Proof of the Theorem - 6.3 | 6. Laplace Transform of an Integral | Mathematics - iii (Differential Calculus) - Vol 1
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Interactive Audio Lesson

Listen to a student-teacher conversation explaining the topic in a relatable way.

Introduction to Laplace Transform of Integrals

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0:00
Teacher
Teacher

Today, we’re discussing the proof of a theorem that simplifies the analysis of differential equations through the Laplace Transform. Can someone remind me what the Laplace Transform does?

Student 1
Student 1

It transforms a function from the time domain into the s-domain.

Teacher
Teacher

Exactly! Now, when we integrate a function, we can actually simplify this process through a theorem we will prove today. What do you think this could help with?

Student 2
Student 2

It could help solve differential equations more easily!

Teacher
Teacher

Great thought! Solving integro-differential equations will be one of the key applications.

Understanding the Theorem

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0:00
Teacher
Teacher

The theorem states that if L{f(t)} = F(s), then the Laplace Transform of the integral from 0 to t of f(t) can be expressed as F(s)/s. Why is dividing by s important here?

Student 3
Student 3

It shows how integrating affects the Laplace Transform.

Teacher
Teacher

Exactly! It provides a direct correlation between operations in the time as compared to the s-domain.

Student 4
Student 4

Is this applicable in real-world problems, like electrical circuits?

Teacher
Teacher

Absolutely! This theorem greatly influences how we analyze systems with memory, like capacitors.

Proof of the Theorem

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0:00
Teacher
Teacher

Let's take a closer look at the proof. We denote g(t) as the integral of f from 0 to t. Can anyone identify the integral we're referring to?

Student 1
Student 1

It's g(t) = ∫f(Ο„)dΟ„ from 0 to t!

Teacher
Teacher

Correct! Now, by applying the Laplace Transform, we work through changing the order of integration. Who remembers why we can do that?

Student 2
Student 2

That's Fubini's Theorem, right? It allows us to switch the order under certain conditions.

Teacher
Teacher

Very well! When we evaluate the integral through this method, we derive that the Laplace Transform can be simplified significantly, showcasing the utility of this theorem.

Student 3
Student 3

So this really does show that integrating makes it easier to manipulate functions!

Teacher
Teacher

Absolutely! Recapping our discussion, we’ve reinforced that the transformation relates integration in time to algebraic manipulation in the s-domain.

Application Insights

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0:00
Teacher
Teacher

Before we conclude, let's discuss how this theorem is applied in engineering. What practical issues can it help?

Student 4
Student 4

It can simplify the analysis of circuits or systems with memory like capacitors!

Teacher
Teacher

Exactly! It allows us to analyze charge accumulation effectively. This makes integral transforms a fascinating area to study.

Student 1
Student 1

Can we use this for signals processing too?

Teacher
Teacher

Yes! Integration plays a vital role there. Understanding these principles thoroughly will enhance your analytical skills.

Introduction & Overview

Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.

Quick Overview

This section outlines the proof of the theorem related to the Laplace Transform of integrals, showcasing its significance in simplifying expressions in engineering mathematics.

Standard

The proof of the theorem shows that the Laplace Transform of an integral can be computed by dividing the Laplace Transform of the original function by the variable s. This property aids in solving complex differential and integral equations, making it crucial for applications in engineering and mathematics.

Detailed

Proof of the Theorem

In this section, we explore the proof of a significant theorem in engineering mathematics, specifically concerning the Laplace Transform of integrals. The theorem states that if L{f(t)} = F(s), then the Laplace Transform of the integral of f(t) from 0 to t is given by:

$$L{\int_0^t f(\tau) d\tau} = \frac{F(s)}{s}$$

To understand this, we define the function g(t) as the integral of f(t) from 0 to t. We then apply the Laplace Transform to g(t) and utilize Fubini's Theorem to interchange the order of integration. By evaluating the inner integral, we arrive at the formula, confirming the relationship between integration in the time domain and manipulation in the Laplace domain. This theorem is foundational for resolving integro-differential equations and analyzing systems with memory, making it particularly relevant in engineering disciplines.

Audio Book

Dive deep into the subject with an immersive audiobook experience.

