Question

show integral of 0 to infinity f(u) is 1/s laplace transform

Answer 1

Do you know that \(\frak{L}\tt^{-1}[ F(s)G(s)]=f(t)*g(t)\)? Where "*" is convolution?

Answer 2

no...

Answer 3

To show the Laplace of \( \tt \int^\infty_0 f(u)~ du = \dfrac{1}{s} \), let \(\tt F(s)=\dfrac{1}{s}\) and \(\tt G(s)=1\) in the formula above. Do you see that?

Answer 4

yes, wheres the 1 c from g(s) coming from

Answer 5

What we are doing is showing that the \(inverse \) of F(s)G(s) is the integral and this is the same as saying that the Laplace of the integral is 1/s.

Answer 6

inverse os 1/s=s
inverse of 1....that = s

Answer 7

Let's take a different approach.

Answer 8

This is convolution

Answer 9

$$f(t)*g(t) = \int^t_0 g(t-u)f(u) du$$
Have you seen this before?

Answer 10

looks a little
like product rule

Answer 11

Very different than the chain rule for derivatives, if that's what you're talking about. Convolution is what happens when you have a linear time-invariant system and you apply an input, here the input is g(t) and the system is f(t). So the operation of "integration" can be a 'system'. So the convolution says what happens in time (t) as you apply an input. For our system, it integrates the input.

That's all convolution is. For different t, you will get a different value in the convolution, but all that's happening is summing (i.e. integrating).

Answer 12

i have 1/s*1=integral 0 to infinity of 1 * t dt

Answer 13

i think I'm understanding the theory

Answer 14

i still don't seem to be able to correctly write it as a prove

Answer 15

Yes, so let g(t) be 1 and f(t) stay as f(t). Then the convolution becomes the integral you are trying to take the Laplace of. But now we can use the convolution theorem that says, the Laplace of a convolution is just the product of the Laplace of each individual function. What is the Laplace of "1"? What is the Laplace of f(t) ?

Answer 16

t/s

Answer 17

multiplied together = t/s, in this case 1/s

Answer 18

If you look at the equation above, with g(t) = 1, the convolution of g(t)*f(t)=1*f(t) = $$\int^t_0 f(u)du$$, which is the integral you want to find the Laplace of? Right?

Answer 19

yes

Answer 20

So finding the Laplace of $$\int^t_0 f(u)du$$ is the same as finding the Laplace of g(t)*f(t). This is the case because both are equal. Do you see that?

Answer 21

i think so...im also fairly new to the lace transform, so sorry for how slow i am

Answer 22

That's ok. It's just important that you see that \(\large \tt g(t)*f(t)= \int^t_0 g(t-u)f(u) du =\int^t_0 f(u)du\), when \(\tt g(t) = 1.\)

Answer 23

This is true because g(t-u) = 1, always in the integrand, when g(t)=1.

Answer 24

Here is the basic result:

$$\frak{L}\tt[f(t)*g(t)]\\
=\frak{L} \tt \left[ \int^t_0 f(u)du\ \right] \tt\\
=F(s)G(s) \\
\text{but}\\
G(s)=\int^\infty_0e^{-st}g(t)dt =\int^\infty_0e^{-st}dt
=\frac{1}{s}\\
So~, \frak{L} \tt \left[ \int^t_0 f(u)du\ \right] \tt=\frac{F(s)}{s}\\
$$
Does this make sense?