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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 8, Problem 8.9.104

102–106. Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by F(s) = ∫[0 to ∞] e^(-st) f(t) dt, where we assume s is a positive real number. For example, to find the Laplace transform of f(t) = e^(-t), the following improper integral is evaluated using integration by parts:
F(s) = ∫[0 to ∞] e^(-st) e^(-t) dt = ∫[0 to ∞] e^(-(s+1)t) dt = 1/(s+1).
Verify the following Laplace transforms, where a is a real number.
104. f(t) = t → F(s) = 1/s²

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1
Start with the definition of the Laplace transform for the function \(f(t) = t\): \[F(s) = \int_0^{\infty} e^{-st} t \, dt\]
Recognize that this integral involves the product of \(t\) and an exponential function, which suggests using integration by parts. Recall the integration by parts formula: \[\int u \, dv = uv - \int v \, du\]
Choose \(u = t\) so that \(du = dt\), and choose \(dv = e^{-st} dt\) so that \(v = \int e^{-st} dt = -\frac{1}{s} e^{-st}\). Substitute these into the integration by parts formula:
Apply the limits of integration from 0 to \(\infty\) to the term \(uv = t \cdot \left(-\frac{1}{s} e^{-st}\right)\) and evaluate the remaining integral \(\int_0^{\infty} \frac{1}{s} e^{-st} dt\).
Simplify the expression carefully, noting that the exponential term \(e^{-st}\) tends to zero as \(t \to \infty\) for \(s > 0\), and evaluate the definite integrals to express \(F(s)\) in terms of \(s\). This will lead to the result \(F(s) = \frac{1}{s^2}\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definition of the Laplace Transform

The Laplace transform converts a time-domain function f(t) into a complex frequency-domain function F(s) using the integral F(s) = ∫₀^∞ e^(-st) f(t) dt. This transformation simplifies solving differential equations by turning them into algebraic equations in terms of s.
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Improper Integrals and Convergence

The Laplace transform integral is improper because it extends to infinity. Understanding when the integral converges (typically for s > 0) is essential to ensure the transform exists. Techniques like evaluating limits and recognizing exponential decay help determine convergence.
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Integration Techniques: Integration by Parts

Integration by parts is a method used to evaluate integrals involving products of functions, such as t·e^(-st). It is based on the formula ∫u dv = uv - ∫v du and is crucial for finding Laplace transforms of functions like f(t) = t, where direct integration is not straightforward.
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