Laplace Transform

**Simplicity** · November 7th 2009, 04:49 AM

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.

**HallsofIvy** · November 7th 2009, 05:33 AM

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.

**Simplicity** · November 8th 2009, 04:27 AM

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.

**calum** · November 8th 2009, 04:54 AM

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.

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

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