6.2. Theorem: Laplace Transform of an Integral

Let f(t) be a piecewise continuous function for t≥0 and of exponential order. Define another function as:

g(t)=∫f(τ)dτ 0

Here, we begin by defining the function f(t), which represents a piecewise continuous function for non-negative values of t and is of exponential order. This essentially means that the function doesn’t grow infinitely fast. Then, we introduce a new function g(t), which is defined as the integral of f(τ) from 0 to t. This new function g(t) is essential as we aim to find its Laplace Transform.

Think of f(t) as the speed of a car over time. Over a certain period, the speed can change (piecewise continuous), but it doesn't accelerate to an infinite speed. The function g(t) represents the total distance covered by the car from the start (t=0) to time t. Thus, calculating g(t) helps us understand the accumulation of distance over time.

✅ Theorem: Laplace Transform of an Integral If L{f(t)}=F(s), then {t } F(s) L ∫f(τ)dτ = s 0

This theorem states that if we know the Laplace Transform of the function f(t) is F(s), then the Laplace Transform of the integral of f(τ) from 0 to t can be computed by dividing F(s) by s. This relationship is significant because it connects the operations of integration and transformation, simplifying the process of solving differential equations involving integrals.

Imagine you're collecting water in a bucket over time, where f(t) represents the flow rate of water into the bucket. The total amount of water in the bucket (g(t)) is the accumulation of all the water collected. The theorem tells us that if we know how to analyze the water flow (f(t) → F(s)), we can easily figure out how much water is in the bucket at any point in time by just adjusting our calculations (F(s)/s).

Let’s denote: g(t)=∫f (τ)dτ 0 Now, taking Laplace Transform of g(t), ∞ t ( ) L{g(t)}=∫e−st ∫f(τ)dτ dt 0 0 Interchanging the order of integration (using Fubini's Theorem), ∞ ∞ ( ) ¿∫f(τ) ∫e−stdt dτ 0 τ

In this chunk, we begin the proof of the theorem by considering the expression of g(t) again. To find its Laplace Transform L{g(t)}, we express it as an integral. When we write L{g(t)}, we replace g(t) with its integral definition and utilize a double integral. Through Fubini's Theorem, we swap the order of integration. This is important as it allows us to address the inner integral separately, making the math tractable.

Consider organizing books on a shelf (g(t)). To find out how many books you've arranged at any time, you count the books repeatedly (the integral). By rearranging your counting method (changing the order of integration), you simplify the process and can tally faster, making calculations easier.

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

e−sτ s s τ τ

In this step, we specifically focus on evaluating the inner integral (the integral of e^(-st) dt). By integrating e^(-st), we arrive at a formula involving exponential decay. This step is crucial because it bridges our mathematical manipulation towards the final form of L{g(t)}, leading to establishing a relationship that includes s in the denominator.

Imagine you're measuring how quickly the scent of a perfume spreads (e^(-st)). As time passes (t increases), the scent becomes weaker. By mathematically determining how much scent is left (evaluating the integral), we can predict the scent's intensity over time, just as we can understand how it affects our calculations.

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

Here we conclude the proof by substituting the evaluated inner integral back into our expression for L{g(t)}. We find that it combines neatly into a single formula involving the Laplace Transform of f(τ). The proof is complete as we derive that L{g(t)} equals the integral of f(τ) multiplied by e^(-sτ), demonstrating the theorem's validity.

Imagine you've collected all the data on a project (g(t)), and you've now translated that data into a usable format (the integral). The completed project (the proof) shows how efficiently you were able to transform raw data into meaningful insights. This final form (L{g(t)}) perfectly represents the value of your gathered information, just as accurately summarizing the real data does.

This formula shows that integrating a function in the time domain corresponds to dividing its Laplace transform by s.

The key takeaway from the theorem is that there's a direct connection between integration in the time domain and transformation in the Laplace domain. Specifically, integrating a function f(t) results in its Laplace Transform F(s) being divided by s. This provides a powerful tool for simplifying problems that involve integration, especially in control and signal processing applications.

Just as you simplify tasks by breaking them down into manageable steps (integrating time), understanding that those steps correspond to a single simplified outcome in the end (transformed function) fosters efficiency and clarity in managing complex projects.

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