6.1. 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 We are interested in finding L{g(t)}. That is,

{t } L ∫f(τ)dτ 0

In this chunk, we define a function g(t) that is based on another function f(t). Here, f(t) is a piecewise continuous function that is defined for all non-negative values of t and has an exponential order, which implies that it does not grow too fast as t increases. The function g(t) represents the integral of f from 0 to t. We seek to find the Laplace transform of this function, denoted as L{g(t)}. The notation indicates that we will take the Laplace transform of the integral of f rather than f itself.

Imagine g(t) as a running total of sales over time. If f(t) is the rate of sales at any given moment, g(t) shows the total sales accumulated from the start up to time t. This relationship will help us analyze how cumulative values are affected in various systems.

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 a crucial relationship between Laplace transforms and integrals. If we know the Laplace transform of f(t), denoted as F(s), then the Laplace transform of the integral of f from 0 to t results in F(s) divided by the variable s. This division by s indicates that integrating a function in the time domain translates to multiplying its Laplace transform by a scaling factor in the frequency domain.

Think of this theorem as determining how much more you can earn (integration) when you know your hourly wage (Laplace transform). If you know how much you earn per hour, dividing by the rate of return (s) helps us find out how much you would earn over a given period.

Let’s denote:

t 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 Evaluating the inner integral:

∞ ∞ [−1 ] 1 ∫e−stdt= e−st

e−sτ s s τ τ

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 provides the proof for the theorem we discussed. We start with the definition of g(t) and take its Laplace transform. The proof involves using Fubini's Theorem to switch the order of integration. This allows us to evaluate the inner integral, which gives us an expression in terms of the variable τ. After computing the integrals, we find that L{g(t)} is equal to the integral over f(τ) times the exponential term, divided by s. The result confirms the theorem we proposed earlier.

Consider this proof like calculating the total amount of water accumulated in a tank over time. By re-arranging our approach (changing the order of integration), we can simplify our calculations and manage to find an efficient method to compute the total water collected at any time, demonstrating that different approaches can lead to the same result.

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

This chunk summarizes the key result obtained from the theorem and proof. It reinforces the concept that when we integrate a function in the time domain, it is represented in the Laplace domain by dividing by s. This principle is foundational for many applications of the Laplace Transform, especially in engineering and physics where we deal with dynamic systems changes.

Imagine you are tracking distance traveled over time while driving. If the speed function is known (the function before integration), integrating that function will provide the total distance. In the Laplace domain, dividing that knowledge by your average speed allows you to predict and analyze distances over different time periods.

This result is extremely useful in:

• Solving integro-differential equations
• Analyzing systems with accumulation or memory, such as charge in capacitors
• Evaluating convolution-type integrals
• Inverse Laplace Transform simplifications

This chunk elaborates on the numerous applications of the theorem established earlier. The ability to transform integrals enhances our capacity to work with systems where memory effects are prevalent, such as in electrical circuits (e.g., capacitors storing charge). Moreover, this ability is beneficial for solving integro-differential equations, a common occurrence in engineering dynamics.

Think about managing your bank account; knowing how much interest accumulates (akin to the accumulation of charge in capacitors) allows you to understand your finances over time and plan for future expenses. This theorem helps engineers in calculating such accumulations accurately in systems.

Example 1: Find the Laplace Transform of

t ∫sin(aτ)dτ 0 Solution: Let f(t)=sin(at), then

a F(s)=L{sin(at)}=

s2 +a2 Using the theorem,

{t } F(s) a L ∫sin(aτ)dτ = =

s s(s2 +a2 ) 0

Example 2: {t } Find L ∫e2τdτ 0 1 Solution: Let f(t)=e2t⇒F(s)= ,s>2 s−2 So, {t } L ∫e2τdτ =

s(s−2) 0

In this chunk, we present example problems to illustrate the application of the theorem in finding Laplace transforms of integrals. In the first example, we determine the Laplace transform of the integral of sin(aτ). By applying the theorem, we find its transform quite straightforwardly. The second example demonstrates a similar process with the exponential function e^(2t), further solidifying the theorem's practical importance.

Consider a baker who needs to manage various baking times for different pies. If each pie recipes gives a complex calculation for both time and temperature settings (like the Laplace transforms), knowing how to integrate their baking times together helps streamline the baking process—a common application of these transforms in practical scenarios!

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