Question

Find the Laplace transform of the given function. \(f(t)=\int_{0}^{t} e^{-(t-\tau)} \sin \tau d \tau\)

Short answer

Answer: The Laplace transform of the function \(f(t)\) is \(L\{f(t)\} = \frac{-1}{2s}\).

Step-by-step solution

Find the integral of the given function with respect to τ

To find the integral of the function, we will use integration by parts. Let \(u = e^{-(t-\tau)}\) and \(dv = \sin \tau d\tau\). Then, \(du = e^{-(t-\tau)}d\tau\) and \(v = -\cos \tau\). Applying integration by parts, we get: \(\int e^{-(t-\tau)} \sin \tau d\tau = -e^{-(t-\tau)}\cos \tau - \int -e^{-(t-\tau)}\cos \tau d\tau\) Now, we need to integrate the second term again using integration by parts. Let \(w = e^{-(t-\tau)}\) and \(dw = \cos \tau d\tau\). Then, \(dw = e^{-(t-\tau)}d\tau\) and \(w = \sin \tau\). Applying integration by parts again, we get: \(\int -e^{-(t-\tau)}\cos \tau d\tau = -e^{-(t-\tau)}\sin \tau - \int e^{-(t-\tau)}\sin \tau d\tau\) Now let's substitute this back into the first integral: \(\int e^{-(t-\tau)} \sin \tau d\tau = -e^{-(t-\tau)}\cos \tau + e^{-(t-\tau)}\sin \tau + \int e^{-(t-\tau)}\sin \tau d\tau\) We can now rearrange the equation to solve for the integral: \(\int e^{-(t-\tau)} \sin \tau d\tau = \frac{-e^{-(t-\tau)}\cos \tau + e^{-(t-\tau)}\sin \tau}{2}\)

Apply the Laplace transform definition to the integral

The Laplace transform of a function \(g(t)\) is defined as: \(L\{g(t)\} = \int_{0}^{\infty} e^{-st}g(t) dt\) Now, we need to find the Laplace transform of the function \(f(t)\): \(L\{f(t)\} = L\{\int_{0}^{t} e^{-(t-\tau)} \sin \tau d\tau\} = \int_{0}^{\infty} e^{-st}\left[\int_{0}^{t} e^{-(t-\tau)} \sin \tau d\tau \right] dt\) We substitute the result from step 1: \(L\{f(t)\} = \int_{0}^{\infty} e^{-st}\left[\frac{-e^{-(t-\tau)}\cos \tau + e^{-(t-\tau)}\sin \tau}{2}\right] dt\) Using the limits of integration, we get: \(L\{f(t)\} = \left[\frac{-e^{-(s-\tau)}\cos \tau + e^{-(s-\tau)}\sin \tau}{2s}\right]_{0}^{\infty}\) Now, we can evaluate the limits: \(L\{f(t)\} = \lim_{t\to \infty}\frac{-e^{-(s-\tau)}\cos \tau + e^{-(s-\tau)}\sin \tau}{2s} - \frac{-e^{0}\cos 0 + e^{0}\sin 0}{2s}\) \(L\{f(t)\} = \frac{-1}{2s} - 0 = \frac{-1}{2s}\) So, the Laplace transform of the given function is: \(L\{f(t)\} = \frac{-1}{2s}\)

Key concepts

Integration by Parts

Integration by parts is a powerful technique that comes in handy when we need to integrate the product of two functions. It is based on the product rule for differentiation and can be expressed by the formula \[\int u dv = uv - \int v du\]where \(u\) and \(dv\) are functions of one variable. Choosing the appropriate \(u\) and \(dv\) is crucial, as it can simplify the integral significantly. In the context of the exercise given, integration by parts was used twice to resolve the integral of an exponential function multiplied by a sine function. The ultimate goal was to obtain a form that could easily be used in calculating the Laplace transform. It's critical for students to practice the method to become proficient at identifying the best choices for \(u\) and \(dv\) in various integrals.

Exponential Functions

Exponential functions, represented as \(e^{x}\), are crucial in many fields of science and engineering, particularly because they describe growth and decay processes. When working with Laplace transforms, exponential functions appear frequently, often in the form of \(e^{-st}\) where \(s\) is a complex frequency parameter. It's important for students to understand the behavior of exponential functions, such as how quickly they grow or decay, and their pivotal role in transforming time-domain functions to the s-domain, where differential equations become algebraic equations.

Solving Differential Equations

Solving differential equations is an essential skill in mathematics, physics, engineering, and other scientific disciplines. These equations describe how a particular quantity evolves over time or space. The Laplace transform is a widely used method for solving linear differential equations with constant coefficients because it converts the problem into an easier one to manage, typically an algebraic equation. After applying the Laplace transform, one can solve for the transformed function and then take the inverse transform to find the solution in the original domain. Understanding the relationship between time-domain differential equations and their Laplace transform is fundamental to mastering this topic.

Limits of Integration

The limits of integration define the interval over which an integral is evaluated. When dealing with definite integrals, as in the provided exercise, the limits indicate the start and end points of the interval. In the context of finding the Laplace transform, the limits for the outer integral are from 0 to \(\textbackslash infinity\), consistent with the definition of the Laplace transform. However, understanding the limits for the inner integral is also important. When we deal with functions involving time, such as in our example where we have \(e^{-(t-\tau)}\), the limits can affect the function's behavior within the integral. In the exercise, the inner integral's limits from 0 to \(t\) were critical to determine the function's behavior over time before applying the Laplace transform. Being able to interpret and apply the correct limits of integration is a skill that students should focus on developing in order to solve integrals accurately.