Problem 86 - Essential Calculus

Question

Laplace Transforms Let \(f(t)\) be a function defined for all positive values of \(t\). The Laplace Transform of \(f(t)\) is defined by \(F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t\) if the improper integral exists. Laplace Transforms are used to solve differential equations. Find the Laplace Transform of the function. $$ f(t)=\sin a t $$

Step-by-Step Solution

Answer

The Laplace Transform of the function \(f(t) = \sin(at)\) is \(F(s) = \frac{a}{a^2 + s^2}\)

Step 1: Write Down the Formula

The Laplace Transform for the function \(f(t)\) is defined as: \(F(s) = \int_{0}^{\infty} e^{-st} f(t) dt\). In our case, our function \(f(t)\) is a sine function, or specifically \(f(t) = \sin(at)\).

Step 2: Apply the Transform

We substitute our function \(f(t) = \sin(at)\) into our transform formula. This gives us: \(F(s) = \int_{0}^{\infty} e^{-st} \sin(at) dt\).

Step 3: Integrate by Parts

Here, we use the technique of integration by parts. We let \(u = e^{-st}\) and \(dv = \sin(at) dt\). Therefore, \(du = -s e^{-st} dt\) and \(v = -\frac{1}{a} \cos(at)\). Integration by parts is defined as \(\int u dv = uv - \int v du\). Substituting \(u, v, du\) and \(dv\) gives us \(-\frac{1}{a} e^{-st} \cos(at) + \frac{s}{a} \int_{0}^{\infty} e^{-st} \cos(at) dt\).

Step 4: Integrate by Parts Again

The remaining integral in the equation also requires integration by parts. We let \(u = e^{-st}\) and \(dv = \cos(at) dt\). Thus, \(du = -s e^{-st} dt\) and \(v = \frac{1}{a} \sin(at)\). Integration by parts gives us \(-\frac{1}{a^2} e^{-st} \sin(at) - \frac{s}{a^2} \int_{0}^{\infty} e^{-st} \sin(at) dt\).

Step 5: Solve the Equation

Finally, we use the boundary conditions for the integral from 0 to infinity to find the Laplace Transform of the function \(f(t) = sin(at)\). This gives us: \(F(s) = \frac{a}{a^2 + s^2}\)

Key Concepts

Integration by Parts

Integration by parts is a powerful mathematical technique used to integrate products of functions. It is based on the product rule for differentiation and can be expressed as the integral of the product of two functions u and dv: \[ \text{\(\int u dv = uv - \int v du\)} \. \]
Essentially, the rule allows for the simplification of an integral by transforming it into a form that is easier to solve. To apply this technique, you must decide which part of the function to set as u and which as dv. This decision can greatly affect the simplicity of the resulting integral.

In the case of Laplace Transforms involving trigonometric functions like sin(at), integration by parts is used twice, due to the cyclical nature of sine and cosine functions, eventually leading to an expression that includes the original integral on both sides of the equation, allowing us to solve for the Laplace Transform.

Improper Integrals

Improper integrals occur when the interval of integration is infinite or the integrand becomes infinity within the interval. The Laplace Transform features an improper integral because it integrates from 0 to infinity. To handle this, the integral is interpreted as the limit of a definite integral as one of its limits approaches infinity.

The process is represented as follows: \[ F(s) = \lim_{b \to \infty} \int_{0}^{b} e^{-st} f(t) dt \. \]
If this limit exists and is finite, the improper integral converges, meaning the Laplace Transform is well-defined. If the limit does not exist or is infinite, the integral diverges and the Laplace Transform does not exist for the function. Ensuring convergence is a key part in solving differential equations with Laplace Transforms.

Solving Differential Equations

Differential equations describe relationships involving rates of change and are omnipresent in the sciences and engineering. Solving differential equations can be quite complex, especially when they involve non-constant coefficients or are non-linear.

The Laplace Transform is a technique for transforming a differential equation in the time domain into an algebraic equation in the Laplace domain. This domain conversion simplifies the process of solving because one can work with algebraic terms instead of differential operators.

After solving the algebraic equation, one employs the inverse Laplace Transform to revert the solution back to the time domain. This technique is particularly useful for linear time-invariant systems and is a fundamental tool in control engineering and signal processing.

Continuous Function

In mathematics, a continuous function is a function that does not have any abrupt changes in value, known as discontinuities. For all points within the domain of the function, a small change in the input of the function results in a small change in the output.

Continuous functions are important in the context of Laplace Transforms because the transforms are typically applied to functions that are piecewise continuous or exhibit exponential growth. In the example problem, f(t) = sin(at) is continuous. This continuity means that any disturbances or inputs into the system that the function describes can be expected to result in smooth outputs, without sudden jumps or gaps, making it easier to predict system behavior.