Question

Suppose that $$ g(t)=\int_{0}^{t} f(\tau) d \tau $$ If \(G(s)\) and \(F(s)\) are the Iaplace transforms of \(g(t)\) and \(f(t),\) respectively, show that $$ G(s)=F(s) / s \text { . } $$

Short answer

Question: Find the Laplace transform of a function g(t), where g(t) is defined as the integral of f(τ) with respect to τ, and f(t) has the Laplace transform F(s). Prove that G(s) = F(s)/s. Answer: We have proven that the Laplace transform of the function g(t), where g(t) is the integral of f(τ) with respect to τ, is G(s) = F(s)/s, where F(s) is the Laplace transform of the function f(t).

Step-by-step solution

Find the Laplace transform of g(t)

Recall the definition of the Laplace transform for a function g(t): $$ G(s) = \mathcal{L}\{g(t)\} = \int_{0}^{\infty} e^{-st}g(t)dt $$ Now substitute the definition of g(t) into the expression for the Laplace transform: $$ G(s) = \int_{0}^{\infty} e^{-st}\Big(\int_{0}^{t} f(\tau) d\tau \Big)dt $$

Use the properties of the Laplace transform

We will use the property $$ \mathcal{L}\{f'(t)\} = sF(s) - f(0) $$ Now, differentiate g(t) with respect to t: $$ g'(t) = \frac{d}{dt}\int_{0}^{t} f(\tau) d\tau = f(t) $$ Next, find the Laplace transform of g'(t) and f(0): $$ \mathcal{L}\{g'(t)\} = \mathcal{L}\{f(t)\} = F(s) $$ and $$ f(0) = 0 $$ because the integral equals to zero when t=0. Now, apply the property: $$ F(s) = sG(s) - 0 \Rightarrow F(s) = sG(s) $$

Prove the given identity

Now we have $$ F(s) = sG(s) $$ Finally, we can show that $$ G(s) = \frac{F(s)}{s} $$ by dividing both sides of the equation by s: $$ G(s) = \frac{F(s)}{s} $$ And we have successfully proven the identity.

Key concepts

Integral Transform

Integral transforms are powerful mathematical tools used to analyze functions in a different domain. One such transform is the Laplace transform. It helps to change functions from the time domain into the frequency domain. This transformation simplifies many complex mathematical problems. For functions like those in the original exercise, applying the Laplace transform makes solving differential equations easier.
The essence of an integral transform is built on integrating the product of a given function and a kernel function. For the Laplace transform, the kernel is \( e^{-st} \). This integrates the original function over time.
Integral transforms, like the Laplace transform, are used extensively in engineering, physics, and applied mathematics. They simplify solving complex equations by transforming them into algebraic equations.

Differential Equations

Differential equations involve functions and their derivatives. Such equations describe relationships between a function and its rate of change.
This is essential in modeling a wide range of real-world phenomena. By using solutions to these equations, we can predict how systems behave.
Many classical differential equations, especially linear ones, benefit from the Laplace transform.

• It converts a differential equation into an algebraic equation, making it more straightforward to solve.
• The transformed equation is usually easier to manipulate and solve for.

This approach leverages the Laplace transform because it deals effectively with initial conditions in differential equations, as demonstrated in the original problem example.

Integration by Parts

Integration by parts is a method of integrals used to compute integrals of products of functions. It stems from the product rule for differentiation. Here's a quick look at how it works:
The formula is:
\[ \int u \, dv = uv - \int v \, du \]
This technique can simplify complex integration tasks, breaking them down into simpler parts.

• It's especially beneficial where direct integration isn't straightforward.
• It helps solve integrals by using derivatives and anti-derivatives.

Although not directly used in the Laplace transform solution presented, understanding integration by parts can help in comprehending the integration process and its relations.

Initial Value Problems

Initial value problems (IVPs) are a type of differential equation accompanied by specific initial conditions. This means they not only look to solve the equation, but also anywhere particular solution that meets given starting points.
When addressed using the Laplace transform, IVPs benefit significantly, especially because:

• They convert derivatives into algebraic terms, which makes solving easier.
• Initial conditions are directly incorporated into the transformed equations.

In the exercise and solution example, encountering \( f(0) = 0 \) directly handles the initial condition, which plays a crucial role in obtaining the final result. IVPs are omnipresent in mathematical modeling across physics, engineering, and other sciences.