1. 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 $\displaystyle 1$ would be $\displaystyle \frac{1}{s}$ for $\displaystyle s>0$. Why only for $\displaystyle s>0$, I don't understand the boundary.

2. 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 $\displaystyle 1$ would be $\displaystyle \frac{1}{s}$ for $\displaystyle s>0$. Why only for $\displaystyle s>0$, I don't understand the boundary.
Then you know that the Laplace transform of f(x) is $\displaystyle \int_0^\infty e^{-st}f(t)dt$. In particular, the Laplace transform of f(x)= 1 is $\displaystyle \int_0^\infty e^{-st}dt$. If s were negative, that integral would not exist. For example, if s= -1, that integral is $\displaystyle \int_0^\infty e^t dt$ which would be $\displaystyle \lim_{t\to\infty} e^t -1$ and that limit does not exist.

3. Originally Posted by HallsofIvy
Then you know that the Laplace transform of f(x) is $\displaystyle \int_0^\infty e^{-st}f(t)dt$. In particular, the Laplace transform of f(x)= 1 is $\displaystyle \int_0^\infty e^{-st}dt$. If s were negative, that integral would not exist. For example, if s= -1, that integral is $\displaystyle \int_0^\infty e^t dt$ which would be $\displaystyle \lim_{t\to\infty} e^t -1$ and that limit does not exist.
But that limit would work. It would just be $\displaystyle 0-1=-1$? Why is that wrong. It's not like we are getting an infinite answer.

4. Originally Posted by Simplicity
But that limit would work. It would just be $\displaystyle 0-1=-1$? Why is that wrong. It's not like we are getting an infinite answer.
Look at it again: $\displaystyle \lim_{t \to \infty} e^t = e^{\infty} = \infty$ so the function diverges. There is no limit.