
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:
$\displaystyle \frac {1} {2} \{L\}(e^bt) + \frac {1} {2} \{\L\}(e^{bt})$
= $\displaystyle \frac {1} {2} \int e^{st} e^{bt} + \frac {1} {2} \int e^{st} e^{bt}$
=$\displaystyle \lim_{A\to\infty} \frac {1} {2} \displaystyle\int^A_0 e^{t(bs)}\,dt + \frac {1} {2} \lim_{A\to\infty} \displaystyle\int^A_0 e^{t(b+s)}\,dt$
However, this is where I'm stuck. I'm not sure what do when I have a nonnegative "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!

Just out of curiosity is the answer
$\displaystyle \frac{1}{2}\big[\frac{1}{sb}+\frac{1}{s+b}\big]$?

It is actually an advantage to having negative ts.
Here is what the integration will look like.
$\displaystyle \frac{1}{2}\bigg[\frac{1}{e^{t(s+b)}(s+b)}\frac{1}{e^{t(s+b)}(sb)}\bigg]_{0}^{\infty}$
Can you see how this is good for the bounds integration?