac{du}{p})

= \int_{0}^{\infty} \frac{1}{n!} {(\frac{1}{p})}^{n + 1} u^{n} e^{- u} du