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

Prove that if f is a function R→R and there exist x_0 >0 , such as L(f)(x_0 ) exist then lim_(t→∞) f(t)e^(−x_0 t) =0 and ∀ x>x_0 L(f)(x) exist. L(f) is the Laplace transformed function

ProvethatiffisafunctionR→Randthereexistx0>0,suchasL(f)(x0)existthenlimt→∞f(t)e−x0t=0and∀x>x0L(f)(x)exist.L(f)istheLaplacetransformedfunction

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

L(f)(s)=∫_0 ^(+∞) e^(−st) f(t)dt L(f)(x_0 )=∫_0 ^(+∞) ∣e^(−x_0 t) f(t)∣dt<∞ ⇒e^(−x_0 t) f(t) is integrabl in+∞ ⇒e^(−x_0 t) f(t)→0 by cv for x>x_0 x=x_0 +n ,n∈R_+ L(f)(x)=∫_0 ^(+∞) e^(−(x_0 +n)t) f(t)dt =∫_0 ^(+∞) e^(−x_0 t) f(t).e^(−nt) dt since L(f)(x_0 ) exist⇒ lim f(t)e^(−x_0 t) =o⇒∃A∈R ∀x>A ∣f(t)e^(−x_0 t) ∣<1⇒ ∣f(t)e^(−x_0 t) .e^(−nt) ∣<e^(−nt) so ∣f(t)e^(−(x_0 +n)t) ∣<e^(−nt) t→e^(−nt) is integral n>0 ∀t>A so ∫_0 ^(+∞) ∣f(t)e^(−(x_0 +n)t) ∣dt exist⇒∫_0 ^(+∞) f(t)e^(−(x_0 +n)t ) dt exist

L(f)(s)=∫0+∞e−stf(t)dtL(f)(x0)=∫0+∞∣e−x0tf(t)∣dt<∞⇒e−x0tf(t)isintegrablin+∞⇒e−x0tf(t)→0bycvforx>x0x=x0+n,n∈R+L(f)(x)=∫0+∞e−(x0+n)tf(t)dt=∫0+∞e−x0tf(t).e−ntdtsinceL(f)(x0)exist⇒limf(t)e−x0t=o⇒∃A∈R∀x>A∣f(t)e−x0t∣<1⇒∣f(t)e−x0t.e−nt∣<e−ntso∣f(t)e−(x0+n)t∣<e−ntt→e−ntisintegraln>0∀t>Aso∫0+∞∣f(t)e−(x0+n)t∣dtexist⇒∫0+∞f(t)e−(x0+n)tdtexist

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

sir L(f)(x_0 ) exist just mean that ∫_0 ^∞ f(t)e^(−x_0 t) dt <+∞ that condition you used is just sufficient cause all if f∈ L^1 (R_+ )(∫_0 ^∞ ∣f(t)∣dt<+∞) then L(f)(s) exist for all s>0

sirL(f)(x0)existjustmeanthat∫0∞f(t)e−x0tdt<+∞thatconditionyouusedisjustsufficientcausealliff∈L1(R+)(∫0∞∣f(t)∣dt<+∞)thenL(f)(s)existforalls>0

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

yes simples cv not absulute

yessimplescvnotabsulute

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