Problem 56 - Calculus Early Transcendentals: Pearson New International Edition

Question

The Laplace transform, named after the French mathematician Pierre-Simon de Laplace (1749-1827), of a function \(f(x)\) is given by \(L\\{f(t)\\}(s)=\int_{0}^{\infty} f(t) e^{-s t} d t\). Laplace transforms are useful for solving differential equations. (a) Show that the Laplace transform of \(t^{\alpha}\) is given by \(\Gamma(\alpha+1) / s^{\alpha+1}\) and is defined for \(s>0\). (b) Show that the Laplace transform of \(e^{\alpha t}\) is given by \(1 /(s-\alpha)\) and is defined for \(s>\alpha\). (c) Show that the Laplace transform of \(\sin (\alpha t)\) is given by \(\alpha /\left(s^{2}+\alpha^{2}\right)\) and is defined for \(s>0\).

Step-by-Step Solution

Answer

(a) \( \Gamma(\alpha+1) / s^{\alpha+1} \), (b) \( 1 / (s-\alpha) \), (c) \( \alpha / (s^2+\alpha^2) \).

Step 1: Apply the Laplace Transform Formula to t^α

To find the Laplace transform of the function \( f(t) = t^{\alpha} \), apply the definition of the Laplace transform:\[L\{t^{\alpha}\}(s) = \int_{0}^{\infty} t^{\alpha} e^{-st} dt.\]This is an improper integral that requires the calculation of Gamma functions.

Step 2: Solve the Integral for t^α

Use integration by parts or the known result \[\int_0^{\infty} x^a e^{-bx} dx = \frac{\Gamma(a+1)}{b^{a+1}}\]where \(a = \alpha\) and \(b = s\). This simplifies our integral to:\[L\{t^{\alpha}\}(s) = \frac{\Gamma(\alpha + 1)}{s^{\alpha + 1}}, \quad s > 0.\]

Step 3: Apply the Laplace Transform Formula to e^{αt}

For the function \( f(t) = e^{\alpha t} \), apply the formula:\[L\{e^{\alpha t}\}(s) = \int_{0}^{\infty} e^{\alpha t} e^{-st} dt.\]Combine the exponents in the integrand to simplify.

Step 4: Solve the Integral for e^{αt}

The integral becomes:\[\int_{0}^{\infty} e^{(\alpha - s)t} dt.\]Since the integral of \(e^{kt}\) is \(1/k\), provided that \(k < 0\), we have\[L\{e^{\alpha t}\}(s) = \frac{1}{s - \alpha}, \quad s > \alpha.\]

Step 5: Apply the Laplace Transform Formula to \\sin(αt)

Apply the Laplace transform to \( f(t) = \sin(\alpha t) \) with the formula:\[L\{\sin(\alpha t)\}(s) = \int_{0}^{\infty} \sin(\alpha t) e^{-st} dt.\]Use integration by parts to solve this integral.

Step 6: Solve the Integral for \\sin(αt)

Using integration by parts or known results:\[\int_{0}^{\infty} e^{-st} \sin(\alpha t) dt = \frac{\alpha}{s^2 + \alpha^2}, \quad s > 0.\]This gives us the Laplace transform: \[L\{\sin(\alpha t)\}(s) = \frac{\alpha}{s^2 + \alpha^2}, \quad s > 0.\]

Key Concepts

Gamma Function

The Gamma Function is a continuation of the factorial function to real and complex numbers, except for negative integers. For a positive integer, it is given by \( \Gamma(n) = (n-1)! \). However, the Gamma function is defined more generally for any complex number with a positive real part by the integral:\[\Gamma(x) = \int_{0}^{\infty} t^{x-1} e^{-t} dt.\]This integral formulation extends the utility of the factorial function beyond just positive integers. It plays an essential role in various fields, notably in Laplace transformations. This is because it allows us to simplify integrals involving powers of \(t\), as seen when determining Laplace transforms involving terms like \(t^\alpha\).

• The Gamma function helps evaluate improper integrals, especially when used alongside the Laplace transform helping simplify complex expressions.
• It is crucial for finding the Laplace transform of functions like \(t^\alpha \), offering a straightforward method to relate algebraic expressions to Laplace domain representations.

Whenever you encounter an integral where a function is expressed as a power of \(t\), and multiplied by an exponential decay term, the Gamma function often provides a direct approach to solving the problem.

Differential Equations

Differential equations describe relationships involving the rates at which quantities change. They are fundamental in modeling real-world phenomena like population growth, heat conduction, and electrical circuits. Laplace transforms are vital for solving differential equations because they transform differential equations into algebraic ones, which are often easier to solve.In practice, the Laplace transform converts functions of time (which are common in differential equations) into functions of a complex variable. This transformation uses the integral:\[L\{f(t)\}(s) = \int_{0}^{\infty} f(t) e^{-st} dt.\]Once in the Laplace domain:- Algebraic techniques apply to solve the transformed equation.- Inverse Laplace transforms bring the solution back to the time domain.This process particularly benefits linear differential equations with constant coefficients, turning a seemingly intractable problem into a manageable algebraic one. The Laplace transform method is most efficient when initial conditions are known, allowing initial value problems to be solved systematically.When given an initial condition boundary, turning it into a Laplace problem simplifies solving complex systems by reducing them to simpler algebraic forms.

Integration by Parts

Integration by Parts is a technique used to solve integrals where the standard methods of integration are not applicable. This method is based on the integration rule derived from the product rule of differentiation. The formula is:\[\int u \, dv = uv - \int v \, du\]Where:

• \(u\) is a function of \(t\),
• \(dv\) is the differential of a function of \(t\),
• \(du\) and \(v\) are the derivatives and antiderivatives, respectively.

This technique helps in transforming complex integrals into simpler ones that can be solved more directly. It is particularly useful in solving terms that arise in Laplace transforms such as the \( \sin(\alpha t) \) integral. Often, using Integration by Parts leads to recursive solutions where repeated application simplifies the integral step by step.By applying integration by parts to solve the integral \( \int_0^{\infty} \sin(\alpha t) e^{-st} dt \),we break it into simpler components. Integration by Parts may often be paired with known results or iterative methods to achieve a solvable form, essential in evaluating complex Laplace transform problems.