Question

Calculate the Laplace transform of $$ f(t)=\int_{0}^{t}\left(u^{2}-u+e^{-u}\right) d u . $$

Short answer

\( \mathcal{L}\{f(t)\} = \frac{2}{s^4} - \frac{1}{s^3} + \frac{1}{s(s+1)} \).

Step-by-step solution

Define the Exercise

We need to calculate the Laplace Transform of the given function \( f(t) = \int_{0}^{t}(u^2 - u + e^{-u}) \, du \). This involves understanding the definition of Laplace Transform and Convolution.

Set Up the Laplace Transform

The Laplace Transform of a function \( f(t) \) is given by \( \mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt \). Here, we start with the integral expression for \( f(t) \), which is a convolution type of problem.

Interpret as a Convolution

Notice that the expression for \( f(t) \) is in a form that suggests a convolution. Recall that \( \mathcal{L}\{f \ast g\} = \mathcal{L}\{f\}\cdot\mathcal{L}\{g\} \). In this case, consider identifying what \( g(t) \) is where \( f(t) = \int_0^t g(u) \, du \).

Identify the Convolution Pair

Recognize that the convolution theorem applies because \( f(t) \) is the convolution of the Heaviside step function \( u(t) \) with \( g(t) = u^2 - u + e^{-u} \). Here, \( g(t) \) is the function inside the integral.

Compute the Laplace Transform of \( g(t) \)

Find \( \mathcal{L}\{g(t)\} = \mathcal{L}\{t^2\} - \mathcal{L}\{t\} + \mathcal{L}\{e^{-t}\} \). This breaks down to:- \( \mathcal{L}\{t^2\} = \frac{2}{s^3} \)- \( \mathcal{L}\{t\} = \frac{1}{s^2} \)- \( \mathcal{L}\{e^{-t}\} = \frac{1}{s+1} \)

Apply the Laplace Transform of Convolution

By the Convolution Theorem, \( \mathcal{L}\{f(t)\} = \frac{1}{s} \cdot \left( \frac{2}{s^3} - \frac{1}{s^2} + \frac{1}{s+1} \right) \). Simplifying this gives the final Laplace Transform of \( f(t) \).

Simplify the Expression

Simplify the expression to get \( \mathcal{L}\{f(t)\} = \frac{2}{s^4} - \frac{1}{s^3} + \frac{1}{s(s+1)} \).

Key concepts

Convolution Theorem

The Convolution Theorem is a powerful tool in the realm of Laplace Transforms. It describes how the Laplace Transform of a convolution is the product of individual Laplace Transforms of the functions involved. In simpler terms, if two functions are convolved in the time domain, their transforms multiply in the frequency domain.

Convolution of two functions, say \( f(t) \) and \( g(t) \), is expressed as:

• \( f \ast g(t) = \int_0^t f(u)g(t-u)\, du \)

When faced with such integrals, the Convolution Theorem simplifies the process greatly:

• \( \mathcal{L}\{f \ast g\} = \mathcal{L}\{f\}\cdot\mathcal{L}\{g\} \)

Applying this to our expression allows us to calculate the Laplace Transform more efficiently, transferring a complex integration problem into simple multiplication.

Integral Transform

Integral transforms like the Laplace Transform are techniques used to switch from one domain to another, typically from the time domain to the frequency domain. This can greatly simplify solving differential equations, convolution integrals, and more.

The Laplace Transform, specifically, takes a given function \( f(t) \) and transforms it into \( F(s) \) using the formula:

• \( \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt \)

This new function, \( F(s) \), is usually easier to handle, especially for linear differential equations. The transformation effectively allows us to convert operations like differentiation into algebraic forms, streamlining solutions for complex equations.

Once solutions in the frequency domain are found, we often need to revert back to the time domain, usually by employing the Inverse Laplace Transform.

Heaviside Step Function

The Heaviside step function, denoted as \( u(t) \), is used frequently in signal processing and control theory. It is a function that jumps from zero to one at a specified value, commonly zero, making it useful for representing on/off signals.

Mathematically, it's defined as:

• \( u(t) = 0 \text{ for } t < 0 \)
• \( u(t) = 1 \text{ for } t \geq 0 \)

In the context of convolution and integral transforms, the Heaviside step function is often used to "turn on" a function. For instance, in our exercise we see \( f(t) = \int_0^t g(u) \, du \) which can be thought of as the convolution of \( g(t) \) with \( u(t) \), allowing for the application of the Convolution Theorem.

The Laplace Transform of the Heaviside step function is crucial, as it simplifies how we handle piecewise functions and initial value problems in more complex differential equations.