Home page for accesible maths 6 Chapter 6 contents 6.41 Existence of Laplace transforms 6.43 Laplace transform of

\cos ax

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6.42 Laplace transform of the derivative

Proposition.

Suppose that $f$ is as in frame 6.41 and has continuous derivative $f^{\prime}$ . Then the Laplace transform converts differentiaton to multiplication, so

s>a

Proof. We integrate by parts:

\int_{0}^{R}e^{-sx}f^{\prime}(x)\,dx=\,{\left[e^{-sx}f(x)\right]_{0}^{R}+s\int% _{0}^{R}e^{-sx}f(x)\,dx}

{=f(R)e^{-sR}-f(0)+s\int_{0}^{R}e^{-sx}f(x)\,dx.}

Now by assumption, $|f(R)e^{-sR}|\leq{Me^{aR}e^{-sR}\rightarrow 0}$ as $R\rightarrow\infty$ .