PlanetMath: Laplace transform of derivative

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Laplace transform of derivative

(Theorem)

Theorem. If the real function $t \mapsto f(t)$ and its derivative are Laplace-transformable and $f$ is continuous for $t > 0$ , then

$\displaystyle \mathcal{L}\{f'(t)\} \;=\; s\,F(s)-\lim_{t\to0+}\!f(t).$

(1)

Proof. By the definition of Laplace transform and using integration by parts, the left hand side of (1) may be written $$\int_0^\infty\!e^{-st}f'(t)\,dt \;=\; \sijoitus{t=0}{\quad \infty}\!e^{-st}f(t)+s\!\int_0^\infty\!e^{-st}f(t)\,dt \;=\; \lim_{t\to\infty}e^{-st}f(t)-\lim_{t\to0}e^{-st}f(t)+s\,F(s).$$ The Laplace-transformability of $f$ implies that $e^{-st}f(t)$ tends to zero as $t$ increases boundlessly. Thus the last expression leads to the right hand side of (1).

"Laplace transform of derivative" is owned by pahio.

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See Also: substitution notation

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Attachments:: Laplace transform of integral (Derivation) by pahio

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Cross-references: right hand side, expression, implies, left hand side, integration by parts, Laplace transform, proof, continuous, derivative, real function, theorem
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This is version 2 of Laplace transform of derivative, born on 2008-09-21, modified 2008-09-23.
Object id is 11065, canonical name is LaplaceTransformOfDerivative.
Accessed 932 times total.

Classification:

AMS MSC:

44A10 (Integral transforms, operational calculus :: Laplace transform)

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