6.3. Proof of the Theorem

Let’s denote: t g(t)=∫f (τ)dτ 0

In this chunk, we introduce the function g(t), defined as the integral of another function f(t) from 0 to t. Here, τ is a dummy variable for integration. This means g(t) accumulates the values of f(τ) over the interval from 0 to t, essentially summing up the outputs of f(τ) as τ progresses from 0 to t.

Think of g(t) as tracking your savings over time. If f(τ) represents the amount you save each day, then g(t) shows the total amount saved by day t. Just like accumulating savings, g(t) aggregates the contributions from f(τ).

Now, taking Laplace Transform of g(t), ∞ t ( ) L{g(t)}=∫e−st ∫f(τ)dτ dt 0 0

In this step, we are applying the Laplace Transform to the function g(t). This involves calculating the double integral: first integrating f(τ) from 0 to t, then applying the e^{-st} factor, which is crucial for transforming the function into the Laplace domain. The outer integral sums the contributions from g(t) over time, while the inner integral sums f(τ) from 0 to t.

Imagine you're calculating your total savings with respect to days and considering the effect of inflation (represented by the e^{-st} factor). The inner integral is like checking how much you've saved during each specific period, while the outer integral considers all those periods together to find a total adjusted for currency value over time.

Interchanging the order of integration (using Fubini's Theorem), ∞ ∞ ( ) ¿∫f(τ) ∫e−stdt dτ 0 τ

Here we apply Fubini's Theorem, which allows us to change the order of integration. By doing this, we first integrate with respect to t, then with respect to τ. This mathematical tool is useful when dealing with double integrals and simplifies our calculations significantly, allowing us to evaluate the inner integral independently.

Think of rearranging a group project where team members can work on their tasks in any order. By switching the tasks (integrations) around, each member can focus on their part more effectively, allowing the overall project to be streamlined. Similarly, changing the order of integrals often makes complex problems easier to manage.

Evaluating the inner integral: ∞ ∞ [−1 ] 1 ∫e−stdt= e−st

e−sτ s s τ τ

In this chunk, we compute the inner integral, which involves integrating e^{-st} with respect to t. This gives us an expression that simplifies further calculations. The result is [−1/s] e^{-st}, evaluated from 0 to ∞, which leads to e^{-sτ}/s. This step is crucial as it translates the time-based integral into a simpler form.

Imagine calculating how much money you'd save over time if you received compound interest. By integrating your savings, we can find how much that account grows over various periods. We evaluate this growth (inner integral) at the limits of time, similar to how we assess our savings at key points.

So, ∞ 1 1 ∞ F(s) L{g(t)}=∫f(τ)⋅ e−sτdτ= ∫f(τ)e−sτdτ= =s s s 0 0 Hence proved.

This chunk wraps up the proof by integrating the simplified expression from the previous step. The final result shows that the Laplace Transform of g(t) yields a new expression that is related to the Laplace Transform of f(t). Specifically, we find that the Laplace Transform of the integral of f(τ) corresponds to the Laplace Transform of f(τ) divided by s. This conclusion is key to understanding the relationship between integration in the time domain and manipulation in the Laplace domain.

When putting together pieces of a puzzle, finding how they fit (proving a theorem) is just like integrating functions. By bringing all the parts (functions) together and seeing how they relate under certain transformations (like Laplace), we achieve a clearer picture or understanding. The proof demonstrates how integral calculations connect with Laplace Transforms, paving the way for more complex problems.

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