Theorem: Laplace Transform of an Integral

Today, we will recap the Laplace Transform. Can anyone tell me its basic definition?

Um, it's the integral of a function multiplied by an exponential decay term?

Exactly! The definition is $L\{f(t)\}=F(s)=\int_0^{\infty} e^{-st} f(t) dt$. This is a powerful tool in engineering mathematics.

What is it typically used for?

Great question! It’s especially useful for solving differential equations in systems like electrical circuits.

Let's dive into the theorem regarding integrals. The theorem states if $L\{f(t)\}=F(s)$, then $L\{ \int_0^t f(\tau)d\tau\} = \frac{F(s)}{s}$. Can someone explain what that means?

It means we can find the Laplace Transform of an integral by dividing the original transform by $s$.

Exactly! This transformation is key in simplifying problems involving integrals in systems analysis.

Why is this useful in applications?

This theorem is vital for solving integro-differential equations and analyzing systems that involve accumulative processes, like capacitor charging. It streamlines calculations.

Now, let's prove the theorem. We start with the integral $g(t) = \int_0^t f(\tau)d\tau$. How do we start to find its Laplace Transform?

We need to take the Laplace Transform of $g(t)$!

Exactly! $L\{g(t)\}=\int_0^{\infty} e^{-st} g(t) dt$. How do we express $g(t)$ in this integral?

By substituting it into the integral, we get $L\{g(t)\}=\int_0^{\infty} e^{-st}\int_0^{t} f(\tau)d\tau dt$.

Perfect! Now we can exchange the order of integration using Fubini's Theorem. This gives us $\,\int_0^{\infty} f(\tau) \int_{\tau}^{\infty} e^{-st} dt d\tau$. What does the inner integral evaluate to?

It evaluates to $\frac{e^{-s\tau}}{s}$?

Great! Thus, we conclude that $L\{g(t)\}=\int_0^{\infty} f(\tau)e^{-s\tau}d\tau = \frac{F(s)}{s}$. Well done!

Let’s apply what we’ve learned! If we need to find the Laplace Transform of the integral $g(t)=\int_0^t sin(a\tau)d\tau$, how do we start?

We start with $f(t)=sin(a t)$, then find $F(s)$.

Exactly! What is $F(s)$?

$F(s)=\frac{a}{s^2 + a^2}$, and then, using the theorem...

You would get $L\{\int_0^t sin(a\tau)d\tau\} = \frac{\frac{a}{s^2 + a^2}}{s}$. Excellent work!