what is the difference between $X(j\omega)$ and $X(\omega)$ notation
Both notations are common and correct. As pointed out by Yuri Nenakhov, the advantage of the argument $j\omega$ is that it coincides with the complex (Laplace transform) variable $s$ when its real-part is zero. Note that in the complex $s$-plane the frequency axis is the imaginary axis. So $j\omega$ has nothing to do with complex frequency (which makes no sense).

So if the Laplace transform $X(s)$ of a signal $x(t)$ exists, and if the imaginary axis is inside its region of convergence, then the Fourier transform is obtained by setting $s=j\omega$.

Note that this does not work in general! In general you can't get the Fourier transform by replacing $s$ with $j\omega$ and vice versa. Two conditions must be satisfied in order for this to lead to a correct result:


Both transforms must exist (in the sense that the corresponding signal $x(t)$ has a Laplace transform and a Fourier transform).
The imaginary axis $s=j\omega$ must be inside the region of convergence of the Laplace transform.


An example where replacing $s$ by $j\omega$ doesn't work, even though both transforms exist, is the step function:

$$\begin{align}&amp;x(t)=u(t)\\\text{Laplace transform: }&amp;X(s)=\frac{1}{s}\\
\text{Fourier transform: }&amp;\hat{X}(j\omega)=\pi\delta(\omega)+\frac{1}{j\omega}\neq X(s)|_{s=j\omega}\end{align}$$