Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.7 Improper Integrals - Exercises - Page 442: 90


$\dfrac{1} {s-\alpha}$

Work Step by Step

The Laplace Transform of $f(x)$ can be found as: $L f(s)=\int_0^{\infty} e^{\alpha x}e^{-sx} \ dx\\=\int_0^{\infty} e^{-(s-\alpha) x } dx\\=\lim\limits_{R \to \infty}\dfrac{-1}{s-\alpha}[e^{-(s-\alpha) x }]_0^R\\=\lim\limits_{R \to \infty} \dfrac{-1} {s-\alpha}(e^{-(s-\alpha) R }-1)\\=\dfrac{-1} {s-\alpha}(0-1) \\=\dfrac{1} {s-\alpha}$
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