prove-that-L-1-s-1-s-and-L-t-n-s-n-s-n-1-L-means-laplace-transform-

Question Number 28991 by abdo imad last updated on 02/Feb/18

p r o v e t h a t L (1) (s) = \frac{1}{s} a n d L (t^{n}) (s) = \frac{n!}{s^{n + 1}} . L m e a n s

l a p l a c e t r a n s f o r m .

Commented by abdo imad last updated on 03/Feb/18

L (1) (s) = \int_{0}^{\infty} e^{- s t} d t = \frac{1}{s} w i t h s > 0 a n d

L (t^{n}) (s) = \int_{0}^{\infty} t^{n} e^{- s t} d t l e t u s e t h e c h . s t = x

L (t^{n}) (s) = \int_{0}^{\infty} {(\frac{x}{s})}^{n} e^{- x} \frac{d x}{s} = \int_{0}^{\infty} \frac{1}{s^{n + 1}} x^{n} e^{- x} d x

= \frac{1}{s^{n + 1}} \int_{0}^{\infty} x^{n} e^{- x} d x l e t p u t A_{n} = \int_{0}^{\infty} x^{n} e^{- x} d x b y p a r t s

A_{n} = {[- x^{n} e^{- x}]}_{0}^{\infty} + \int_{0}^{\infty} {n x}^{n - 1} e^{- x} d x = n A_{n - 1}

A_{n} = n (n - 1) A_{n - 2} = \dots . = n (n - 1) \dots . (n - p + 1) {L L A}_{n - p}

= n! A_{0} = n! (A_{o} = 1) \Rightarrow

L (t^{n}) (s) = \frac{n!}{s^{n + 1}} .

Answered by sma3l2996 last updated on 03/Feb/18

L (1) (s) = \int_{0}^{\infty} e^{- s t} d t = - \frac{1}{s} {[e^{- s t}]}_{0}^{\infty} = \frac{1}{s}

L (t^{n}) (s) = \int_{0}^{\infty} t^{n} e^{- s t} d t

u = t^{n} \Rightarrow u^{'} = {n t}^{n - 1}

v^{'} = e^{- s t} \Rightarrow v = - \frac{1}{s} e^{- s t}

L (t^{n}) (s) = \frac{n}{s} \int_{0}^{\infty} t^{n - 1} e^{- s t} d t

s o L (t^{n}) (s) = \frac{n}{s} L (t^{n - 1}) (s)

L (t^{n - 1}) (s) = \frac{n - 1}{s} L (t^{n - 2}) (s)

s o :

L (t^{n}) (s) = \frac{n}{s} (\frac{n - 1}{s} (\frac{n - 2}{s} \dots (\frac{n - (n - 1)}{s} L (t^{n - (n - 1)}) \dots))

= \frac{n!}{s^{n}} L (t) (s) = \frac{n!}{s^{n - 1}} (\frac{1}{s} L (1) (s)) = \frac{n!}{s^{n}} (\frac{1}{s})

L (t^{n}) = \frac{n!}{s^{n + 1}}