Example the function e at is of exponential order a

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Example. The function e at is of exponential order a as t → ∞ because (8.5) holds with K σ = 1 and T σ = 0 for every σ > a . Example. The function cos( bt ) is of exponential order 0 as t → ∞ because (8.5) holds with K σ = 1 and T σ = 0 for every σ > 0.
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7 Example. For every p > 0 the function t p is of exponential order 0 as t → ∞ . Indeed, for every σ > 0 the function e σt t p takes on its maximum over [0 , ) at t = p/σ , whereby e σt t p parenleftbigg p parenrightbigg p for every t 0 . It follows that (8.5) holds with K σ = ( p ) p and T σ = 0 for every σ > 0. One can show that if functions f and g are of exponential orders α and β respectively as t → ∞ then the function f + g is of exponential order max { α, β } as t → ∞ , while the function fg is of exponential order α + β as t → ∞ . Example. For every real a the function e at + e at is of exponential order | a | as t → ∞ . This is because the functions e at and e at are exponential orders a and a respectively as t → ∞ , and because | a | = max { a, a } . Example. For every p > 0 and every real a and b the function t p e at cos( bt ) is of exponential order a as t → ∞ . This is because the functions t p , e at , and cos( bt ) are of exponential orders 0, a , and 0 respectively as t → ∞ . The fact you should know about the existence of the Laplace transform for certain s is the following. Theorem. Let f ( t ) be piecewise continuous over every [0 , T ], of exponential order α as t → ∞ . Then for every positive integer k the function t k f ( t ) has these same properties. The function F ( s ) = L [ f ]( s ) is defined for every s > α . Moreover, F ( s ) is infinitely differentiable over s > α with L [ t k f ( t )]( s ) = ( 1) k d k d s k F ( s ) for s > α . (8.6) Proof. Formula (8.6) can be derived formally by differentiating the integrands: F ( s ) = integraldisplay 0 e st f ( t ) d t , d d s F ( s ) = integraldisplay 0 t e st f ( t ) d t , d 2 d s 2 F ( s ) = integraldisplay 0 t 2 e st f ( t ) d t , . . . d k d s k F ( s ) = ( 1) k integraldisplay 0 t k e st f ( t ) d t .
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8 A correct proof would require some justification of taking the derivatives inside the above imporper integrals. We will not go into those details here. We will however give an easier proof of the fact that F ( s ) is defined for s > α . The proof uses the direct comparison test for the convergence of improper integrals. That test implies that if g ( t ) and G ( t ) are piecewise continuous over every [0 , T ] such that | g ( t ) | ≤ G ( t ) for every t 0 then integraldisplay 0 G ( t ) d t converges = integraldisplay 0 g ( t ) d t converges . Let s > α and apply this test to g ( t ) = e st f ( t ). Pick σ so that α < σ < s . Because f ( t ) is of exponential order α as t → ∞ and σ > α there exist K σ and T σ such that (8.5) holds. Because g ( t ) = e st f ( t ) is bounded over [0 , T σ ] there exists B σ such that | g ( t ) | ≤ B σ over [0 , T σ ]. It thereby follows that | g ( t ) | = e st | f ( t ) | ≤ G ( t ) braceleftbigg B σ for 0 t < T σ K σ e ( σ s ) t for t T σ .
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  • Spring '10
  • LEVERMORE
  • Laplace

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