Why we take Laplace Transform of functions which converged using Fourier Transform
There is a large class of functions for which both the Fourier transform and the Laplace transform exist, and for which one can be obtained from the other by setting $s=j\omega$. (Note that even when both exist, the latter need not be the case). So for this class of functions, obtaining the Laplace transform from the Fourier transform (or vice versa) does not require any additional work.

Example: $$\begin{align}x(t)&amp;=e^{-at}u(t),\quad a&gt;0\\
\text{Fourier transform:}\quad X(j\omega)&amp;=\frac{1}{j\omega +a}\\
\text{Laplace transform:}\quad X(s)&amp;=\frac{1}{s +a},\quad |s|&gt; -a\end{align}$$

What the Laplace transform offers is a description in the complex $s$-plane, such as poles and zeros of transfer functions, from which many system properties can be easily deduced. The Fourier transform only shows the behavior on the frequency axis (i.e., the spectrum).