Given $f(t)= \sin (at)$ I want to calculate the Laplace transform of $f(t)$.
I have determined by using integration by parts twice, that the answer should be $$F(s)= \frac{a}{s^2+a^2}$$
Now I want to recalculate it by using just that $$\sin (at)= \frac{1}{2i} \left( {e^{ati}-e^{-ati}}\right) $$
So the integral becomes
$$\frac{1}{2i}\int_{0}^{\infty} e^{-st}\left( {e^{ati}-e^{-ati}}\right) = $$
$$ \frac{1}{2i}\int_{0}^{\infty} e^{t(ai-s)}-e^{-t(ai+s)}dt =$$
$$ \frac{1}{2i} \left[\frac{1}{ai-s}e^{t(ai-s)} \right]_0^{\infty} +\frac{1}{2i} \left[\frac{1}{ai+s}e^{-t(ai+s)} \right]_0^{\infty} $$
My question now is: How to determine the limit of $e^{t(ai-s)}$ as $t$ goes to $\infty$ ?
Best Answer
The traditional way to handle this is as follows:
In order to proceed, we need to have some more information about $s$; otherwise, we can't be sure that the integral will converge. So, we assume that $s$ is in the "region of convergence" (ROC) for this function. In this particular integral, we suppose that $\text{Re} \{s\}>0$, so that $\lim_{t\to\infty}e^{t(-s\pm ai)}=0$. You should be able to handle the rest.
$$ %\begin{align*} %\left|\lim_{t\to \infty} e^{t(-s\pm ai)}\right| &= %\lim_{t\to \infty} \left|e^{t(-s\pm ai)}\right|\\&= %\lim_{t\to \infty} \left|e^{-st}\right|\cdot\left|e^{\pm ai}\right|\\&= %\lim_{t\to \infty} \left|e^{-st}\right|\cdot 1=0 %\end{align*} %$$