# Laplace transform - Hypberbolic cosine

• December 13th 2010, 10:27 PM
mukmar
Laplace transform - Hypberbolic cosine
Problem:
Find the laplace transform of $\cosh bt$ where $b \in \mathbb{R}$

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

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

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

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

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

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

So I'm wondering where my error is. Any hints?
• December 13th 2010, 11:02 PM
snowtea
$s > |b|$ means $s > b > -s$
$b > -s$ is the same as $-b < s$ right?

You have the right answer, and the book does too.
• December 13th 2010, 11:20 PM
mukmar
Thanks, that was one of those dumb oversights on my part. (Doh)