Question

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

Short answer

Question: Determine the Laplace transform of the function f(t) defined by the following integral: \[f(t) = \int_{0}^{t} (t-\tau)e^{\tau} d\tau\] Solution: The Laplace transform of the given function f(t) is given by \[F(s) = \frac{1}{(s-1)^3}\].

Step-by-step solution

Recall the definition of Laplace transform

To find the Laplace transform of the given function, we first recall the definition of the Laplace transform. Given a function f(t), its Laplace transform F(s) is defined as: \[F(s) = \mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) dt\]

Substitute the given function into the definition

Now, we'll substitute the given function f(t) into the Laplace transform definition to obtain the expression we need to evaluate: \[F(s) = \int_{0}^{\infty} e^{-st} \left(\int_{0}^{t}(t-\tau) e^{\tau} d\tau\right) dt\]

Exchange the order of integration

To make this integral easier to evaluate, we can exchange the order of integration using Fubini's theorem. The original order was to integrate with respect to τ and then with respect to t. We will now change it to integrate with respect to t first and then with respect to τ. Note that the limits of integration will also change accordingly: \[F(s) = \int_{0}^{\infty} \left(\int_{\tau}^{\infty} e^{-st}(t-\tau) e^{\tau} dt\right) d\tau\]

Combine the exponential terms

Now, we can simplify the expression inside the integral by combining the exponential terms: \[F(s) = \int_{0}^{\infty} \left(\int_{\tau}^{\infty} e^{-(s-1)t} (t - \tau) dt\right) d\tau\]

Integrate with respect to t

Next, we need to compute the inner integral with respect to t. We'll use integration by parts to solve this: Let u = t - \tau, dv = e^{-(s-1)t} dt Then du = dt, v = -\frac{1}{s-1} e^{-(s-1)t} Using integration by parts: \[\int u dv = uv - \int v du\] \[F(s)=-\int_{0}^{\infty} \left[u v\bigg|_{t=\tau}^{t=\infty} - \int_{\tau}^{\infty} v du\right] d\tau\] Plugging u and v and simplifying the limit expressions, we get: \[F(s)=\int_{0}^{\infty}\left[\frac{1}{s-1} e^{-(s-1)\tau}+\frac{1}{(s-1)^2} e^{-(s-1)t}\bigg|_{t=\tau}^{t=\infty}\right] d\tau\] Simplify the inner bracket expression: \[F(s)=\int_{0}^{\infty}\left[\frac{1}{s-1} e^{-(s-1)\tau}-\frac{1}{(s-1)^2} (1-e^{-(s-1)\tau})\right] d\tau\]

Find the Laplace transform

Finally, we compute the remaining integral to find the Laplace transform: \[F(s)=\left[\frac{1}{(s-1)^2} \tau e^{-(s-1)\tau} +\frac{1}{(s-1)^3}(1-e^{-(s-1)\tau}) \right]_{\tau=0}^{\tau=\infty}\] After evaluating the limits of the expression, we get: \[F(s) = \frac{1}{(s-1)^3}\] Thus, the Laplace transform of the given function is: \[F(s) = \frac{1}{(s-1)^3}\]

Key concepts

Integration by Parts

When you encounter an integral that is the product of two functions, integration by parts is a useful technique to simplify the problem. This method is particularly valuable when dealing with products of polynomial and exponential functions, which is often the case within the realm of differential equations. The basic formula for integration by parts stems from the product rule for differentiation and is given by: \br[\br\text{Let } u \text{ and } dv \text{ be differentiable functions, then} \ \br\text{Integration by parts formula:} \ \br\text{If } u = f(t), \text{ and } dv = g(t) dt, \text{ then } \ \br\text{Integrate } dv \text{ to obtain } v = \text{the antiderivative of } g(t), \ \br\text{Then } \ \br\text{Use the formula } \ \br\text{\(\int u dv = uv - \int v du\)}. \br\text{This process is key to solving many differential equations and is exemplified in the step-by-step solution when finding the Laplace transform of a product of functions.}

Fubini's Theorem

Fubini's theorem is a powerful tool in calculus, particularly when dealing with double integrals, as seen in the finding of Laplace transforms. This theorem allows us to switch the order of integration when dealing with integrals over two variables — a necessity in cases where the inner integral is challenging to compute directly. Essentially, Fubini's theorem states:\brIf we have a function f(x, y) that is continuous on a rectangle \( [a, b] \times [c, d] \), then the double integral of f over this rectangle can be computed as an iterated integral in either order.
Using Fubini's theorem can simplify the computation process by making it possible to first perform the easier integration and subsequently tackle the more complex one. In the exercise at hand, this theorem has been utilized to change the order of integration to a more manageable form, thereby facilitating the calculation of the Laplace transform.

Exponential Functions

Exponential functions are essential elements in various areas of mathematics, including differential equations and the calculus of transformations, as demonstrated by the Laplace transform. These functions are readily recognizable by the form \( e^{kt} \) where \( e \) is the base of the natural logarithm, \( t \) generally represents time, and \( k \) is a constant which affects the rate of growth or decay.
Properties of exponential functions, such as \(\br\text{\(e^{kt}e^{lt} = e^{(k+l)t}\)}\br\), are frequently applied when combining terms within the Laplace transform calculations. They can determine how a system behaves either in growth or decay scenarios, making them incredibly useful in the study of differential equations. In our example, the exponential term \( e^{\tau} \) and its properties facilitate the integration process.

Differential Equations

Differential equations are mathematical equations that describe relationships involving functions and their derivatives. They are used extensively in various scientific fields to model phenomena where the rate of change is a crucial component. Solutions to differential equations often involve integrating functions, which may require methods like integration by parts or the application of the Laplace transform.
In our exercise, solving the differential equation would involve finding a function of time \( t \) that corresponds to the integral expression given. However, by taking the Laplace transform, we convert the problem into an algebraic one, which is often easier to handle. The resulting function of \( s \), the Laplace variable, captures all the necessary information to understand and manipulate the system described by the original differential equation. The Laplace transform's power lies in its ability to simplify the solution of complex differential equations by transforming them into a solvable algebraic format.