Is the Laplace transform a special case of Fourier transform? (Not the other way around) Always had a thought about why Laplace transform reveals the transient properties of the system?
My doubt is based on the following fact,
Fourier transform is given as 

\begin{equation}
\mathscr{F}\left\lbrace f(t)\right\rbrace = \int_{-\infty}^\infty f(t) e^{ -j \omega t}  dt
\end{equation}

Where Mathematically and intuitively we believe that the angular frequency $\omega$
takes only real value. 

What if, instead of taking real angular frequencies, if the variable 
$\omega$ 
assumes a complex angular frequency in the form
$\beta - j \alpha$
, then,

$$
j \omega t = j (\beta - j  \alpha) t = (\alpha + j \beta ) t = s t
$$

While taking Fourier transform w.r.t $\omega$, the quantity  $\beta$ will be real angular frequency in radians per second and $\alpha$ will be the $\textbf{imaginary angular }$ frequency in radians per second.

\begin{equation}
\int_{-\infty}^\infty f(t) e^{ -j \omega t}  dt = \int_{-\infty}^\infty f(t) e^{ - s t}  dt = \mathscr{L}\left\lbrace f(t)\right\rbrace
\end{equation}

Hence is it mathematically correct to consider bilateral Laplace transform as a special case of Fourier transform (not the other way around) when $\omega$ takes a complex angular form $\beta - j  \alpha$ ?  I believe the fact that $\omega$ can take complex values is the reason why we get transient properties of the system when using Laplace transform.
The Fourier Transform is the Laplace Transform with the complex variable s restricted to be the imaginary axis on the s plane. For this reason the Fourier Transform only exists when the imaginary axis is within the region of convergence. The variable s is called a "complex frequency" as it is the frequency variable that can take on real ($\sigma$) and imaginary ($\omega$) components. That said, I would view the Fourier Transform as a subset of the Laplace Transform, or the Laplace Transform as an expansion on the Fourier Transform that provides a lot more functionality and can exist when the Fourier Transform can't. 

This is also the reason that the frequency response for a system with a general transfer function $H(s)$ is given as $H(j\omega)$. 

When a system is restricted to $s= j\omega$ as the input, then the input is restricted to be only sinusoids or signals given by $e^{st}$ with $s = j\omega$ which maintain a constant magnitude with time. By allowing s to have real and imaginary components as in $s = \sigma + j\omega$ then we also allow the input to grow or shrink with time, depending on which point in the s plane is used as the input to the system.