Lets look at the Laplace Transform of a \emph{Toneburst}, $x(t) = \cos(\omega_0 t) u(t)$:


\begin{align*}
x(t) &= \cos(\omega_0 t) u(t) \\
\hat{X}(s) &= \int_{-\infty}^{\infty} \cos(\omega_0 t) u(t) e^{-st} dt \\
\hat{X}(s) &= \int_{0}^{\infty} \cos(\omega_0 t) e^{-st} dt \\
\hat{X}(s) &= \int_{0}^{\infty} \lr{ \frac{1}{2} e^{i\omega_0 t} + \frac{1}{2}e^{-i\omega_0 t}  } e^{-st} dt \\
\hat{X}(s) &= \frac{1}{2} \int_{0}^{\infty}  e^{i\omega_0 t}  e^{-st} dt  + \frac{1}{2} \int_{0}^{\infty} e^{-i\omega_0 t}   e^{-st} dt \\
\hat{X}(s) &= \frac{1}{2} \int_{0}^{\infty}  e^{-(s - i\omega_0) t} dt  + \frac{1}{2} \int_{0}^{\infty} e^{-(s + i\omega_0) t} dt \\
\hat{X}(s) &= \left. \frac{1}{2} \frac{e^{-(s - i\omega_0) t}}{-(s - i\omega_0)} \right|_0^\infty + \left. \frac{1}{2} \frac{e^{-(s + i\omega_0) t}}{-(s + i\omega_0)} \right|_0^\infty \\ 
\hat{X}(s) &= \lim_{t \to \infty} \frac{1}{2} \frac{e^{-(s - i\omega_0) t}}{-(s - i\omega_0)} - \frac{1}{2} \frac{1}{-(s - i\omega_0)} + \lim_{t \to \infty} \frac{1}{2} \frac{e^{-(s + i\omega_0) t}}{-(s + i\omega_0)} - \frac{1}{2} \frac{1}{-(s + i\omega_0)} \\
\end{align*}

Before evaluating, we must note conditions for the limits to converge for the evaluation at infinity. The limit $\lim_{t \to \infty} \frac{1}{2} \frac{e^{-(s - i\omega_0) t}}{-(s - i\omega_0)} $ converges to 0 if and only if $\Real{s - i\omega_0} \gt 0$. The limit $\lim_{t \to \infty} \frac{1}{2} \frac{e^{-(s + i\omega_0) t}}{-(s + i\omega_0)}$ converges to 0 if and only if $\Real{s +i\omega_0} \gt 0$. Both of these simplify to $\Real{s} \gt 0$. Thus, for $\Real{s} \gt 0$ we have

\begin{align*}
\hat{X}(s) &= 0 - \frac{1}{2} \frac{1}{-(s - i\omega_0)} + 0 - \frac{1}{2} \frac{1}{-(s + i\omega_0)} \\
\hat{X}(s) &=\frac{1}{2} \frac{1}{(s - i\omega_0)} +  \frac{1}{2} \frac{1}{(s + i\omega_0)} \\
\hat{X}(s) &=\frac{1}{2} \lr{ \frac{1}{(s - i\omega_0)} +  \frac{1}{(s + i\omega_0)}} \\
\hat{X}(s) &=\frac{1}{2} \lr{ \frac{2s}{(s^2 + \omega_0^2)} } \\
\hat{X}(s) &=\frac{s}{(s^2 + \omega_0^2)}  \\
\end{align*}

This gives us the transform pair:

\begin{nicebox}
$$
\cos(\omega_0 t) u(t) \ltp \frac{s}{(s^2 + \omega_0^2)} \quad \Real{s} \gt 0
$$
\end{nicebox}

Note that we have two finite poles as $\pm i\omega$. The zeros are at 0 and $\infty$. The $RoC$ is to the right of the right-most pole, which is on the $i\omega$-axis. Here is a question: What is the Laplace Transform of $\cos(\omega_0 t)$? This doesn't have a Laplace transform, although it does have a Fourier Transform. 


Trigonometric Toneburst