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Question Number 75066 by ~blr237~ last updated on 06/Dec/19

Prove that if f is a function R \to R

and there exist x_{0} > 0, such as L (f) (x_{0}) exist

then \lim_{t \to \infty} f (t) e^{- x_{0} t} = 0 and \forall x > x_{0} L (f) (x) exist .

L (f) is the Laplace transformed function

Answered by mind is power last updated on 07/Dec/19

L (f) (s) = \int_{0}^{+ \infty} e^{- st} f (t) dt

L (f) (x_{0}) = \int_{0}^{+ \infty} ∣ e^{- x_{0} t} f (t) ∣ dt < \infty

\Rightarrow e^{- x_{0} t} f (t) is integrabl in + \infty

\Rightarrow e^{- x_{0} t} f (t) \to 0 by cv

for x > x_{0}

x = x_{0} + n, n \in R_{+}

L (f) (x) = \int_{0}^{+ \infty} e^{- (x_{0} + n) t} f (t) dt

= \int_{0}^{+ \infty} e^{- x_{0} t} f (t) . e^{- nt} dt

since L (f) (x_{0}) exist \Rightarrow \lim f (t) e^{- x_{0} t} = o \Rightarrow \exists A \in R \forall x > A

∣ f (t) e^{- x_{0} t} ∣< 1 \Rightarrow ∣ f (t) e^{- x_{0} t} . e^{- nt} ∣< e^{- nt}

so ∣ f (t) e^{- (x_{0} + n) t} ∣< e^{- nt} t \to e^{- nt} is integral n > 0

\forall t > A

so \int_{0}^{+ \infty} ∣ f (t) e^{- (x_{0} + n) t} ∣ dt exist \Rightarrow \int_{0}^{+ \infty} f (t) e^{- (x_{0} + n) t} dt exist

Commented by ~blr237~ last updated on 06/Dec/19

sir L (f) (x_{0}) exist just mean that \int_{0}^{\infty} f (t) e^{- x_{0} t} dt < + \infty

that condition you used is just sufficient

cause all if f \in L^{1} (R_{+}) (\int_{0}^{\infty} ∣ f (t) ∣ dt < + \infty) then L (f) (s) exist for all s > 0

Commented by mind is power last updated on 07/Dec/19

yes simples cv not absulute