# Laplace Transform

• Apr 14th 2010, 09:48 AM
TheBerkeleyBoss
Laplace Transform
I've been working on a problem, this one in particular:

$\displaystyle f(t) = cosh(bt)$

where $\displaystyle (e^{bt} + e^{-bt})/2$ is equal to cosh bt

I attempt to take the Laplace transform:

$\displaystyle \frac {1} {2} \{L\}(e^bt) + \frac {1} {2} \{\L\}(e^-{bt})$

= $\displaystyle \frac {1} {2} \int e^{-st} e^{bt} + \frac {1} {2} \int e^{-st} e^{-bt}$

=$\displaystyle \lim_{A\to\infty} \frac {1} {2} \displaystyle\int^A_0 e^{t(b-s)}\,dt + \frac {1} {2} \lim_{A\to\infty} \displaystyle\int^A_0 e^{-t(b+s)}\,dt$

However, this is where I'm stuck. I'm not sure what do when I have a non-negative "t" (cause if it's negative, I can assume it's going to equal zero as A goes towards infinity). Can anyone show me where I went wrong, or what step is next? Thank you!
• Apr 14th 2010, 09:58 AM
dwsmith
Just out of curiosity is the answer

$\displaystyle \frac{1}{2}\big[\frac{1}{s-b}+\frac{1}{s+b}\big]$?
• Apr 14th 2010, 07:18 PM
dwsmith
It is actually an advantage to having negative ts.

Here is what the integration will look like.

$\displaystyle \frac{1}{2}\bigg[\frac{-1}{e^{t(s+b)}(s+b)}-\frac{1}{e^{t(s+b)}(s-b)}\bigg]_{0}^{\infty}$

Can you see how this is good for the bounds integration?