Defining the Function g(t)

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Let’s denote:
t
g(t)=∫f (Ο„)dΟ„
0

Detailed Explanation

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

Examples & Analogies

Think of g(t) as tracking your savings over time. If f(Ο„) 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(Ο„).

Taking the Laplace Transform of g(t)

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Now, taking Laplace Transform of g(t),
∞ t
( )
L{g(t)}=∫eβˆ’st ∫f(Ο„)dΟ„ dt
0 0

Detailed Explanation

In this step, we are applying the Laplace Transform to the function g(t). This involves calculating the double integral: first integrating f(Ο„) 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(Ο„) from 0 to t.

Examples & Analogies

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

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Interchanging the order of integration (using Fubini's Theorem),
∞ ∞
( )
¿∫f(Ο„) ∫eβˆ’stdt dΟ„
0 Ο„

Detailed Explanation

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 Ο„. This mathematical tool is useful when dealing with double integrals and simplifies our calculations significantly, allowing us to evaluate the inner integral independently.

Examples & Analogies

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

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Evaluating the inner integral:
∞ ∞
[βˆ’1 ] 1
∫eβˆ’stdt= eβˆ’st
=
eβˆ’sΟ„
s s
Ο„ Ο„

Detailed Explanation

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 [βˆ’1/s] e^{-st}, evaluated from 0 to ∞, which leads to e^{-sΟ„}/s. This step is crucial as it translates the time-based integral into a simpler form.

Examples & Analogies

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.

Final Steps and Conclusion of the Proof

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So,
∞ 1 1 ∞ F(s)
L{g(t)}=∫f(Ο„)β‹… eβˆ’sΟ„dΟ„= ∫f(Ο„)eβˆ’sΟ„dΟ„=
=s s s
0 0
Hence proved.

Detailed Explanation

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(Ο„) corresponds to the Laplace Transform of f(Ο„) divided by s. This conclusion is key to understanding the relationship between integration in the time domain and manipulation in the Laplace domain.

Examples & Analogies

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.

Definitions & Key Concepts

Learn essential terms and foundational ideas that form the basis of the topic.

Key Concepts

  • Laplace Transform of an Integral: The theorem L{∫f(Ο„)dΟ„} = F(s)/s connects integration in time to algebraic manipulation in the s-domain.

  • Fubini's Theorem: Allows interchanging the order of integration, essential for deriving the theorem.

  • Application in Engineering: The theorem facilitates solving integro-differential equations crucial in engineering applications.

Examples & Real-Life Applications

See how the concepts apply in real-world scenarios to understand their practical implications.

Examples

  • Example 1 illustrates finding the Laplace Transform of ∫sin(aΟ„)dΟ„, demonstrating how to use the theorem in practice.

  • Example 2 shows calculations for L{∫e^(2Ο„)dΟ„}, applying the theorem to find results.

Memory Aids

Use mnemonics, acronyms, or visual cues to help remember key information more easily.

🎡 Rhymes Time

  • When functions fold under integral's sway, In Laplace they play without delay!

πŸ“– Fascinating Stories

  • Once upon a time, a troublesome integral couldn't reveal its secrets until the wizard Laplace showed how to transform it into the simpler realm of s-domain. With its newfound powers, the integral became a straightforward equation to solve!

🧠 Other Memory Gems

  • Remember β€˜TEM’ for theorem: Transformation, Evaluate, Manipulate to help remember the steps of proving.

🎯 Super Acronyms

Use β€˜LIFT’ for Laplace Integral Function Transformation to recall the main purpose of the theorem.

Flash Cards

Review key concepts with flashcards.

Glossary of Terms

Review the Definitions for terms.

  • Term: Laplace Transform

    Definition:

    A mathematical transformation that converts a time function into a complex frequency domain.

  • Term: Fubini's Theorem

    Definition:

    A theorem that allows the interchange of the order of integration in double integrals under certain conditions.

  • Term: Integral

    Definition:

    A mathematical concept representing the area under the curve of a function.

  • Term: Differential Equation

    Definition:

    An equation involving derivatives of a function or functions.

  • Term: Piecewise Continuous Function

    Definition:

    A function that is continuous on intervals, except for a finite number of jump discontinuities.