# Laplace Transform

• Nov 7th 2009, 05:49 AM
Simplicity
Laplace Transform
Hello MHF!

I've been learning Laplace Transform and I understand the method but I don't understand the boundary.

For example, Laplace transform on $1$ would be $\frac{1}{s}$ for $s>0$. Why only for $s>0$, I don't understand the boundary.
• Nov 7th 2009, 06:33 AM
HallsofIvy
Quote:

Originally Posted by Simplicity
Hello MHF!

I've been learning Laplace Transform and I understand the method but I don't understand the boundary.

For example, Laplace transform on $1$ would be $\frac{1}{s}$ for $s>0$. Why only for $s>0$, I don't understand the boundary.

Then you know that the Laplace transform of f(x) is $\int_0^\infty e^{-st}f(t)dt$. In particular, the Laplace transform of f(x)= 1 is $\int_0^\infty e^{-st}dt$. If s were negative, that integral would not exist. For example, if s= -1, that integral is $\int_0^\infty e^t dt$ which would be $\lim_{t\to\infty} e^t -1$ and that limit does not exist.
• Nov 8th 2009, 05:27 AM
Simplicity
Quote:

Originally Posted by HallsofIvy
Then you know that the Laplace transform of f(x) is $\int_0^\infty e^{-st}f(t)dt$. In particular, the Laplace transform of f(x)= 1 is $\int_0^\infty e^{-st}dt$. If s were negative, that integral would not exist. For example, if s= -1, that integral is $\int_0^\infty e^t dt$ which would be $\lim_{t\to\infty} e^t -1$ and that limit does not exist.

But that limit would work. It would just be $0-1=-1$? Why is that wrong. It's not like we are getting an infinite answer. (Worried)
• Nov 8th 2009, 05:54 AM
calum
Quote:

Originally Posted by Simplicity
But that limit would work. It would just be $0-1=-1$? Why is that wrong. It's not like we are getting an infinite answer. (Worried)

Look at it again: $\lim_{t \to \infty} e^t = e^{\infty} = \infty$ so the function diverges. There is no limit.