# Laplace transform - Hypberbolic cosine

• Dec 13th 2010, 09:27 PM
mukmar
Laplace transform - Hypberbolic cosine
Problem:
Find the laplace transform of $\displaystyle \cosh bt$ where $\displaystyle b \in \mathbb{R}$

Attempt:
$\displaystyle \mathcal{L}\{\cosh bt\} = F(s) = \frac{1}{2}\left[\displaystyle\int_0^{\infty} e^{(b - s)t}dt + \displaystyle\int_0^{\infty} e^{(-b - s)t}dt \right]$

$\displaystyle = \frac{1}{2}\left[\displaystyle\lim_{A \to \infty} \frac{1}{b - s}e^{(b - s)t}\big|^A_0 + \displaystyle\lim_{A \to \infty} \frac{1}{- b - s}e^{(-b - s)t}\big|^A_0 \right]$

So the first limit converges when $\displaystyle t = A$ if $\displaystyle b - s < 0$, therefore $\displaystyle s > b$.

The second limit converges when $\displaystyle t = A$ if $\displaystyle - b - s < 0$, therefore $\displaystyle s > -b$.

So the solution is $\displaystyle F(s) = \frac{s}{s^2 - b^2 }$, with the conditions $\displaystyle s > -b$ and $\displaystyle s > b$.

But the textbook solution requires $\displaystyle s > |b|$, which is the same as $\displaystyle s > b$ and $\displaystyle s < -b$.

So I'm wondering where my error is. Any hints?
• Dec 13th 2010, 10:02 PM
snowtea
$\displaystyle s > |b|$ means $\displaystyle s > b > -s$
$\displaystyle b > -s$ is the same as $\displaystyle -b < s$ right?

You have the right answer, and the book does too.
• Dec 13th 2010, 10:20 PM
mukmar
Thanks, that was one of those dumb oversights on my part. (Doh)